Optimal. Leaf size=236 \[ -\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt{c d^2-a e^2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}} \]
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Rubi [A] time = 0.151789, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {662, 660, 205} \[ -\frac{5 c^2 d^2 \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 e^3 (d+e x)^{3/2}}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{\sqrt{d+e x} \sqrt{c d^2-a e^2}}\right )}{8 e^{7/2} \sqrt{c d^2-a e^2}}-\frac{5 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{12 e^2 (d+e x)^{7/2}}-\frac{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}} \]
Antiderivative was successfully verified.
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Rule 662
Rule 660
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)^{13/2}} \, dx &=-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\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)^{9/2}} \, dx}{6 e}\\ &=-\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 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\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}}{(d+e x)^{5/2}} \, dx}{8 e^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 e^3 (d+e x)^{3/2}}-\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 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{\left (5 c^3 d^3\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{16 e^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 e^3 (d+e x)^{3/2}}-\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 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{\left (5 c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 c d^2 e-e \left (c d^2+a e^2\right )+e^2 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 e^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 e^3 (d+e x)^{3/2}}-\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 e^2 (d+e x)^{7/2}}-\frac{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{3 e (d+e x)^{11/2}}+\frac{5 c^3 d^3 \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{c d^2-a e^2} \sqrt{d+e x}}\right )}{8 e^{7/2} \sqrt{c d^2-a e^2}}\\ \end{align*}
Mathematica [A] time = 0.258554, size = 220, normalized size = 0.93 \[ \frac{15 c^3 d^3 (d+e x)^3 \sqrt{a e+c d x} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d^2-a e^2}}\right )-\sqrt{e} \sqrt{c d^2-a e^2} \left (2 a^2 c d e^3 (5 d+17 e x)+8 a^3 e^5+a c^2 d^2 e \left (15 d^2+50 d e x+59 e^2 x^2\right )+c^3 d^3 x \left (15 d^2+40 d e x+33 e^2 x^2\right )\right )}{24 e^{7/2} (d+e x)^{5/2} \sqrt{c d^2-a e^2} \sqrt{(d+e x) (a e+c d x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.247, size = 443, normalized size = 1.9 \begin{align*} -{\frac{1}{24\,{e}^{3}}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( 15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{3}{c}^{3}{d}^{3}{e}^{3}+45\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){x}^{2}{c}^{3}{d}^{4}{e}^{2}+45\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ) x{c}^{3}{d}^{5}e+15\,{\it Artanh} \left ({\frac{e\sqrt{cdx+ae}}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}} \right ){c}^{3}{d}^{6}+33\,{x}^{2}{c}^{2}{d}^{2}{e}^{2}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+26\,xacd{e}^{3}\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+40\,x{c}^{2}{d}^{3}e\sqrt{cdx+ae}\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}+8\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{a}^{2}{e}^{4}+10\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}ac{d}^{2}{e}^{2}+15\,\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}\sqrt{cdx+ae}{c}^{2}{d}^{4} \right ) \left ( ex+d \right ) ^{-{\frac{7}{2}}}{\frac{1}{\sqrt{cdx+ae}}}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) e}}}} \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{13}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.99794, size = 1709, normalized size = 7.24 \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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