Optimal. Leaf size=162 \[ -\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (a+b x) (b d-a e)}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}+\frac{5 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4} \]
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Rubi [A] time = 0.0892755, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {27, 47, 51, 63, 208} \[ -\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (a+b x) (b d-a e)}-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}+\frac{5 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4} \]
Antiderivative was successfully verified.
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Rule 27
Rule 47
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b x) (d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{(d+e x)^{5/2}}{(a+b x)^5} \, dx\\ &=-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac{(5 e) \int \frac{(d+e x)^{3/2}}{(a+b x)^4} \, dx}{8 b}\\ &=-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac{\left (5 e^2\right ) \int \frac{\sqrt{d+e x}}{(a+b x)^3} \, dx}{16 b^2}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac{\left (5 e^3\right ) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{64 b^3}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}-\frac{\left (5 e^4\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{128 b^3 (b d-a e)}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}-\frac{\left (5 e^3\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{64 b^3 (b d-a e)}\\ &=-\frac{5 e^2 \sqrt{d+e x}}{32 b^3 (a+b x)^2}-\frac{5 e^3 \sqrt{d+e x}}{64 b^3 (b d-a e) (a+b x)}-\frac{5 e (d+e x)^{3/2}}{24 b^2 (a+b x)^3}-\frac{(d+e x)^{5/2}}{4 b (a+b x)^4}+\frac{5 e^4 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{64 b^{7/2} (b d-a e)^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.0188115, size = 52, normalized size = 0.32 \[ \frac{2 e^4 (d+e x)^{7/2} \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};-\frac{b (d+e x)}{a e-b d}\right )}{7 (a e-b d)^5} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.014, size = 246, normalized size = 1.5 \begin{align*}{\frac{5\,{e}^{4}}{64\, \left ( bex+ae \right ) ^{4} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{7}{2}}}}-{\frac{73\,{e}^{4}}{192\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{5}{2}}}}-{\frac{55\,{e}^{5}a}{192\, \left ( bex+ae \right ) ^{4}{b}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{55\,{e}^{4}d}{192\, \left ( bex+ae \right ) ^{4}b} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{e}^{6}{a}^{2}}{64\, \left ( bex+ae \right ) ^{4}{b}^{3}}\sqrt{ex+d}}+{\frac{5\,{e}^{5}ad}{32\, \left ( bex+ae \right ) ^{4}{b}^{2}}\sqrt{ex+d}}-{\frac{5\,{e}^{4}{d}^{2}}{64\, \left ( bex+ae \right ) ^{4}b}\sqrt{ex+d}}+{\frac{5\,{e}^{4}}{ \left ( 64\,ae-64\,bd \right ){b}^{3}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Fricas [B] time = 1.12195, size = 1859, normalized size = 11.48 \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 [A] time = 1.20026, size = 358, normalized size = 2.21 \begin{align*} -\frac{5 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{4}}{64 \,{\left (b^{4} d - a b^{3} e\right )} \sqrt{-b^{2} d + a b e}} - \frac{15 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{3} e^{4} + 73 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{3} d e^{4} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d^{2} e^{4} + 15 \, \sqrt{x e + d} b^{3} d^{3} e^{4} - 73 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{2} e^{5} + 110 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} d e^{5} - 45 \, \sqrt{x e + d} a b^{2} d^{2} e^{5} - 55 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b e^{6} + 45 \, \sqrt{x e + d} a^{2} b d e^{6} - 15 \, \sqrt{x e + d} a^{3} e^{7}}{192 \,{\left (b^{4} d - a b^{3} e\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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