Optimal. Leaf size=160 \[ -\frac{b B g \text{PolyLog}\left (2,\frac{d (a+b x)}{b (c+d x)}\right )}{d^2 i^2}-\frac{b g \log \left (\frac{b c-a d}{b (c+d x)}\right ) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2}-\frac{A g (a+b x)}{d i^2 (c+d x)}-\frac{B g (a+b x) \log \left (\frac{e (a+b x)}{c+d x}\right )}{d i^2 (c+d x)}+\frac{B g (a+b x)}{d i^2 (c+d x)} \]
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Rubi [A] time = 0.403279, antiderivative size = 222, normalized size of antiderivative = 1.39, number of steps used = 15, number of rules used = 11, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {2528, 2525, 12, 44, 2524, 2418, 2394, 2393, 2391, 2390, 2301} \[ -\frac{b B g \text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )}{d^2 i^2}+\frac{b g \log (c+d x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2}+\frac{g (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{d^2 i^2 (c+d x)}-\frac{B g (b c-a d)}{d^2 i^2 (c+d x)}-\frac{b B g \log (c+d x) \log \left (-\frac{d (a+b x)}{b c-a d}\right )}{d^2 i^2}-\frac{b B g \log (a+b x)}{d^2 i^2}+\frac{b B g \log ^2(c+d x)}{2 d^2 i^2}+\frac{b B g \log (c+d x)}{d^2 i^2} \]
Antiderivative was successfully verified.
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Rule 2528
Rule 2525
Rule 12
Rule 44
Rule 2524
Rule 2418
Rule 2394
Rule 2393
Rule 2391
Rule 2390
Rule 2301
Rubi steps
\begin{align*} \int \frac{(a g+b g x) \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{(41 c+41 d x)^2} \, dx &=\int \left (\frac{(-b c+a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d (c+d x)^2}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d (c+d x)}\right ) \, dx\\ &=\frac{(b g) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{c+d x} \, dx}{1681 d}-\frac{((b c-a d) g) \int \frac{A+B \log \left (\frac{e (a+b x)}{c+d x}\right )}{(c+d x)^2} \, dx}{1681 d}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac{(b B g) \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (c+d x)}{e (a+b x)} \, dx}{1681 d^2}-\frac{(B (b c-a d) g) \int \frac{b c-a d}{(a+b x) (c+d x)^2} \, dx}{1681 d^2}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac{\left (B (b c-a d)^2 g\right ) \int \frac{1}{(a+b x) (c+d x)^2} \, dx}{1681 d^2}-\frac{(b B g) \int \frac{(c+d x) \left (-\frac{d e (a+b x)}{(c+d x)^2}+\frac{b e}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{1681 d^2 e}\\ &=\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac{\left (B (b c-a d)^2 g\right ) \int \left (\frac{b^2}{(b c-a d)^2 (a+b x)}-\frac{d}{(b c-a d) (c+d x)^2}-\frac{b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{1681 d^2}-\frac{(b B g) \int \left (\frac{b e \log (c+d x)}{a+b x}-\frac{d e \log (c+d x)}{c+d x}\right ) \, dx}{1681 d^2 e}\\ &=-\frac{B (b c-a d) g}{1681 d^2 (c+d x)}-\frac{b B g \log (a+b x)}{1681 d^2}+\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b B g \log (c+d x)}{1681 d^2}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}-\frac{\left (b^2 B g\right ) \int \frac{\log (c+d x)}{a+b x} \, dx}{1681 d^2}+\frac{(b B g) \int \frac{\log (c+d x)}{c+d x} \, dx}{1681 d}\\ &=-\frac{B (b c-a d) g}{1681 d^2 (c+d x)}-\frac{b B g \log (a+b x)}{1681 d^2}+\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b B g \log (c+d x)}{1681 d^2}-\frac{b B g \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac{(b B g) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,c+d x\right )}{1681 d^2}+\frac{(b B g) \int \frac{\log \left (\frac{d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{1681 d}\\ &=-\frac{B (b c-a d) g}{1681 d^2 (c+d x)}-\frac{b B g \log (a+b x)}{1681 d^2}+\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b B g \log (c+d x)}{1681 d^2}-\frac{b B g \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac{b B g \log ^2(c+d x)}{3362 d^2}+\frac{(b B g) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{1681 d^2}\\ &=-\frac{B (b c-a d) g}{1681 d^2 (c+d x)}-\frac{b B g \log (a+b x)}{1681 d^2}+\frac{(b c-a d) g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right )}{1681 d^2 (c+d x)}+\frac{b B g \log (c+d x)}{1681 d^2}-\frac{b B g \log \left (-\frac{d (a+b x)}{b c-a d}\right ) \log (c+d x)}{1681 d^2}+\frac{b g \left (A+B \log \left (\frac{e (a+b x)}{c+d x}\right )\right ) \log (c+d x)}{1681 d^2}+\frac{b B g \log ^2(c+d x)}{3362 d^2}-\frac{b B g \text{Li}_2\left (\frac{b (c+d x)}{b c-a d}\right )}{1681 d^2}\\ \end{align*}
Mathematica [A] time = 0.162783, size = 175, normalized size = 1.09 \[ \frac{g \left (-b B \left (2 \text{PolyLog}\left (2,\frac{b (c+d x)}{b c-a d}\right )+\log (c+d x) \left (2 \log \left (\frac{d (a+b x)}{a d-b c}\right )-\log (c+d x)\right )\right )+2 b \log (c+d x) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )+\frac{2 (b c-a d) \left (B \log \left (\frac{e (a+b x)}{c+d x}\right )+A\right )}{c+d x}-2 B \left (\frac{b c-a d}{c+d x}+b \log (a+b x)-b \log (c+d x)\right )\right )}{2 d^2 i^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.061, size = 978, normalized size = 6.1 \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, B b g{\left (\frac{{\left (d x + c\right )} \log \left (d x + c\right )^{2} + 2 \, c \log \left (d x + c\right )}{d^{3} i^{2} x + c d^{2} i^{2}} - 2 \, \int \frac{d x \log \left (b x + a\right ) + d x \log \left (e\right ) + c}{d^{3} i^{2} x^{2} + 2 \, c d^{2} i^{2} x + c^{2} d i^{2}}\,{d x}\right )} + A b g{\left (\frac{c}{d^{3} i^{2} x + c d^{2} i^{2}} + \frac{\log \left (d x + c\right )}{d^{2} i^{2}}\right )} - B a g{\left (\frac{\log \left (\frac{b e x}{d x + c} + \frac{a e}{d x + c}\right )}{d^{2} i^{2} x + c d i^{2}} - \frac{1}{d^{2} i^{2} x + c d i^{2}} - \frac{b \log \left (b x + a\right )}{{\left (b c d - a d^{2}\right )} i^{2}} + \frac{b \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} i^{2}}\right )} - \frac{A a g}{d^{2} i^{2} x + c d i^{2}} \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 b g x + A a g +{\left (B b g x + B a g\right )} \log \left (\frac{b e x + a e}{d x + c}\right )}{d^{2} i^{2} x^{2} + 2 \, c d i^{2} x + c^{2} i^{2}}, 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 (b g x + a g\right )}{\left (B \log \left (\frac{{\left (b x + a\right )} e}{d x + c}\right ) + A\right )}}{{\left (d i x + c i\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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