Optimal. Leaf size=196 \[ \frac{63 b^2 e^2}{4 \sqrt{d+e x} (b d-a e)^5}-\frac{63 b^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{11/2}}+\frac{21 b e^2}{4 (d+e x)^{3/2} (b d-a e)^4}+\frac{63 e^2}{20 (d+e x)^{5/2} (b d-a e)^3}+\frac{9 e}{4 (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac{1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]
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Rubi [A] time = 0.139166, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ \frac{63 b^2 e^2}{4 \sqrt{d+e x} (b d-a e)^5}-\frac{63 b^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{11/2}}+\frac{21 b e^2}{4 (d+e x)^{3/2} (b d-a e)^4}+\frac{63 e^2}{20 (d+e x)^{5/2} (b d-a e)^3}+\frac{9 e}{4 (a+b x) (d+e x)^{5/2} (b d-a e)^2}-\frac{1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)} \]
Antiderivative was successfully verified.
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Rule 27
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{1}{(a+b x)^3 (d+e x)^{7/2}} \, dx\\ &=-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}-\frac{(9 e) \int \frac{1}{(a+b x)^2 (d+e x)^{7/2}} \, dx}{4 (b d-a e)}\\ &=-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{\left (63 e^2\right ) \int \frac{1}{(a+b x) (d+e x)^{7/2}} \, dx}{8 (b d-a e)^2}\\ &=\frac{63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{\left (63 b e^2\right ) \int \frac{1}{(a+b x) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^3}\\ &=\frac{63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{21 b e^2}{4 (b d-a e)^4 (d+e x)^{3/2}}+\frac{\left (63 b^2 e^2\right ) \int \frac{1}{(a+b x) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^4}\\ &=\frac{63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{21 b e^2}{4 (b d-a e)^4 (d+e x)^{3/2}}+\frac{63 b^2 e^2}{4 (b d-a e)^5 \sqrt{d+e x}}+\frac{\left (63 b^3 e^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 (b d-a e)^5}\\ &=\frac{63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{21 b e^2}{4 (b d-a e)^4 (d+e x)^{3/2}}+\frac{63 b^2 e^2}{4 (b d-a e)^5 \sqrt{d+e x}}+\frac{\left (63 b^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 (b d-a e)^5}\\ &=\frac{63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac{1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac{21 b e^2}{4 (b d-a e)^4 (d+e x)^{3/2}}+\frac{63 b^2 e^2}{4 (b d-a e)^5 \sqrt{d+e x}}-\frac{63 b^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 (b d-a e)^{11/2}}\\ \end{align*}
Mathematica [C] time = 0.0177296, size = 52, normalized size = 0.27 \[ -\frac{2 e^2 \, _2F_1\left (-\frac{5}{2},3;-\frac{3}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (d+e x)^{5/2} (a e-b d)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.019, size = 231, normalized size = 1.2 \begin{align*} -{\frac{2\,{e}^{2}}{5\, \left ( ae-bd \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-12\,{\frac{{b}^{2}{e}^{2}}{ \left ( ae-bd \right ) ^{5}\sqrt{ex+d}}}+2\,{\frac{b{e}^{2}}{ \left ( ae-bd \right ) ^{4} \left ( ex+d \right ) ^{3/2}}}-{\frac{15\,{e}^{2}{b}^{4}}{4\, \left ( ae-bd \right ) ^{5} \left ( bex+ae \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{17\,{b}^{3}{e}^{3}a}{4\, \left ( ae-bd \right ) ^{5} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{17\,{e}^{2}{b}^{4}d}{4\, \left ( ae-bd \right ) ^{5} \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}-{\frac{63\,{b}^{3}{e}^{2}}{4\, \left ( ae-bd \right ) ^{5}}\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 = 1.22795, size = 3753, normalized size = 19.15 \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 [B] time = 1.17055, size = 512, normalized size = 2.61 \begin{align*} \frac{63 \, b^{3} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt{-b^{2} d + a b e}} + \frac{15 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} e^{2} - 17 \, \sqrt{x e + d} b^{4} d e^{2} + 17 \, \sqrt{x e + d} a b^{3} e^{3}}{4 \,{\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} + \frac{2 \,{\left (30 \,{\left (x e + d\right )}^{2} b^{2} e^{2} + 5 \,{\left (x e + d\right )} b^{2} d e^{2} + b^{2} d^{2} e^{2} - 5 \,{\left (x e + d\right )} a b e^{3} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}}{5 \,{\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\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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