Optimal. Leaf size=200 \[ \frac{e \sqrt{d+e x} \left (3 b^2 e^2-2 b c d e+2 c^2 d^2\right )}{b^2 c^2}-\frac{(c d-b e)^{5/2} (3 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{5/2}}+\frac{d^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{5/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e (d+e x)^{3/2} (2 c d-b e)}{b^2 c} \]
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Rubi [A] time = 0.395153, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {738, 824, 826, 1166, 208} \[ \frac{e \sqrt{d+e x} \left (3 b^2 e^2-2 b c d e+2 c^2 d^2\right )}{b^2 c^2}-\frac{(c d-b e)^{5/2} (3 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{5/2}}+\frac{d^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(d+e x)^{5/2} (x (2 c d-b e)+b d)}{b^2 \left (b x+c x^2\right )}+\frac{e (d+e x)^{3/2} (2 c d-b e)}{b^2 c} \]
Antiderivative was successfully verified.
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Rule 738
Rule 824
Rule 826
Rule 1166
Rule 208
Rubi steps
\begin{align*} \int \frac{(d+e x)^{7/2}}{\left (b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{(d+e x)^{3/2} \left (\frac{1}{2} d (4 c d-7 b e)-\frac{3}{2} e (2 c d-b e) x\right )}{b x+c x^2} \, dx}{b^2}\\ &=\frac{e (2 c d-b e) (d+e x)^{3/2}}{b^2 c}-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\sqrt{d+e x} \left (\frac{1}{2} c d^2 (4 c d-7 b e)-\frac{1}{2} e \left (2 c^2 d^2-2 b c d e+3 b^2 e^2\right ) x\right )}{b x+c x^2} \, dx}{b^2 c}\\ &=\frac{e \left (2 c^2 d^2-2 b c d e+3 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{3/2}}{b^2 c}-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\int \frac{\frac{1}{2} c^2 d^3 (4 c d-7 b e)+\frac{1}{2} e (2 c d-b e) \left (c^2 d^2-b c d e-3 b^2 e^2\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{b^2 c^2}\\ &=\frac{e \left (2 c^2 d^2-2 b c d e+3 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{3/2}}{b^2 c}-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} c^2 d^3 e (4 c d-7 b e)-\frac{1}{2} d e (2 c d-b e) \left (c^2 d^2-b c d e-3 b^2 e^2\right )+\frac{1}{2} e (2 c d-b e) \left (c^2 d^2-b c d e-3 b^2 e^2\right ) x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{b^2 c^2}\\ &=\frac{e \left (2 c^2 d^2-2 b c d e+3 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{3/2}}{b^2 c}-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}-\frac{\left (c d^3 (4 c d-7 b e)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3}+\frac{\left ((c d-b e)^3 (4 c d+3 b e)\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b^3 c^2}\\ &=\frac{e \left (2 c^2 d^2-2 b c d e+3 b^2 e^2\right ) \sqrt{d+e x}}{b^2 c^2}+\frac{e (2 c d-b e) (d+e x)^{3/2}}{b^2 c}-\frac{(d+e x)^{5/2} (b d+(2 c d-b e) x)}{b^2 \left (b x+c x^2\right )}+\frac{d^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3}-\frac{(c d-b e)^{5/2} (4 c d+3 b e) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b^3 c^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.311376, size = 167, normalized size = 0.84 \[ \frac{\frac{b \sqrt{d+e x} \left (b^2 c e^2 x (2 e x-3 d)+3 b^3 e^3 x-b c^2 d^2 (d-3 e x)-2 c^3 d^3 x\right )}{c^2 x (b+c x)}-\frac{(c d-b e)^{5/2} (3 b e+4 c d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{c^{5/2}}+d^{5/2} (4 c d-7 b e) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.229, size = 403, normalized size = 2. \begin{align*} 2\,{\frac{{e}^{3}\sqrt{ex+d}}{{c}^{2}}}+{\frac{{e}^{4}b}{{c}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-3\,{\frac{{e}^{3}\sqrt{ex+d}d}{c \left ( cex+be \right ) }}+3\,{\frac{{e}^{2}\sqrt{ex+d}{d}^{2}}{b \left ( cex+be \right ) }}-{\frac{ce{d}^{3}}{{b}^{2} \left ( cex+be \right ) }\sqrt{ex+d}}-3\,{\frac{{e}^{4}b}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+5\,{\frac{{e}^{3}d}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+3\,{\frac{{d}^{2}{e}^{2}}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-9\,{\frac{ce{d}^{3}}{{b}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+4\,{\frac{{c}^{2}{d}^{4}}{{b}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-{\frac{{d}^{3}}{{b}^{2}x}\sqrt{ex+d}}-7\,{\frac{e{d}^{5/2}}{{b}^{2}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+4\,{\frac{{d}^{7/2}c}{{b}^{3}}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \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 [A] time = 6.24772, size = 2628, normalized size = 13.14 \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 = 1.38429, size = 464, normalized size = 2.32 \begin{align*} \frac{2 \, \sqrt{x e + d} e^{3}}{c^{2}} - \frac{{\left (4 \, c d^{4} - 7 \, b d^{3} e\right )} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b^{3} \sqrt{-d}} + \frac{{\left (4 \, c^{4} d^{4} - 9 \, b c^{3} d^{3} e + 3 \, b^{2} c^{2} d^{2} e^{2} + 5 \, b^{3} c d e^{3} - 3 \, b^{4} e^{4}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b^{3} c^{2}} - \frac{2 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{3} d^{3} e - 2 \, \sqrt{x e + d} c^{3} d^{4} e - 3 \,{\left (x e + d\right )}^{\frac{3}{2}} b c^{2} d^{2} e^{2} + 4 \, \sqrt{x e + d} b c^{2} d^{3} e^{2} + 3 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{2} c d e^{3} - 3 \, \sqrt{x e + d} b^{2} c d^{2} e^{3} -{\left (x e + d\right )}^{\frac{3}{2}} b^{3} e^{4} + \sqrt{x e + d} b^{3} d e^{4}}{{\left ({\left (x e + d\right )}^{2} c - 2 \,{\left (x e + d\right )} c d + c d^{2} +{\left (x e + d\right )} b e - b d e\right )} b^{2} c^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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