Optimal. Leaf size=332 \[ \frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {PolyLog}\left (3,-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {PolyLog}\left (3,-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c} \]
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Rubi [A]
time = 0.40, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5783, 5846,
5827, 5680, 2221, 2611, 2320, 6724} \begin {gather*} \frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c} \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2320
Rule 2611
Rule 5680
Rule 5783
Rule 5827
Rule 5846
Rule 6724
Rubi steps
\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right ) \log \left (h (f+g x)^m\right )}{\sqrt {1+c^2 x^2}} \, dx &=\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {(g m) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{f+g x} \, dx}{2 b c}\\ &=\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {(g m) \text {Subst}\left (\int \frac {(a+b x)^2 \cosh (x)}{c f+g \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {(g m) \text {Subst}\left (\int \frac {e^x (a+b x)^2}{c f+e^x g-\sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}-\frac {(g m) \text {Subst}\left (\int \frac {e^x (a+b x)^2}{c f+e^x g+\sqrt {c^2 f^2+g^2}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {m \text {Subst}\left (\int (a+b x) \log \left (1+\frac {e^x g}{c f-\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac {m \text {Subst}\left (\int (a+b x) \log \left (1+\frac {e^x g}{c f+\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {(b m) \text {Subst}\left (\int \text {Li}_2\left (-\frac {e^x g}{c f-\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}+\frac {(b m) \text {Subst}\left (\int \text {Li}_2\left (-\frac {e^x g}{c f+\sqrt {c^2 f^2+g^2}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {(b m) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {g x}{-c f+\sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}+\frac {(b m) \text {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {g x}{c f+\sqrt {c^2 f^2+g^2}}\right )}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c}\\ &=\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}\\ \end {align*}
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Mathematica [A]
time = 0.18, size = 304, normalized size = 0.92 \begin {gather*} \frac {\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b}-m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )-m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )+\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )+2 b m \left (-\left (\left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {e^{\sinh ^{-1}(c x)} g}{-c f+\sqrt {c^2 f^2+g^2}}\right )\right )+b \text {PolyLog}\left (3,\frac {e^{\sinh ^{-1}(c x)} g}{-c f+\sqrt {c^2 f^2+g^2}}\right )\right )+2 b m \left (-\left (\left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )\right )+b \text {PolyLog}\left (3,-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )\right )}{2 b c} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right ) \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {c^{2} x^{2}+1}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {c^{2} x^{2} + 1}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {c^2\,x^2+1}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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