Integrand size = 18, antiderivative size = 211 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {a b \sqrt {x}}{2 c^7}+\frac {71 b^2 x}{420 c^6}+\frac {3 b^2 x^2}{70 c^4}+\frac {b^2 x^3}{84 c^2}+\frac {b^2 \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )}{2 c^7}+\frac {b x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{6 c^5}+\frac {b x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{10 c^3}+\frac {b x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{14 c}-\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{4 c^8}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2+\frac {44 b^2 \log \left (1-c^2 x\right )}{105 c^8} \]
71/420*b^2*x/c^6+3/70*b^2*x^2/c^4+1/84*b^2*x^3/c^2+1/6*b*x^(3/2)*(a+b*arct anh(c*x^(1/2)))/c^5+1/10*b*x^(5/2)*(a+b*arctanh(c*x^(1/2)))/c^3+1/14*b*x^( 7/2)*(a+b*arctanh(c*x^(1/2)))/c-1/4*(a+b*arctanh(c*x^(1/2)))^2/c^8+1/4*x^4 *(a+b*arctanh(c*x^(1/2)))^2+44/105*b^2*ln(-c^2*x+1)/c^8+1/2*a*b*x^(1/2)/c^ 7+1/2*b^2*arctanh(c*x^(1/2))*x^(1/2)/c^7
Time = 0.08 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.06 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {210 a b c \sqrt {x}+71 b^2 c^2 x+70 a b c^3 x^{3/2}+18 b^2 c^4 x^2+42 a b c^5 x^{5/2}+5 b^2 c^6 x^3+30 a b c^7 x^{7/2}+105 a^2 c^8 x^4+2 b c \sqrt {x} \left (105 a c^7 x^{7/2}+b \left (105+35 c^2 x+21 c^4 x^2+15 c^6 x^3\right )\right ) \text {arctanh}\left (c \sqrt {x}\right )+105 b^2 \left (-1+c^8 x^4\right ) \text {arctanh}\left (c \sqrt {x}\right )^2+b (105 a+176 b) \log \left (1-c \sqrt {x}\right )-105 a b \log \left (1+c \sqrt {x}\right )+176 b^2 \log \left (1+c \sqrt {x}\right )}{420 c^8} \]
(210*a*b*c*Sqrt[x] + 71*b^2*c^2*x + 70*a*b*c^3*x^(3/2) + 18*b^2*c^4*x^2 + 42*a*b*c^5*x^(5/2) + 5*b^2*c^6*x^3 + 30*a*b*c^7*x^(7/2) + 105*a^2*c^8*x^4 + 2*b*c*Sqrt[x]*(105*a*c^7*x^(7/2) + b*(105 + 35*c^2*x + 21*c^4*x^2 + 15*c ^6*x^3))*ArcTanh[c*Sqrt[x]] + 105*b^2*(-1 + c^8*x^4)*ArcTanh[c*Sqrt[x]]^2 + b*(105*a + 176*b)*Log[1 - c*Sqrt[x]] - 105*a*b*Log[1 + c*Sqrt[x]] + 176* b^2*Log[1 + c*Sqrt[x]])/(420*c^8)
Time = 2.04 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.45, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.111, Rules used = {6454, 6452, 6542, 6452, 243, 49, 2009, 6542, 6452, 243, 49, 2009, 6542, 6452, 243, 49, 2009, 6542, 2009, 6510}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle 2 \int x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2d\sqrt {x}\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \int \frac {x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{7} b c \int \frac {x^{7/2}}{1-c^2 x}d\sqrt {x}}{c^2}\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \int \frac {x^{3/2}}{1-c^2 x}dx}{c^2}\right )\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \int \left (-\frac {x}{c^2}-\frac {x}{c^4}-\frac {1}{c^6 \left (c^2 x-1\right )}-\frac {1}{c^6}\right )dx}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{5} b c \int \frac {x^{5/2}}{1-c^2 x}d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \int \frac {x}{1-c^2 x}dx}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \int \left (-\frac {x}{c^2}-\frac {1}{c^4 \left (c^2 x-1\right )}-\frac {1}{c^4}\right )dx}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\int \frac {x^2 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{3} b c \int \frac {x^{3/2}}{1-c^2 x}d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \int \frac {x}{1-c^2 x}dx}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x-1\right )}\right )dx}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\int \frac {x \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\frac {\int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )d\sqrt {x}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\frac {\int \frac {a+b \text {arctanh}\left (c \sqrt {x}\right )}{1-c^2 x}d\sqrt {x}}{c^2}-\frac {a \sqrt {x}+b \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )+\frac {b \log \left (1-c^2 x\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle 2 \left (\frac {1}{8} x^4 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2-\frac {1}{4} b c \left (\frac {\frac {\frac {\frac {\left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2}{2 b c^3}-\frac {a \sqrt {x}+b \sqrt {x} \text {arctanh}\left (c \sqrt {x}\right )+\frac {b \log \left (1-c^2 x\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^{3/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{6} b c \left (-\frac {x}{c^2}-\frac {\log \left (1-c^2 x\right )}{c^4}\right )}{c^2}}{c^2}-\frac {\frac {1}{5} x^{5/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{10} b c \left (-\frac {x}{c^4}-\frac {x}{2 c^2}-\frac {\log \left (1-c^2 x\right )}{c^6}\right )}{c^2}}{c^2}-\frac {\frac {1}{7} x^{7/2} \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )-\frac {1}{14} b c \left (-\frac {x}{c^6}-\frac {x}{2 c^4}-\frac {x^{3/2}}{3 c^2}-\frac {\log \left (1-c^2 x\right )}{c^8}\right )}{c^2}\right )\right )\) |
2*((x^4*(a + b*ArcTanh[c*Sqrt[x]])^2)/8 - (b*c*(-(((x^(7/2)*(a + b*ArcTanh [c*Sqrt[x]]))/7 - (b*c*(-(x/c^6) - x/(2*c^4) - x^(3/2)/(3*c^2) - Log[1 - c ^2*x]/c^8))/14)/c^2) + (-(((x^(5/2)*(a + b*ArcTanh[c*Sqrt[x]]))/5 - (b*c*( -(x/c^4) - x/(2*c^2) - Log[1 - c^2*x]/c^6))/10)/c^2) + (-(((x^(3/2)*(a + b *ArcTanh[c*Sqrt[x]]))/3 - (b*c*(-(x/c^2) - Log[1 - c^2*x]/c^4))/6)/c^2) + ((a + b*ArcTanh[c*Sqrt[x]])^2/(2*b*c^3) - (a*Sqrt[x] + b*Sqrt[x]*ArcTanh[c *Sqrt[x]] + (b*Log[1 - c^2*x])/(2*c))/c^2)/c^2)/c^2)/c^2))/4)
3.2.95.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x ], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl ify[(m + 1)/n]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTanh[c* x])^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 7.76 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.46
method | result | size |
parts | \(\frac {a^{2} x^{4}}{4}+\frac {2 b^{2} \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{8}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{7} x^{\frac {7}{2}}}{28}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{20}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{12}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{4}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{8}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{8}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{32}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{32}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {c^{6} x^{3}}{168}+\frac {3 c^{4} x^{2}}{140}+\frac {71 c^{2} x}{840}+\frac {22 \ln \left (c \sqrt {x}-1\right )}{105}+\frac {22 \ln \left (1+c \sqrt {x}\right )}{105}\right )}{c^{8}}+\frac {4 a b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\) | \(309\) |
derivativedivides | \(\frac {\frac {a^{2} c^{8} x^{4}}{4}+2 b^{2} \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{8}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{7} x^{\frac {7}{2}}}{28}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{20}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{12}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{4}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{8}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{8}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{32}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{32}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {c^{6} x^{3}}{168}+\frac {3 c^{4} x^{2}}{140}+\frac {71 c^{2} x}{840}+\frac {22 \ln \left (c \sqrt {x}-1\right )}{105}+\frac {22 \ln \left (1+c \sqrt {x}\right )}{105}\right )+4 a b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\) | \(310\) |
default | \(\frac {\frac {a^{2} c^{8} x^{4}}{4}+2 b^{2} \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )^{2}}{8}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{7} x^{\frac {7}{2}}}{28}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{5} x^{\frac {5}{2}}}{20}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c^{3} x^{\frac {3}{2}}}{12}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) c \sqrt {x}}{4}+\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{8}-\frac {\operatorname {arctanh}\left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{8}+\frac {\ln \left (c \sqrt {x}-1\right )^{2}}{32}-\frac {\ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {\ln \left (1+c \sqrt {x}\right )^{2}}{32}-\frac {\left (\ln \left (1+c \sqrt {x}\right )-\ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )\right ) \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{16}+\frac {c^{6} x^{3}}{168}+\frac {3 c^{4} x^{2}}{140}+\frac {71 c^{2} x}{840}+\frac {22 \ln \left (c \sqrt {x}-1\right )}{105}+\frac {22 \ln \left (1+c \sqrt {x}\right )}{105}\right )+4 a b \left (\frac {c^{8} x^{4} \operatorname {arctanh}\left (c \sqrt {x}\right )}{8}+\frac {c^{7} x^{\frac {7}{2}}}{56}+\frac {c^{5} x^{\frac {5}{2}}}{40}+\frac {c^{3} x^{\frac {3}{2}}}{24}+\frac {c \sqrt {x}}{8}+\frac {\ln \left (c \sqrt {x}-1\right )}{16}-\frac {\ln \left (1+c \sqrt {x}\right )}{16}\right )}{c^{8}}\) | \(310\) |
1/4*a^2*x^4+2*b^2/c^8*(1/8*c^8*x^4*arctanh(c*x^(1/2))^2+1/28*arctanh(c*x^( 1/2))*c^7*x^(7/2)+1/20*arctanh(c*x^(1/2))*c^5*x^(5/2)+1/12*arctanh(c*x^(1/ 2))*c^3*x^(3/2)+1/4*arctanh(c*x^(1/2))*c*x^(1/2)+1/8*arctanh(c*x^(1/2))*ln (c*x^(1/2)-1)-1/8*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))+1/32*ln(c*x^(1/2)-1)^ 2-1/16*ln(c*x^(1/2)-1)*ln(1/2*c*x^(1/2)+1/2)+1/32*ln(1+c*x^(1/2))^2-1/16*( ln(1+c*x^(1/2))-ln(1/2*c*x^(1/2)+1/2))*ln(-1/2*c*x^(1/2)+1/2)+1/168*c^6*x^ 3+3/140*c^4*x^2+71/840*c^2*x+22/105*ln(c*x^(1/2)-1)+22/105*ln(1+c*x^(1/2)) )+4*a*b/c^8*(1/8*c^8*x^4*arctanh(c*x^(1/2))+1/56*c^7*x^(7/2)+1/40*c^5*x^(5 /2)+1/24*c^3*x^(3/2)+1/8*c*x^(1/2)+1/16*ln(c*x^(1/2)-1)-1/16*ln(1+c*x^(1/2 )))
Time = 0.27 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.29 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {420 \, a^{2} c^{8} x^{4} + 20 \, b^{2} c^{6} x^{3} + 72 \, b^{2} c^{4} x^{2} + 284 \, b^{2} c^{2} x + 105 \, {\left (b^{2} c^{8} x^{4} - b^{2}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right )^{2} + 4 \, {\left (105 \, a b c^{8} - 105 \, a b + 176 \, b^{2}\right )} \log \left (c \sqrt {x} + 1\right ) - 4 \, {\left (105 \, a b c^{8} - 105 \, a b - 176 \, b^{2}\right )} \log \left (c \sqrt {x} - 1\right ) + 4 \, {\left (105 \, a b c^{8} x^{4} - 105 \, a b c^{8} + {\left (15 \, b^{2} c^{7} x^{3} + 21 \, b^{2} c^{5} x^{2} + 35 \, b^{2} c^{3} x + 105 \, b^{2} c\right )} \sqrt {x}\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 8 \, {\left (15 \, a b c^{7} x^{3} + 21 \, a b c^{5} x^{2} + 35 \, a b c^{3} x + 105 \, a b c\right )} \sqrt {x}}{1680 \, c^{8}} \]
1/1680*(420*a^2*c^8*x^4 + 20*b^2*c^6*x^3 + 72*b^2*c^4*x^2 + 284*b^2*c^2*x + 105*(b^2*c^8*x^4 - b^2)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1))^2 + 4*(105*a*b*c^8 - 105*a*b + 176*b^2)*log(c*sqrt(x) + 1) - 4*(105*a*b*c^8 - 105*a*b - 176*b^2)*log(c*sqrt(x) - 1) + 4*(105*a*b*c^8*x^4 - 105*a*b*c^8 + (15*b^2*c^7*x^3 + 21*b^2*c^5*x^2 + 35*b^2*c^3*x + 105*b^2*c)*sqrt(x))*log (-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 8*(15*a*b*c^7*x^3 + 21*a*b*c^5* x^2 + 35*a*b*c^3*x + 105*a*b*c)*sqrt(x))/c^8
\[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\int x^{3} \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )^{2}\, dx \]
Time = 0.20 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.26 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{420} \, {\left (210 \, x^{4} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )}\right )} a b + \frac {1}{1680} \, {\left (4 \, c {\left (\frac {2 \, {\left (15 \, c^{6} x^{\frac {7}{2}} + 21 \, c^{4} x^{\frac {5}{2}} + 35 \, c^{2} x^{\frac {3}{2}} + 105 \, \sqrt {x}\right )}}{c^{8}} - \frac {105 \, \log \left (c \sqrt {x} + 1\right )}{c^{9}} + \frac {105 \, \log \left (c \sqrt {x} - 1\right )}{c^{9}}\right )} \operatorname {artanh}\left (c \sqrt {x}\right ) + \frac {20 \, c^{6} x^{3} + 72 \, c^{4} x^{2} + 284 \, c^{2} x - 2 \, {\left (105 \, \log \left (c \sqrt {x} - 1\right ) - 352\right )} \log \left (c \sqrt {x} + 1\right ) + 105 \, \log \left (c \sqrt {x} + 1\right )^{2} + 105 \, \log \left (c \sqrt {x} - 1\right )^{2} + 704 \, \log \left (c \sqrt {x} - 1\right )}{c^{8}}\right )} b^{2} \]
1/4*b^2*x^4*arctanh(c*sqrt(x))^2 + 1/4*a^2*x^4 + 1/420*(210*x^4*arctanh(c* sqrt(x)) + c*(2*(15*c^6*x^(7/2) + 21*c^4*x^(5/2) + 35*c^2*x^(3/2) + 105*sq rt(x))/c^8 - 105*log(c*sqrt(x) + 1)/c^9 + 105*log(c*sqrt(x) - 1)/c^9))*a*b + 1/1680*(4*c*(2*(15*c^6*x^(7/2) + 21*c^4*x^(5/2) + 35*c^2*x^(3/2) + 105* sqrt(x))/c^8 - 105*log(c*sqrt(x) + 1)/c^9 + 105*log(c*sqrt(x) - 1)/c^9)*ar ctanh(c*sqrt(x)) + (20*c^6*x^3 + 72*c^4*x^2 + 284*c^2*x - 2*(105*log(c*sqr t(x) - 1) - 352)*log(c*sqrt(x) + 1) + 105*log(c*sqrt(x) + 1)^2 + 105*log(c *sqrt(x) - 1)^2 + 704*log(c*sqrt(x) - 1))/c^8)*b^2
\[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c \sqrt {x}\right ) + a\right )}^{2} x^{3} \,d x } \]
Time = 6.55 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.15 \[ \int x^3 \left (a+b \text {arctanh}\left (c \sqrt {x}\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}+\frac {44\,b^2\,\ln \left (c\,\sqrt {x}-1\right )}{105\,c^8}+\frac {44\,b^2\,\ln \left (c\,\sqrt {x}+1\right )}{105\,c^8}+\frac {71\,b^2\,x}{420\,c^6}-\frac {b^2\,{\ln \left (c\,\sqrt {x}+1\right )}^2}{16\,c^8}-\frac {b^2\,{\ln \left (1-c\,\sqrt {x}\right )}^2}{16\,c^8}+\frac {b^2\,x^3}{84\,c^2}+\frac {3\,b^2\,x^2}{70\,c^4}+\frac {b^2\,x^4\,{\ln \left (c\,\sqrt {x}+1\right )}^2}{16}+\frac {b^2\,x^4\,{\ln \left (1-c\,\sqrt {x}\right )}^2}{16}+\frac {b^2\,x^{7/2}\,\ln \left (c\,\sqrt {x}+1\right )}{28\,c}+\frac {b^2\,x^{5/2}\,\ln \left (c\,\sqrt {x}+1\right )}{20\,c^3}+\frac {b^2\,x^{3/2}\,\ln \left (c\,\sqrt {x}+1\right )}{12\,c^5}+\frac {b^2\,\sqrt {x}\,\ln \left (c\,\sqrt {x}+1\right )}{4\,c^7}-\frac {b^2\,x^{7/2}\,\ln \left (1-c\,\sqrt {x}\right )}{28\,c}-\frac {b^2\,x^{5/2}\,\ln \left (1-c\,\sqrt {x}\right )}{20\,c^3}-\frac {b^2\,x^{3/2}\,\ln \left (1-c\,\sqrt {x}\right )}{12\,c^5}-\frac {b^2\,\sqrt {x}\,\ln \left (1-c\,\sqrt {x}\right )}{4\,c^7}+\frac {a\,b\,\ln \left (c\,\sqrt {x}-1\right )}{4\,c^8}-\frac {a\,b\,\ln \left (c\,\sqrt {x}+1\right )}{4\,c^8}+\frac {a\,b\,x^4\,\ln \left (c\,\sqrt {x}+1\right )}{4}-\frac {a\,b\,x^4\,\ln \left (1-c\,\sqrt {x}\right )}{4}+\frac {b^2\,\ln \left (c\,\sqrt {x}+1\right )\,\ln \left (1-c\,\sqrt {x}\right )}{8\,c^8}+\frac {a\,b\,x^{7/2}}{14\,c}+\frac {a\,b\,x^{5/2}}{10\,c^3}+\frac {a\,b\,x^{3/2}}{6\,c^5}+\frac {a\,b\,\sqrt {x}}{2\,c^7}-\frac {b^2\,x^4\,\ln \left (c\,\sqrt {x}+1\right )\,\ln \left (1-c\,\sqrt {x}\right )}{8} \]
(a^2*x^4)/4 + (44*b^2*log(c*x^(1/2) - 1))/(105*c^8) + (44*b^2*log(c*x^(1/2 ) + 1))/(105*c^8) + (71*b^2*x)/(420*c^6) - (b^2*log(c*x^(1/2) + 1)^2)/(16* c^8) - (b^2*log(1 - c*x^(1/2))^2)/(16*c^8) + (b^2*x^3)/(84*c^2) + (3*b^2*x ^2)/(70*c^4) + (b^2*x^4*log(c*x^(1/2) + 1)^2)/16 + (b^2*x^4*log(1 - c*x^(1 /2))^2)/16 + (b^2*x^(7/2)*log(c*x^(1/2) + 1))/(28*c) + (b^2*x^(5/2)*log(c* x^(1/2) + 1))/(20*c^3) + (b^2*x^(3/2)*log(c*x^(1/2) + 1))/(12*c^5) + (b^2* x^(1/2)*log(c*x^(1/2) + 1))/(4*c^7) - (b^2*x^(7/2)*log(1 - c*x^(1/2)))/(28 *c) - (b^2*x^(5/2)*log(1 - c*x^(1/2)))/(20*c^3) - (b^2*x^(3/2)*log(1 - c*x ^(1/2)))/(12*c^5) - (b^2*x^(1/2)*log(1 - c*x^(1/2)))/(4*c^7) + (a*b*log(c* x^(1/2) - 1))/(4*c^8) - (a*b*log(c*x^(1/2) + 1))/(4*c^8) + (a*b*x^4*log(c* x^(1/2) + 1))/4 - (a*b*x^4*log(1 - c*x^(1/2)))/4 + (b^2*log(c*x^(1/2) + 1) *log(1 - c*x^(1/2)))/(8*c^8) + (a*b*x^(7/2))/(14*c) + (a*b*x^(5/2))/(10*c^ 3) + (a*b*x^(3/2))/(6*c^5) + (a*b*x^(1/2))/(2*c^7) - (b^2*x^4*log(c*x^(1/2 ) + 1)*log(1 - c*x^(1/2)))/8