Optimal. Leaf size=253 \[ -\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 g^3 \sqrt{d+e x} (f+g x)}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d f-a e g}}\right )}{8 g^{7/2} \sqrt{c d f-a e g}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3} \]
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Rubi [A] time = 0.336287, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {862, 874, 205} \[ -\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 g^3 \sqrt{d+e x} (f+g x)}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d f-a e g}}\right )}{8 g^{7/2} \sqrt{c d f-a e g}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3} \]
Antiderivative was successfully verified.
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Rule 862
Rule 874
Rule 205
Rubi steps
\begin{align*} \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^{5/2} (f+g x)^4} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac{(5 c d) \int \frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2} (f+g x)^3} \, dx}{6 g}\\ &=-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac{\left (5 c^2 d^2\right ) \int \frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x} (f+g x)^2} \, dx}{8 g^2}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt{d+e x} (f+g x)}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac{\left (5 c^3 d^3\right ) \int \frac{\sqrt{d+e x}}{(f+g x) \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 g^3}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt{d+e x} (f+g x)}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac{\left (5 c^3 d^3 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{-e \left (c d^2+a e^2\right ) g+c d e (e f+d g)+e^2 g x^2} \, dx,x,\frac{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}}\right )}{8 g^3}\\ &=-\frac{5 c^2 d^2 \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 g^3 \sqrt{d+e x} (f+g x)}-\frac{5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{12 g^2 (d+e x)^{3/2} (f+g x)^2}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 g (d+e x)^{5/2} (f+g x)^3}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d f-a e g} \sqrt{d+e x}}\right )}{8 g^{7/2} \sqrt{c d f-a e g}}\\ \end{align*}
Mathematica [A] time = 0.390151, size = 171, normalized size = 0.68 \[ \frac{\sqrt{(d+e x) (a e+c d x)} \left (\frac{15 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c d f-a e g}}\right )}{\sqrt{a e+c d x} \sqrt{c d f-a e g}}-\frac{\sqrt{g} \left (8 a^2 e^2 g^2+2 a c d e g (5 f+13 g x)+c^2 d^2 \left (15 f^2+40 f g x+33 g^2 x^2\right )\right )}{(f+g x)^3}\right )}{24 g^{7/2} \sqrt{d+e x}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.328, size = 441, normalized size = 1.7 \begin{align*} -{\frac{1}{24\,{g}^{3} \left ( gx+f \right ) ^{3}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 15\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ){x}^{3}{c}^{3}{d}^{3}{g}^{3}+45\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ){x}^{2}{c}^{3}{d}^{3}f{g}^{2}+45\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ) x{c}^{3}{d}^{3}{f}^{2}g+15\,{\it Artanh} \left ({\frac{\sqrt{cdx+ae}g}{\sqrt{ \left ( aeg-cdf \right ) g}}} \right ){c}^{3}{d}^{3}{f}^{3}+33\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}{x}^{2}{c}^{2}{d}^{2}{g}^{2}+26\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}xacde{g}^{2}+40\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}x{c}^{2}{d}^{2}fg+8\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}{a}^{2}{e}^{2}{g}^{2}+10\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}acdefg+15\,\sqrt{ \left ( aeg-cdf \right ) g}\sqrt{cdx+ae}{c}^{2}{d}^{2}{f}^{2} \right ){\frac{1}{\sqrt{ex+d}}}{\frac{1}{\sqrt{cdx+ae}}}{\frac{1}{\sqrt{ \left ( aeg-cdf \right ) g}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac{5}{2}}}{{\left (e x + d\right )}^{\frac{5}{2}}{\left (g x + f\right )}^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.90201, size = 2349, normalized size = 9.28 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [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.
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