Optimal. Leaf size=202 \[ \frac{\left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e^2}+\frac{\left (c+d x^2\right ) (3 b c-7 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b^3 e^2}+\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{7/2} \sqrt{d} e^{3/2}}+\frac{a (b c-a d)}{b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.238669, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1960, 456, 453, 208} \[ \frac{\left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 b^2 e^2}+\frac{\left (c+d x^2\right ) (3 b c-7 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b^3 e^2}+\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{7/2} \sqrt{d} e^{3/2}}+\frac{a (b c-a d)}{b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 1960
Rule 456
Rule 453
Rule 208
Rubi steps
\begin{align*} \int \frac{x^3}{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{-a e+c x^2}{x^2 \left (b e-d x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}-\frac{1}{4} ((b c-a d) e) \operatorname{Subst}\left (\int \frac{\frac{4 a}{b}-\frac{3 (b c-a d) x^2}{b^2 e}}{x^2 \left (b e-d x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{(3 b c-7 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^3 e^2}+\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}+\frac{1}{8} ((b c-a d) e) \operatorname{Subst}\left (\int \frac{-\frac{8 a}{b^2 e}+\frac{(3 b c-7 a d) x^2}{b^3 e^2}}{x^2 \left (b e-d x^2\right )} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{a (b c-a d)}{b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{(3 b c-7 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^3 e^2}+\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}+\frac{(3 (b c-5 a d) (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{b e-d x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{8 b^3 e}\\ &=\frac{a (b c-a d)}{b^3 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{(3 b c-7 a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b^3 e^2}+\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 b^2 e^2}+\frac{3 (b c-5 a d) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{7/2} \sqrt{d} e^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.331849, size = 190, normalized size = 0.94 \[ \frac{\sqrt{d} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \left (-15 a^2 d+a b \left (13 c-5 d x^2\right )+b^2 x^2 \left (5 c+2 d x^2\right )\right )+3 \sqrt{a+b x^2} (b c-5 a d) \sqrt{b c-a d} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{8 b^3 \sqrt{d} e \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.013, size = 679, normalized size = 3.4 \begin{align*}{\frac{b{x}^{2}+a}{16\,{b}^{3} \left ( d{x}^{2}+c \right ) } \left ( 4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{4}{b}^{2}d+15\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{a}^{2}b{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}a{b}^{2}cd+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{3}{c}^{2}-10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}abd+10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}{b}^{2}c+15\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}{d}^{2}-18\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}bcd+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{2}{c}^{2}-14\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{a}^{2}d+10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}abc-16\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{a}^{2}d+16\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }abc \right ){\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}} \left ({\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}} \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 7.55153, size = 1239, normalized size = 6.13 \begin{align*} \left [\frac{3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt{b d e} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} e x^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \,{\left (2 \, b d^{2} x^{4} + b c^{2} + a c d +{\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt{b d e} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}\right ) + 4 \,{\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} +{\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} +{\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{32 \,{\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}, -\frac{3 \,{\left (a b^{2} c^{2} - 6 \, a^{2} b c d + 5 \, a^{3} d^{2} +{\left (b^{3} c^{2} - 6 \, a b^{2} c d + 5 \, a^{2} b d^{2}\right )} x^{2}\right )} \sqrt{-b d e} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{-b d e} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{2 \,{\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \,{\left (2 \, b^{3} d^{3} x^{6} + 13 \, a b^{2} c^{2} d - 15 \, a^{2} b c d^{2} +{\left (7 \, b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{4} +{\left (5 \, b^{3} c^{2} d + 8 \, a b^{2} c d^{2} - 15 \, a^{2} b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{16 \,{\left (b^{5} d e^{2} x^{2} + a b^{4} d e^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]