Optimal. Leaf size=214 \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{A e^2-B d e+C d^2}{e (d+e x) \left (a e^2+c d^2\right )}-\frac{\log (d+e x) \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{\sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]
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Rubi [A] time = 0.355451, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1629, 635, 205, 260} \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )}{2 \left (a e^2+c d^2\right )^2}-\frac{A e^2-B d e+C d^2}{e (d+e x) \left (a e^2+c d^2\right )}-\frac{\log (d+e x) \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )}{\left (a e^2+c d^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right )}{\sqrt{a} \sqrt{c} \left (a e^2+c d^2\right )^2} \]
Antiderivative was successfully verified.
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Rule 1629
Rule 635
Rule 205
Rule 260
Rubi steps
\begin{align*} \int \frac{A+B x+C x^2}{(d+e x)^2 \left (a+c x^2\right )} \, dx &=\int \left (\frac{C d^2-B d e+A e^2}{\left (c d^2+a e^2\right ) (d+e x)^2}+\frac{e \left (-B c d^2+2 A c d e-2 a C d e+a B e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac{A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )+c \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) x}{\left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}\right ) \, dx\\ &=-\frac{C d^2-B d e+A e^2}{e \left (c d^2+a e^2\right ) (d+e x)}-\frac{\left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac{\int \frac{A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )+c \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}\\ &=-\frac{C d^2-B d e+A e^2}{e \left (c d^2+a e^2\right ) (d+e x)}-\frac{\left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac{\left (c \left (B c d^2-2 A c d e+2 a C d e-a B e^2\right )\right ) \int \frac{x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}+\frac{\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) \int \frac{1}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^2}\\ &=-\frac{C d^2-B d e+A e^2}{e \left (c d^2+a e^2\right ) (d+e x)}+\frac{\left (A c \left (c d^2-a e^2\right )+a \left (a C e^2-c d (C d-2 B e)\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{c} \left (c d^2+a e^2\right )^2}-\frac{\left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^2}+\frac{\left (B c d^2-2 A c d e+2 a C d e-a B e^2\right ) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^2}\\ \end{align*}
Mathematica [A] time = 0.358212, size = 188, normalized size = 0.88 \[ \frac{\log \left (a+c x^2\right ) \left (-a B e^2+2 a C d e-2 A c d e+B c d^2\right )-\frac{2 \left (a e^2+c d^2\right ) \left (e (A e-B d)+C d^2\right )}{e (d+e x)}+\log (d+e x) \left (2 a B e^2-4 a C d e+4 A c d e-2 B c d^2\right )+\frac{2 \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (c d^2-a e^2\right )+a \left (a C e^2+c d (2 B e-C d)\right )\right )}{\sqrt{a} \sqrt{c}}}{2 \left (a e^2+c d^2\right )^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.058, size = 462, normalized size = 2.2 \begin{align*} -{\frac{c\ln \left ( c{x}^{2}+a \right ) Ade}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( c{x}^{2}+a \right ) Ba{e}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{c\ln \left ( c{x}^{2}+a \right ) B{d}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( c{x}^{2}+a \right ) Cade}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{aA{e}^{2}c}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{A{c}^{2}{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+2\,{\frac{Bacde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{\frac{{a}^{2}C{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{Cac{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{Ae}{ \left ( a{e}^{2}+c{d}^{2} \right ) \left ( ex+d \right ) }}+{\frac{Bd}{ \left ( a{e}^{2}+c{d}^{2} \right ) \left ( ex+d \right ) }}-{\frac{C{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) e \left ( ex+d \right ) }}+2\,{\frac{\ln \left ( ex+d \right ) Acde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}+{\frac{\ln \left ( ex+d \right ) aB{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-{\frac{\ln \left ( ex+d \right ) Bc{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( ex+d \right ) Cade}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2}}} \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 [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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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 = 1.15432, size = 365, normalized size = 1.71 \begin{align*} -\frac{{\left (C a c d^{2} e^{2} - A c^{2} d^{2} e^{2} - 2 \, B a c d e^{3} - C a^{2} e^{4} + A a c e^{4}\right )} \arctan \left (\frac{{\left (c d - \frac{c d^{2}}{x e + d} - \frac{a e^{2}}{x e + d}\right )} e^{\left (-1\right )}}{\sqrt{a c}}\right ) e^{\left (-2\right )}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt{a c}} + \frac{{\left (B c d^{2} + 2 \, C a d e - 2 \, A c d e - B a e^{2}\right )} \log \left (c - \frac{2 \, c d}{x e + d} + \frac{c d^{2}}{{\left (x e + d\right )}^{2}} + \frac{a e^{2}}{{\left (x e + d\right )}^{2}}\right )}{2 \,{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )}} - \frac{\frac{C d^{2} e}{x e + d} - \frac{B d e^{2}}{x e + d} + \frac{A e^{3}}{x e + d}}{c d^{2} e^{2} + a e^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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