Integrand size = 19, antiderivative size = 109 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=-\frac {c \sqrt {b x^2+c x^4}}{8 x^5}-\frac {c^2 \sqrt {b x^2+c x^4}}{16 b x^3}-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}+\frac {c^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{16 b^{3/2}} \]
-1/6*(c*x^4+b*x^2)^(3/2)/x^9+1/16*c^3*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2 ))/b^(3/2)-1/8*c*(c*x^4+b*x^2)^(1/2)/x^5-1/16*c^2*(c*x^4+b*x^2)^(1/2)/b/x^ 3
Time = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (-\sqrt {b} \sqrt {b+c x^2} \left (8 b^2+14 b c x^2+3 c^2 x^4\right )+3 c^3 x^6 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{48 b^{3/2} x^7 \sqrt {b+c x^2}} \]
(Sqrt[x^2*(b + c*x^2)]*(-(Sqrt[b]*Sqrt[b + c*x^2]*(8*b^2 + 14*b*c*x^2 + 3* c^2*x^4)) + 3*c^3*x^6*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(48*b^(3/2)*x^7*S qrt[b + c*x^2])
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1425, 1425, 1430, 1400, 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.
\(\displaystyle \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {1}{2} c \int \frac {\sqrt {c x^4+b x^2}}{x^6}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\) |
\(\Big \downarrow \) 1425 |
\(\displaystyle \frac {1}{2} c \left (\frac {1}{4} c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\) |
\(\Big \downarrow \) 1430 |
\(\displaystyle \frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\) |
\(\Big \downarrow \) 1400 |
\(\displaystyle \frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\) |
-1/6*(b*x^2 + c*x^4)^(3/2)/x^9 + (c*(-1/4*Sqrt[b*x^2 + c*x^4]/x^5 + (c*(-1 /2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4 ]])/(2*b^(3/2))))/4))/2
3.3.57.3.1 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_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[m + 2*p + 1, 0 ]
Time = 0.80 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {\left (3 c^{2} x^{4}+14 b c \,x^{2}+8 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{48 x^{7} b}+\frac {c^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{16 b^{\frac {3}{2}} x \sqrt {c \,x^{2}+b}}\) | \(100\) |
default | \(\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{3} x^{6}-\left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{3} x^{6}+\left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{2} x^{4}-3 \sqrt {c \,x^{2}+b}\, b \,c^{3} x^{6}+2 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b c \,x^{2}-8 \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2}\right )}{48 x^{9} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{3}}\) | \(145\) |
-1/48*(3*c^2*x^4+14*b*c*x^2+8*b^2)/x^7/b*(x^2*(c*x^2+b))^(1/2)+1/16*c^3/b^ (3/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1/2))/x)*(x^2*(c*x^2+b))^(1/2)/x/(c*x^2 +b)^(1/2)
Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.70 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\left [\frac {3 \, \sqrt {b} c^{3} x^{7} \log \left (-\frac {c x^{3} + 2 \, b x + 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) - 2 \, {\left (3 \, b c^{2} x^{4} + 14 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{96 \, b^{2} x^{7}}, -\frac {3 \, \sqrt {-b} c^{3} x^{7} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (3 \, b c^{2} x^{4} + 14 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{48 \, b^{2} x^{7}}\right ] \]
[1/96*(3*sqrt(b)*c^3*x^7*log(-(c*x^3 + 2*b*x + 2*sqrt(c*x^4 + b*x^2)*sqrt( b))/x^3) - 2*(3*b*c^2*x^4 + 14*b^2*c*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^ 2*x^7), -1/48*(3*sqrt(-b)*c^3*x^7*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x ^3 + b*x)) + (3*b*c^2*x^4 + 14*b^2*c*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2))/(b^ 2*x^7)]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{10}}\, dx \]
\[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{10}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=-\frac {\frac {3 \, c^{4} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-b} b} + \frac {3 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} c^{4} \mathrm {sgn}\left (x\right ) + 8 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b c^{4} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {c x^{2} + b} b^{2} c^{4} \mathrm {sgn}\left (x\right )}{b c^{3} x^{6}}}{48 \, c} \]
-1/48*(3*c^4*arctan(sqrt(c*x^2 + b)/sqrt(-b))*sgn(x)/(sqrt(-b)*b) + (3*(c* x^2 + b)^(5/2)*c^4*sgn(x) + 8*(c*x^2 + b)^(3/2)*b*c^4*sgn(x) - 3*sqrt(c*x^ 2 + b)*b^2*c^4*sgn(x))/(b*c^3*x^6))/c
Timed out. \[ \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^{10}} \, dx=\int \frac {{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{10}} \,d x \]