Optimal. Leaf size=94 \[ -\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \]
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Rubi [A]
time = 0.13, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {5790, 5819,
5556, 3379} \begin {gather*} -\frac {c^3 \left (a^2 x^2+1\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a} \end {gather*}
Antiderivative was successfully verified.
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Rule 3379
Rule 5556
Rule 5790
Rule 5819
Rubi steps
\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^3}{\sinh ^{-1}(a x)^2} \, dx &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\left (7 a c^3\right ) \int \frac {x \left (1+a^2 x^2\right )^{5/2}}{\sinh ^{-1}(a x)} \, dx\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \frac {\cosh ^6(x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \left (\frac {5 \sinh (x)}{64 x}+\frac {9 \sinh (3 x)}{64 x}+\frac {5 \sinh (5 x)}{64 x}+\frac {\sinh (7 x)}{64 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {\left (7 c^3\right ) \text {Subst}\left (\int \frac {\sinh (7 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int \frac {\sinh (5 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}+\frac {\left (63 c^3\right ) \text {Subst}\left (\int \frac {\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{64 a}\\ &=-\frac {c^3 \left (1+a^2 x^2\right )^{7/2}}{a \sinh ^{-1}(a x)}+\frac {35 c^3 \text {Shi}\left (\sinh ^{-1}(a x)\right )}{64 a}+\frac {63 c^3 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{64 a}+\frac {35 c^3 \text {Shi}\left (5 \sinh ^{-1}(a x)\right )}{64 a}+\frac {7 c^3 \text {Shi}\left (7 \sinh ^{-1}(a x)\right )}{64 a}\\ \end {align*}
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Mathematica [A]
time = 0.25, size = 82, normalized size = 0.87 \begin {gather*} \frac {c^3 \left (-64 \left (1+a^2 x^2\right )^{7/2}+35 \sinh ^{-1}(a x) \text {Shi}\left (\sinh ^{-1}(a x)\right )+63 \sinh ^{-1}(a x) \text {Shi}\left (3 \sinh ^{-1}(a x)\right )+35 \sinh ^{-1}(a x) \text {Shi}\left (5 \sinh ^{-1}(a x)\right )+7 \sinh ^{-1}(a x) \text {Shi}\left (7 \sinh ^{-1}(a x)\right )\right )}{64 a \sinh ^{-1}(a x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 3.36, size = 106, normalized size = 1.13
method | result | size |
derivativedivides | \(\frac {c^{3} \left (7 \hyperbolicSineIntegral \left (7 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+63 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-21 \cosh \left (3 \arcsinh \left (a x \right )\right )-7 \cosh \left (5 \arcsinh \left (a x \right )\right )-\cosh \left (7 \arcsinh \left (a x \right )\right )-35 \sqrt {a^{2} x^{2}+1}\right )}{64 a \arcsinh \left (a x \right )}\) | \(106\) |
default | \(\frac {c^{3} \left (7 \hyperbolicSineIntegral \left (7 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+63 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (5 \arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )+35 \hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right ) \arcsinh \left (a x \right )-21 \cosh \left (3 \arcsinh \left (a x \right )\right )-7 \cosh \left (5 \arcsinh \left (a x \right )\right )-\cosh \left (7 \arcsinh \left (a x \right )\right )-35 \sqrt {a^{2} x^{2}+1}\right )}{64 a \arcsinh \left (a x \right )}\) | \(106\) |
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*} c^{3} \left (\int \frac {3 a^{2} x^{2}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {3 a^{4} x^{4}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {a^{6} x^{6}}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx + \int \frac {1}{\operatorname {asinh}^{2}{\left (a x \right )}}\, dx\right ) \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 = -0.01 \begin {gather*} \int \frac {{\left (c\,a^2\,x^2+c\right )}^3}{{\mathrm {asinh}\left (a\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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