Optimal. Leaf size=386 \[ -\frac{2 b^2 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
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Rubi [A] time = 0.531256, antiderivative size = 386, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.27, Rules used = {5712, 5690, 5687, 5714, 3718, 2190, 2279, 2391, 5717, 191} \[ -\frac{2 b^2 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (c^2 x^2+1\right )^{5/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b^2 x \left (c^2 x^2+1\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
Antiderivative was successfully verified.
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Rule 5712
Rule 5690
Rule 5687
Rule 5714
Rule 3718
Rule 2190
Rule 2279
Rule 2391
Rule 5717
Rule 191
Rubi steps
\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (2 b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (4 b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (4 b \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (8 b \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (4 b^2 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 b^2 \left (1+c^2 x^2\right )^{5/2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 b \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 b^2 \left (1+c^2 x^2\right )^{5/2} \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end{align*}
Mathematica [A] time = 7.82093, size = 642, normalized size = 1.66 \[ \frac{-b^2 \left (-16 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-16 \left (c^2 x^2+1\right )^{3/2} \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+2 \sqrt{c^2 x^2+1} \left (\left (6 \sinh ^{-1}(c x)-3 i \pi \right ) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+i \left (-3 i \sinh ^{-1}(c x)^2+6 \pi \sinh ^{-1}(c x)+2 i \sinh ^{-1}(c x)+3 \left (\pi -2 i \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-12 \pi \log \left (e^{\sinh ^{-1}(c x)}+1\right )+3 \pi \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-3 \pi \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+12 \pi \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )+c x-6 c x \sinh ^{-1}(c x)^2-2 \sinh \left (3 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)^2+\sinh \left (3 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x)^2 \cosh \left (3 \sinh ^{-1}(c x)\right )+4 i \pi \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )+4 \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+4 \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )-2 i \pi \log \left (1-i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+2 i \pi \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )-8 i \pi \log \left (e^{\sinh ^{-1}(c x)}+1\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+8 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )-2 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (-\cos \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+2 i \pi \cosh \left (3 \sinh ^{-1}(c x)\right ) \log \left (\sin \left (\frac{1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )\right )+4 a^2 c x \left (2 c^2 x^2+3\right )+2 a b \left (\sqrt{c^2 x^2+1} \left (2-3 \log \left (c^2 x^2+1\right )\right )-\log \left (c^2 x^2+1\right ) \cosh \left (3 \sinh ^{-1}(c x)\right )+2 \sinh ^{-1}(c x) \left (3 c x+\sinh \left (3 \sinh ^{-1}(c x)\right )\right )\right )}{12 d^2 f^2 \left (c^3 x^2+c\right ) \sqrt{d+i c d x} \sqrt{f-i c f x}} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.278, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{-{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{2}}}}\, dx \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}{3} \, a b c{\left (\frac{1}{c^{4} d^{\frac{5}{2}} f^{\frac{5}{2}} x^{2} + c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac{5}{2}} f^{\frac{5}{2}}}\right )} + \frac{2}{3} \, a b{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a^{2}{\left (\frac{x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac{3}{2}} d f} + \frac{2 \, x}{\sqrt{c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} + b^{2} \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{{\left (i \, c d x + d\right )}^{\frac{5}{2}}{\left (-i \, c f x + f\right )}^{\frac{5}{2}}}\,{d x} \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*} \frac{{\left (2 \, b^{2} c^{2} x^{3} + 3 \, b^{2} x\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} + 3 \,{\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}{\rm integral}\left (\frac{3 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a^{2} + 2 \,{\left (3 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a b -{\left (2 \, b^{2} c^{3} x^{3} + 3 \, b^{2} c x\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{3 \,{\left (c^{6} d^{3} f^{3} x^{6} + 3 \, c^{4} d^{3} f^{3} x^{4} + 3 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}}, x\right )}{3 \,{\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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