Optimal. Leaf size=332 \[ \frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )}{c} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.56, 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 = {5675, 5838, 5799, 5561, 2190, 2531, 2282, 6589} \[ -\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}+\frac {b m \text {PolyLog}\left (3,-\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}\right )}{c}+\frac {b m \text {PolyLog}\left (3,-\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}\right )}{c}+\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b^2 c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )}{2 b c}-\frac {m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )}{2 b c}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{2 b c} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2190
Rule 2282
Rule 2531
Rule 5561
Rule 5675
Rule 5799
Rule 5838
Rule 6589
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) \operatorname {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) \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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*}
________________________________________________________________________________________
Mathematica [A] time = 0.23, size = 304, normalized size = 0.92 \[ \frac {2 b m \left (b \text {Li}_3\left (\frac {e^{\sinh ^{-1}(c x)} g}{\sqrt {c^2 f^2+g^2}-c f}\right )-\left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (\frac {e^{\sinh ^{-1}(c x)} g}{\sqrt {c^2 f^2+g^2}-c f}\right )\right )+2 b m \left (b \text {Li}_3\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )-\left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-\frac {e^{\sinh ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2+g^2}}\right )\right )-m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{c f-\sqrt {c^2 f^2+g^2}}+1\right )-m \left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (\frac {g e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 f^2+g^2}+c f}+1\right )+\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )+\frac {m \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b}}{2 b c} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 2.13, 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.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {c^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {c^2\,x^2+1}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {c^{2} x^{2} + 1}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________