Integrand size = 19, antiderivative size = 81 \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^4}{c \sqrt {b x^2+c x^4}}+\frac {3 \sqrt {b x^2+c x^4}}{2 c^2}-\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{2 c^{5/2}} \]
-3/2*b*arctanh(x^2*c^(1/2)/(c*x^4+b*x^2)^(1/2))/c^(5/2)-x^4/c/(c*x^4+b*x^2 )^(1/2)+3/2*(c*x^4+b*x^2)^(1/2)/c^2
Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x \left (\sqrt {c} x \left (3 b+c x^2\right )+6 b \sqrt {b+c x^2} \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b}-\sqrt {b+c x^2}}\right )\right )}{2 c^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \]
(x*(Sqrt[c]*x*(3*b + c*x^2) + 6*b*Sqrt[b + c*x^2]*ArcTanh[(Sqrt[c]*x)/(Sqr t[b] - Sqrt[b + c*x^2])]))/(2*c^(5/2)*Sqrt[x^2*(b + c*x^2)])
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1424, 1124, 25, 1160, 1091, 219}
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.
\(\displaystyle \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1424 |
\(\displaystyle \frac {1}{2} \int \frac {x^6}{\left (c x^4+b x^2\right )^{3/2}}dx^2\) |
\(\Big \downarrow \) 1124 |
\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {b-c x^2}{\sqrt {c x^4+b x^2}}dx^2}{c^2}+\frac {2 b x^2}{c^2 \sqrt {b x^2+c x^4}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {2 b x^2}{c^2 \sqrt {b x^2+c x^4}}-\frac {\int \frac {b-c x^2}{\sqrt {c x^4+b x^2}}dx^2}{c^2}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{2} \left (\frac {2 b x^2}{c^2 \sqrt {b x^2+c x^4}}-\frac {\frac {3}{2} b \int \frac {1}{\sqrt {c x^4+b x^2}}dx^2-\sqrt {b x^2+c x^4}}{c^2}\right )\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {1}{2} \left (\frac {2 b x^2}{c^2 \sqrt {b x^2+c x^4}}-\frac {3 b \int \frac {1}{1-c x^4}d\frac {x^2}{\sqrt {c x^4+b x^2}}-\sqrt {b x^2+c x^4}}{c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {2 b x^2}{c^2 \sqrt {b x^2+c x^4}}-\frac {\frac {3 b \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {b x^2+c x^4}}\right )}{\sqrt {c}}-\sqrt {b x^2+c x^4}}{c^2}\right )\) |
((2*b*x^2)/(c^2*Sqrt[b*x^2 + c*x^4]) - (-Sqrt[b*x^2 + c*x^4] + (3*b*ArcTan h[(Sqrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/Sqrt[c])/c^2)/2
3.3.74.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x _Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + b*x + c*x^2])), x] + Simp[e^2/c^(m - 1) Int[(1/Sqrt[a + b*x + c*x^2])*Exp andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e ^2, 0] && IGtQ[m, 0]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(x_)^(m_.)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{b, c, m, p}, x] && !IntegerQ[p] && IntegerQ[(m - 1)/2]
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90
method | result | size |
default | \(\frac {x^{3} \left (c \,x^{2}+b \right ) \left (c^{\frac {5}{2}} x^{3}+3 c^{\frac {3}{2}} b x -3 \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right ) \sqrt {c \,x^{2}+b}\, b c \right )}{2 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{\frac {7}{2}}}\) | \(73\) |
risch | \(\frac {x^{2} \left (c \,x^{2}+b \right )}{2 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {\left (\frac {b x}{c^{2} \sqrt {c \,x^{2}+b}}-\frac {3 b \ln \left (x \sqrt {c}+\sqrt {c \,x^{2}+b}\right )}{2 c^{\frac {5}{2}}}\right ) x \sqrt {c \,x^{2}+b}}{\sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(92\) |
pseudoelliptic | \(-\frac {3 \left (-\frac {2 c^{\frac {3}{2}} x^{4}}{3}-2 b \,x^{2} \sqrt {c}+\ln \left (\frac {2 c \,x^{2}+2 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \sqrt {c}+b}{\sqrt {c}}\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}-\ln \left (2\right ) b \sqrt {x^{2} \left (c \,x^{2}+b \right )}\right )}{4 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\) | \(101\) |
1/2*x^3*(c*x^2+b)*(c^(5/2)*x^3+3*c^(3/2)*b*x-3*ln(x*c^(1/2)+(c*x^2+b)^(1/2 ))*(c*x^2+b)^(1/2)*b*c)/(c*x^4+b*x^2)^(3/2)/c^(7/2)
Time = 0.26 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.22 \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b c x^{2} + b^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right ) + 2 \, \sqrt {c x^{4} + b x^{2}} {\left (c^{2} x^{2} + 3 \, b c\right )}}{4 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {3 \, {\left (b c x^{2} + b^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-c}}{c x^{2} + b}\right ) + \sqrt {c x^{4} + b x^{2}} {\left (c^{2} x^{2} + 3 \, b c\right )}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \]
[1/4*(3*(b*c*x^2 + b^2)*sqrt(c)*log(-2*c*x^2 - b + 2*sqrt(c*x^4 + b*x^2)*s qrt(c)) + 2*sqrt(c*x^4 + b*x^2)*(c^2*x^2 + 3*b*c))/(c^4*x^2 + b*c^3), 1/2* (3*(b*c*x^2 + b^2)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2 + b )) + sqrt(c*x^4 + b*x^2)*(c^2*x^2 + 3*b*c))/(c^4*x^2 + b*c^3)]
\[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{4}}{2 \, \sqrt {c x^{4} + b x^{2}} c} + \frac {3 \, b x^{2}}{2 \, \sqrt {c x^{4} + b x^{2}} c^{2}} - \frac {3 \, b \log \left (2 \, c x^{2} + b + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {c}\right )}{4 \, c^{\frac {5}{2}}} \]
1/2*x^4/(sqrt(c*x^4 + b*x^2)*c) + 3/2*b*x^2/(sqrt(c*x^4 + b*x^2)*c^2) - 3/ 4*b*log(2*c*x^2 + b + 2*sqrt(c*x^4 + b*x^2)*sqrt(c))/c^(5/2)
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91 \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x {\left (\frac {x^{2}}{c \mathrm {sgn}\left (x\right )} + \frac {3 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )}}{2 \, \sqrt {c x^{2} + b}} - \frac {3 \, b \log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{4 \, c^{\frac {5}{2}}} + \frac {3 \, b \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + b} \right |}\right )}{2 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]
1/2*x*(x^2/(c*sgn(x)) + 3*b/(c^2*sgn(x)))/sqrt(c*x^2 + b) - 3/4*b*log(abs( b))*sgn(x)/c^(5/2) + 3/2*b*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/(c^(5/2) *sgn(x))
Timed out. \[ \int \frac {x^7}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \]