Optimal. Leaf size=151 \[ -\frac{7 b^2 e}{\sqrt{d+e x} (b d-a e)^4}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}-\frac{7 b e}{3 (d+e x)^{3/2} (b d-a e)^3}-\frac{1}{(a+b x) (d+e x)^{5/2} (b d-a e)}-\frac{7 e}{5 (d+e x)^{5/2} (b d-a e)^2} \]
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Rubi [A] time = 0.0996385, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \[ -\frac{7 b^2 e}{\sqrt{d+e x} (b d-a e)^4}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}-\frac{7 b e}{3 (d+e x)^{3/2} (b d-a e)^3}-\frac{1}{(a+b x) (d+e x)^{5/2} (b d-a e)}-\frac{7 e}{5 (d+e x)^{5/2} (b d-a e)^2} \]
Antiderivative was successfully verified.
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Rule 27
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx &=\int \frac{1}{(a+b x)^2 (d+e x)^{7/2}} \, dx\\ &=-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{(7 e) \int \frac{1}{(a+b x) (d+e x)^{7/2}} \, dx}{2 (b d-a e)}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{(7 b e) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{2 (b d-a e)^2}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{\left (7 b^2 e\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{2 (b d-a e)^3}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}-\frac{\left (7 b^3 e\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 (b d-a e)^4}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}-\frac{\left (7 b^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 )}{(b d-a e)^4}\\ &=-\frac{7 e}{5 (b d-a e)^2 (d+e x)^{5/2}}-\frac{1}{(b d-a e) (a+b x) (d+e x)^{5/2}}-\frac{7 b e}{3 (b d-a e)^3 (d+e x)^{3/2}}-\frac{7 b^2 e}{(b d-a e)^4 \sqrt{d+e x}}+\frac{7 b^{5/2} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{9/2}}\\ \end{align*}
Mathematica [C] time = 0.0171591, size = 50, normalized size = 0.33 \[ -\frac{2 e \, _2F_1\left (-\frac{5}{2},2;-\frac{3}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (d+e x)^{5/2} (a e-b d)^2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.207, size = 149, normalized size = 1. \begin{align*} -{\frac{2\,e}{5\, \left ( ae-bd \right ) ^{2}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-6\,{\frac{{b}^{2}e}{ \left ( ae-bd \right ) ^{4}\sqrt{ex+d}}}+{\frac{4\,be}{3\, \left ( ae-bd \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}}-{\frac{e{b}^{3}}{ \left ( ae-bd \right ) ^{4} \left ( bxe+ae \right ) }\sqrt{ex+d}}-7\,{\frac{e{b}^{3}}{ \left ( ae-bd \right ) ^{4}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \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 = 2.13549, size = 2471, normalized size = 16.36 \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac{7}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.20258, size = 410, normalized size = 2.72 \begin{align*} -\frac{7 \, b^{3} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt{-b^{2} d + a b e}} - \frac{\sqrt{x e + d} b^{3} e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}} - \frac{2 \,{\left (45 \,{\left (x e + d\right )}^{2} b^{2} e + 10 \,{\left (x e + d\right )} b^{2} d e + 3 \, b^{2} d^{2} e - 10 \,{\left (x e + d\right )} a b e^{2} - 6 \, a b d e^{2} + 3 \, a^{2} e^{3}\right )}}{15 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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