Optimal. Leaf size=162 \[ \frac{b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{-c f+d e-f}\right )}{2 f (m+1) (m+2) (d e-(c+1) f)}-\frac{b d (e+f x)^{m+2} \text{Hypergeometric2F1}\left (1,m+2,m+3,\frac{d (e+f x)}{-c f+d e+f}\right )}{2 f (m+1) (m+2) (-c f+d e+f)}+\frac{(e+f x)^{m+1} \left (a+b \coth ^{-1}(c+d x)\right )}{f (m+1)} \]
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Rubi [A] time = 0.246004, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {6112, 5927, 712, 68} \[ \frac{(e+f x)^{m+1} \left (a+b \coth ^{-1}(c+d x)\right )}{f (m+1)}+\frac{b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{d (e+f x)}{d e-c f-f}\right )}{2 f (m+1) (m+2) (d e-(c+1) f)}-\frac{b d (e+f x)^{m+2} \, _2F_1\left (1,m+2;m+3;\frac{d (e+f x)}{d e-c f+f}\right )}{2 f (m+1) (m+2) (-c f+d e+f)} \]
Antiderivative was successfully verified.
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Rule 6112
Rule 5927
Rule 712
Rule 68
Rubi steps
\begin{align*} \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^m \left (a+b \coth ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^{1+m}}{1-x^2} \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac{(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac{b \operatorname{Subst}\left (\int \left (\frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^{1+m}}{2 (1-x)}+\frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^{1+m}}{2 (1+x)}\right ) \, dx,x,c+d x\right )}{f (1+m)}\\ &=\frac{(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^{1+m}}{1-x} \, dx,x,c+d x\right )}{2 f (1+m)}-\frac{b \operatorname{Subst}\left (\int \frac{\left (\frac{d e-c f}{d}+\frac{f x}{d}\right )^{1+m}}{1+x} \, dx,x,c+d x\right )}{2 f (1+m)}\\ &=\frac{(e+f x)^{1+m} \left (a+b \coth ^{-1}(c+d x)\right )}{f (1+m)}+\frac{b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac{d (e+f x)}{d e-f-c f}\right )}{2 f (d e-(1+c) f) (1+m) (2+m)}-\frac{b d (e+f x)^{2+m} \, _2F_1\left (1,2+m;3+m;\frac{d (e+f x)}{d e+f-c f}\right )}{2 f (d e+f-c f) (1+m) (2+m)}\\ \end{align*}
Mathematica [F] time = 2.35383, size = 0, normalized size = 0. \[ \int (e+f x)^m \left (a+b \coth ^{-1}(c+d x)\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 1.493, size = 0, normalized size = 0. \begin{align*} \int \left ( fx+e \right ) ^{m} \left ( a+b{\rm arccoth} \left (dx+c\right ) \right ) \, dx \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] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \operatorname{arcoth}\left (d x + c\right ) + a\right )}{\left (f x + e\right )}^{m}, 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{\left (b \operatorname{arcoth}\left (d x + c\right ) + a\right )}{\left (f x + e\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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