Optimal. Leaf size=148 \[ \frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2}}-\frac {b c x}{6 \pi ^{5/2} \left (c^2 x^2+1\right )}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}-\frac {7 b \tan ^{-1}(c x)}{6 \pi ^{5/2}} \]
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Rubi [A] time = 0.34, antiderivative size = 187, normalized size of antiderivative = 1.26, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5755, 5760, 4182, 2279, 2391, 203, 199} \[ -\frac {b \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {b \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi c^2 x^2+\pi }}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {2 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2}}-\frac {b c x}{6 \pi ^2 \sqrt {c^2 x^2+1} \sqrt {\pi c^2 x^2+\pi }}-\frac {7 b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi c^2 x^2+\pi }} \]
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 2279
Rule 2391
Rule 4182
Rule 5755
Rule 5760
Rubi steps
\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \left (\pi +c^2 \pi x^2\right )^{3/2}} \, dx}{\pi }-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{3 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b c x}{6 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {\pi +c^2 \pi x^2}} \, dx}{\pi ^2}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {\left (b c \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}\\ &=-\frac {b c x}{6 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}+\frac {\operatorname {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{5/2}}\\ &=-\frac {b c x}{6 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}-\frac {b \operatorname {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{5/2}}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{\pi ^{5/2}}\\ &=-\frac {b c x}{6 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}-\frac {b \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {b \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}\\ &=-\frac {b c x}{6 \pi ^2 \sqrt {1+c^2 x^2} \sqrt {\pi +c^2 \pi x^2}}+\frac {a+b \sinh ^{-1}(c x)}{3 \pi \left (\pi +c^2 \pi x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{\pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {7 b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{6 \pi ^2 \sqrt {\pi +c^2 \pi x^2}}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}-\frac {b \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}+\frac {b \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\pi ^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.77, size = 209, normalized size = 1.41 \[ \frac {\frac {6 a}{\sqrt {c^2 x^2+1}}+\frac {2 a}{\left (c^2 x^2+1\right )^{3/2}}-6 a \log \left (\pi \left (\sqrt {c^2 x^2+1}+1\right )\right )+6 a \log (x)-\frac {b c x}{c^2 x^2+1}+\frac {6 b c^2 x^2 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+\frac {8 b \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+6 b \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-6 b \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+6 b \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-6 b \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-14 b \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{6 \pi ^{5/2}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {\pi + \pi c^{2} x^{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\pi ^{3} c^{6} x^{7} + 3 \, \pi ^{3} c^{4} x^{5} + 3 \, \pi ^{3} c^{2} x^{3} + \pi ^{3} x}, x\right ) \]
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 \[ \int \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.26, size = 220, normalized size = 1.49 \[ \frac {a}{3 \pi \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {a}{\pi ^{2} \sqrt {\pi \,c^{2} x^{2}+\pi }}-\frac {a \arctanh \left (\frac {\sqrt {\pi }}{\sqrt {\pi \,c^{2} x^{2}+\pi }}\right )}{\pi ^{\frac {5}{2}}}+\frac {b \arcsinh \left (c x \right ) x^{2} c^{2}}{\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {b c x}{6 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}+\frac {4 b \arcsinh \left (c x \right )}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}}}-\frac {7 b \arctan \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{3 \pi ^{\frac {5}{2}}}-\frac {b \dilog \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {5}{2}}}-\frac {b \dilog \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {5}{2}}}-\frac {b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\pi ^{\frac {5}{2}}} \]
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 \[ -\frac {1}{3} \, a {\left (\frac {3 \, \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{\pi ^{\frac {5}{2}}} - \frac {1}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}}} - \frac {3}{\pi ^{2} \sqrt {\pi + \pi c^{2} x^{2}}}\right )} + b \int \frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{x\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2}} \,d x \]
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 \[ \frac {\int \frac {a}{c^{4} x^{5} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{5} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{3} \sqrt {c^{2} x^{2} + 1} + x \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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