Optimal. Leaf size=103 \[ \frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]
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Rubi [A] time = 0.137605, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {1142, 1114, 638, 618, 206} \[ \frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 e \left (b^2-4 a c\right ) \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{e \left (b^2-4 a c\right )^{3/2}} \]
Antiderivative was successfully verified.
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Rule 1142
Rule 1114
Rule 638
Rule 618
Rule 206
Rubi steps
\begin{align*} \int \frac{(d f+e f x)^3}{\left (a+b (d+e x)^2+c (d+e x)^4\right )^2} \, dx &=\frac{f^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b x^2+c x^4\right )^2} \, dx,x,d+e x\right )}{e}\\ &=\frac{f^3 \operatorname{Subst}\left (\int \frac{x}{\left (a+b x+c x^2\right )^2} \, dx,x,(d+e x)^2\right )}{2 e}\\ &=\frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}+\frac{\left (b f^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,(d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e}\\ &=\frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{\left (b f^3\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c (d+e x)^2\right )}{\left (b^2-4 a c\right ) e}\\ &=\frac{f^3 \left (2 a+b (d+e x)^2\right )}{2 \left (b^2-4 a c\right ) e \left (a+b (d+e x)^2+c (d+e x)^4\right )}-\frac{b f^3 \tanh ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} e}\\ \end{align*}
Mathematica [A] time = 0.134716, size = 103, normalized size = 1. \[ \frac{f^3 \left (\frac{2 a+b (d+e x)^2}{\left (b^2-4 a c\right ) \left (a+(d+e x)^2 \left (b+c (d+e x)^2\right )\right )}-\frac{2 b \tan ^{-1}\left (\frac{b+2 c (d+e x)^2}{\sqrt{4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}\right )}{2 e} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.017, size = 500, normalized size = 4.9 \begin{align*} -{\frac{{f}^{3}be{x}^{2}}{ \left ( 2\,c{e}^{4}{x}^{4}+8\,cd{e}^{3}{x}^{3}+12\,c{d}^{2}{e}^{2}{x}^{2}+8\,c{d}^{3}ex+2\,b{e}^{2}{x}^{2}+2\,c{d}^{4}+4\,bdex+2\,b{d}^{2}+2\,a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{f}^{3}bdx}{ \left ( c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{f}^{3}b{d}^{2}}{ \left ( 2\,c{e}^{4}{x}^{4}+8\,cd{e}^{3}{x}^{3}+12\,c{d}^{2}{e}^{2}{x}^{2}+8\,c{d}^{3}ex+2\,b{e}^{2}{x}^{2}+2\,c{d}^{4}+4\,bdex+2\,b{d}^{2}+2\,a \right ) e \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{f}^{3}a}{ \left ( c{e}^{4}{x}^{4}+4\,cd{e}^{3}{x}^{3}+6\,c{d}^{2}{e}^{2}{x}^{2}+4\,c{d}^{3}ex+b{e}^{2}{x}^{2}+c{d}^{4}+2\,bdex+b{d}^{2}+a \right ) e \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{{f}^{3}b}{ \left ( 8\,ac-2\,{b}^{2} \right ) e}\sum _{{\it \_R}={\it RootOf} \left ( c{e}^{4}{{\it \_Z}}^{4}+4\,cd{e}^{3}{{\it \_Z}}^{3}+ \left ( 6\,c{d}^{2}{e}^{2}+b{e}^{2} \right ){{\it \_Z}}^{2}+ \left ( 4\,c{d}^{3}e+2\,bde \right ){\it \_Z}+c{d}^{4}+b{d}^{2}+a \right ) }{\frac{ \left ( -{\it \_R}\,e-d \right ) \ln \left ( x-{\it \_R} \right ) }{2\,c{e}^{3}{{\it \_R}}^{3}+6\,cd{e}^{2}{{\it \_R}}^{2}+6\,c{d}^{2}e{\it \_R}+2\,c{d}^{3}+be{\it \_R}+bd}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -b f^{3} \int -\frac{e x + d}{{\left (b^{2} c - 4 \, a c^{2}\right )} e^{4} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{3} x^{3} +{\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{2} x^{2} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e x}\,{d x} + \frac{b e^{2} f^{3} x^{2} + 2 \, b d e f^{3} x +{\left (b d^{2} + 2 \, a\right )} f^{3}}{2 \,{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} e^{5} x^{4} + 4 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d e^{4} x^{3} +{\left (b^{3} - 4 \, a b c + 6 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{2}\right )} e^{3} x^{2} + 2 \,{\left (2 \,{\left (b^{2} c - 4 \, a c^{2}\right )} d^{3} +{\left (b^{3} - 4 \, a b c\right )} d\right )} e^{2} x +{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} d^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a b c\right )} d^{2}\right )} e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.67859, size = 2290, normalized size = 22.23 \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 [B] time = 39.0261, size = 554, normalized size = 5.38 \begin{align*} \frac{b f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{- 16 a^{2} b c^{2} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{3} c f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - b^{5} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} f^{3} + 2 b c d^{2} f^{3}}{2 b c e^{2} f^{3}} \right )}}{2 e} - \frac{b f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} \log{\left (\frac{2 d x}{e} + x^{2} + \frac{16 a^{2} b c^{2} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{3} c f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{5} f^{3} \sqrt{- \frac{1}{\left (4 a c - b^{2}\right )^{3}}} + b^{2} f^{3} + 2 b c d^{2} f^{3}}{2 b c e^{2} f^{3}} \right )}}{2 e} - \frac{2 a f^{3} + b d^{2} f^{3} + 2 b d e f^{3} x + b e^{2} f^{3} x^{2}}{8 a^{2} c e - 2 a b^{2} e + 8 a b c d^{2} e + 8 a c^{2} d^{4} e - 2 b^{3} d^{2} e - 2 b^{2} c d^{4} e + x^{4} \left (8 a c^{2} e^{5} - 2 b^{2} c e^{5}\right ) + x^{3} \left (32 a c^{2} d e^{4} - 8 b^{2} c d e^{4}\right ) + x^{2} \left (8 a b c e^{3} + 48 a c^{2} d^{2} e^{3} - 2 b^{3} e^{3} - 12 b^{2} c d^{2} e^{3}\right ) + x \left (16 a b c d e^{2} + 32 a c^{2} d^{3} e^{2} - 4 b^{3} d e^{2} - 8 b^{2} c d^{3} e^{2}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 2.37564, size = 522, normalized size = 5.07 \begin{align*} \frac{{\left (b^{3} f^{3} e - 4 \, a b c f^{3} e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left |{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} + 2 \,{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d x e +{\left (b + \sqrt{b^{2} - 4 \, a c}\right )} d^{2} + 2 \, a \right |}\right )}{2 \,{\left (b^{6} e^{2} - 12 \, a b^{4} c e^{2} + 48 \, a^{2} b^{2} c^{2} e^{2} - 64 \, a^{3} c^{3} e^{2}\right )}} - \frac{{\left (b^{3} f^{3} e - 4 \, a b c f^{3} e\right )} \sqrt{b^{2} - 4 \, a c} \log \left ({\left | -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} x^{2} e^{2} - 2 \,{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d x e -{\left (b - \sqrt{b^{2} - 4 \, a c}\right )} d^{2} - 2 \, a \right |}\right )}{2 \,{\left (b^{6} e^{2} - 12 \, a b^{4} c e^{2} + 48 \, a^{2} b^{2} c^{2} e^{2} - 64 \, a^{3} c^{3} e^{2}\right )}} + \frac{b f^{3} x^{2} e^{2} + 2 \, b d f^{3} x e + b d^{2} f^{3} + 2 \, a f^{3}}{2 \,{\left (c x^{4} e^{4} + 4 \, c d x^{3} e^{3} + 6 \, c d^{2} x^{2} e^{2} + 4 \, c d^{3} x e + c d^{4} + b x^{2} e^{2} + 2 \, b d x e + b d^{2} + a\right )}{\left (b^{2} e - 4 \, a c e\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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