Optimal. Leaf size=276 \[ \frac{B i^2 (b c-a d)^2 \text{PolyLog}\left (2,\frac{b (c+d x)}{d (a+b x)}\right )}{b^3 g}+\frac{d i^2 (a+b x) (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b^3 g}-\frac{i^2 (b c-a d)^2 \log \left (1-\frac{b (c+d x)}{d (a+b x)}\right ) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b^3 g}+\frac{i^2 (c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 b g}-\frac{B d i^2 x (b c-a d)}{2 b^2 g}-\frac{B i^2 (b c-a d)^2 \log \left (\frac{a+b x}{c+d x}\right )}{2 b^3 g}-\frac{3 B i^2 (b c-a d)^2 \log (c+d x)}{2 b^3 g} \]
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Rubi [A] time = 0.490415, antiderivative size = 354, normalized size of antiderivative = 1.28, number of steps used = 19, number of rules used = 13, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.325, Rules used = {2528, 2486, 31, 2524, 12, 2418, 2390, 2301, 2394, 2393, 2391, 2525, 43} \[ \frac{B i^2 (b c-a d)^2 \text{PolyLog}\left (2,-\frac{d (a+b x)}{b c-a d}\right )}{b^3 g}+\frac{i^2 (b c-a d)^2 \log (a g+b g x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{b^3 g}+\frac{A d i^2 x (b c-a d)}{b^2 g}+\frac{i^2 (c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{2 b g}+\frac{B d i^2 (a+b x) (b c-a d) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}-\frac{B d i^2 x (b c-a d)}{2 b^2 g}-\frac{B i^2 (b c-a d)^2 \log ^2(g (a+b x))}{2 b^3 g}-\frac{B i^2 (b c-a d)^2 \log (a+b x)}{2 b^3 g}-\frac{B i^2 (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{B i^2 (b c-a d)^2 \log (a g+b g x) \log \left (\frac{b (c+d x)}{b c-a d}\right )}{b^3 g} \]
Antiderivative was successfully verified.
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Rule 2528
Rule 2486
Rule 31
Rule 2524
Rule 12
Rule 2418
Rule 2390
Rule 2301
Rule 2394
Rule 2393
Rule 2391
Rule 2525
Rule 43
Rubi steps
\begin{align*} \int \frac{(14 c+14 d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{a g+b g x} \, dx &=\int \left (\frac{196 d (b c-a d) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^2 g}+\frac{14 d (14 c+14 d x) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b^2 (a g+b g x)}\right ) \, dx\\ &=\frac{\left (196 (b c-a d)^2\right ) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{a g+b g x} \, dx}{b^2}+\frac{(14 d) \int (14 c+14 d x) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \, dx}{b g}+\frac{(196 d (b c-a d)) \int \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \, dx}{b^2 g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac{B \int \frac{196 (b c-a d) (c+d x)}{a+b x} \, dx}{2 b g}+\frac{(196 B d (b c-a d)) \int \log \left (\frac{e (a+b x)}{c+d x}\right ) \, dx}{b^2 g}-\frac{\left (196 B (b c-a d)^2\right ) \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (a g+b g x)}{e (a+b x)} \, dx}{b^3 g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac{(98 B (b c-a d)) \int \frac{c+d x}{a+b x} \, dx}{b g}-\frac{\left (196 B d (b c-a d)^2\right ) \int \frac{1}{c+d x} \, dx}{b^3 g}-\frac{\left (196 B (b c-a d)^2\right ) \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (a g+b g x)}{a+b x} \, dx}{b^3 e g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}-\frac{196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac{(98 B (b c-a d)) \int \left (\frac{d}{b}+\frac{b c-a d}{b (a+b x)}\right ) \, dx}{b g}-\frac{\left (196 B (b c-a d)^2\right ) \int \left (\frac{b e \log (a g+b g x)}{a+b x}-\frac{d e \log (a g+b g x)}{c+d x}\right ) \, dx}{b^3 e g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}-\frac{98 B d (b c-a d) x}{b^2 g}-\frac{98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}-\frac{196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}-\frac{\left (196 B (b c-a d)^2\right ) \int \frac{\log (a g+b g x)}{a+b x} \, dx}{b^2 g}+\frac{\left (196 B d (b c-a d)^2\right ) \int \frac{\log (a g+b g x)}{c+d x} \, dx}{b^3 g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}-\frac{98 B d (b c-a d) x}{b^2 g}-\frac{98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}-\frac{196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac{196 B (b c-a d)^2 \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}-\frac{\left (196 B (b c-a d)^2\right ) \int \frac{\log \left (\frac{b g (c+d x)}{b c g-a d g}\right )}{a g+b g x} \, dx}{b^2}-\frac{\left (196 B (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{g \log (x)}{x} \, dx,x,a g+b g x\right )}{b^3 g^2}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}-\frac{98 B d (b c-a d) x}{b^2 g}-\frac{98 B (b c-a d)^2 \log (a+b x)}{b^3 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}-\frac{196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac{196 B (b c-a d)^2 \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}-\frac{\left (196 B (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,a g+b g x\right )}{b^3 g}-\frac{\left (196 B (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{d x}{b c g-a d g}\right )}{x} \, dx,x,a g+b g x\right )}{b^3 g}\\ &=\frac{196 A d (b c-a d) x}{b^2 g}-\frac{98 B d (b c-a d) x}{b^2 g}-\frac{98 B (b c-a d)^2 \log (a+b x)}{b^3 g}-\frac{98 B (b c-a d)^2 \log ^2(g (a+b x))}{b^3 g}+\frac{196 B d (b c-a d) (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{b^3 g}+\frac{98 (c+d x)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{b g}-\frac{196 B (b c-a d)^2 \log (c+d x)}{b^3 g}+\frac{196 (b c-a d)^2 \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (a g+b g x)}{b^3 g}+\frac{196 B (b c-a d)^2 \log \left (\frac{b (c+d x)}{b c-a d}\right ) \log (a g+b g x)}{b^3 g}+\frac{196 B (b c-a d)^2 \text{Li}_2\left (-\frac{d (a+b x)}{b c-a d}\right )}{b^3 g}\\ \end{align*}
Mathematica [A] time = 0.182786, size = 252, normalized size = 0.91 \[ \frac{i^2 \left (B (b c-a d)^2 \left (2 \text{PolyLog}\left (2,\frac{d (a+b x)}{a d-b c}\right )-\log (g (a+b x)) \left (\log (g (a+b x))-2 \log \left (\frac{b (c+d x)}{b c-a d}\right )\right )\right )+b^2 (c+d x)^2 \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )+2 (b c-a d)^2 \log (g (a+b x)) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )+2 A b d x (b c-a d)+2 B d (a+b x) (b c-a d) \log \left (\frac{e (a+b x)}{c+d x}\right )-2 B (b c-a d)^2 \log (c+d x)-B (b c-a d) ((b c-a d) \log (a+b x)+b d x)\right )}{2 b^3 g} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.176, size = 2538, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.61365, size = 699, normalized size = 2.53 \begin{align*} 2 \, A c d i^{2}{\left (\frac{x}{b g} - \frac{a \log \left (b x + a\right )}{b^{2} g}\right )} + \frac{1}{2} \, A d^{2} i^{2}{\left (\frac{2 \, a^{2} \log \left (b x + a\right )}{b^{3} g} + \frac{b x^{2} - 2 \, a x}{b^{2} g}\right )} + \frac{A c^{2} i^{2} \log \left (b g x + a g\right )}{b g} - \frac{{\left (3 \, b c^{2} i^{2} - 2 \, a c d i^{2}\right )} B \log \left (d x + c\right )}{2 \, b^{2} g} + \frac{{\left (b^{2} c^{2} i^{2} - 2 \, a b c d i^{2} + a^{2} d^{2} i^{2}\right )}{\left (\log \left (b x + a\right ) \log \left (\frac{b d x + a d}{b c - a d} + 1\right ) +{\rm Li}_2\left (-\frac{b d x + a d}{b c - a d}\right )\right )} B}{b^{3} g} + \frac{B b^{2} d^{2} i^{2} x^{2} \log \left (e\right ) +{\left (b^{2} c^{2} i^{2} - 2 \, a b c d i^{2} + a^{2} d^{2} i^{2}\right )} B \log \left (b x + a\right )^{2} +{\left ({\left (4 \, i^{2} \log \left (e\right ) - i^{2}\right )} b^{2} c d -{\left (2 \, i^{2} \log \left (e\right ) - i^{2}\right )} a b d^{2}\right )} B x +{\left (B b^{2} d^{2} i^{2} x^{2} + 2 \,{\left (2 \, b^{2} c d i^{2} - a b d^{2} i^{2}\right )} B x +{\left (2 \, b^{2} c^{2} i^{2} \log \left (e\right ) - 4 \,{\left (i^{2} \log \left (e\right ) - i^{2}\right )} a b c d +{\left (2 \, i^{2} \log \left (e\right ) - 3 \, i^{2}\right )} a^{2} d^{2}\right )} B\right )} \log \left (b x + a\right ) -{\left (B b^{2} d^{2} i^{2} x^{2} + 2 \,{\left (2 \, b^{2} c d i^{2} - a b d^{2} i^{2}\right )} B x + 2 \,{\left (b^{2} c^{2} i^{2} - 2 \, a b c d i^{2} + a^{2} d^{2} i^{2}\right )} B \log \left (b x + a\right )\right )} \log \left (d x + c\right )}{2 \, b^{3} g} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{A d^{2} i^{2} x^{2} + 2 \, A c d i^{2} x + A c^{2} i^{2} +{\left (B d^{2} i^{2} x^{2} + 2 \, B c d i^{2} x + B c^{2} i^{2}\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{b g x + a g}, x\right ) \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d i x + c i\right )}^{2}{\left (B \log \left (\frac{{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}}{b g x + a g}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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