Optimal. Leaf size=206 \[ \frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {5 \log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c} \]
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Rubi [A]
time = 0.21, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5791, 3393,
3384, 3379, 3382} \begin {gather*} \frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{32 b c}+\frac {5 \log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c} \end {gather*}
Antiderivative was successfully verified.
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Rule 3379
Rule 3382
Rule 3384
Rule 3393
Rule 5791
Rubi steps
\begin {align*} \int \frac {\left (1+c^2 x^2\right )^{5/2}}{a+b \sinh ^{-1}(c x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\cosh ^6(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {\text {Subst}\left (\int \left (\frac {5}{16 (a+b x)}+\frac {15 \cosh (2 x)}{32 (a+b x)}+\frac {3 \cosh (4 x)}{16 (a+b x)}+\frac {\cosh (6 x)}{32 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c}\\ &=\frac {5 \log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c}+\frac {\text {Subst}\left (\int \frac {\cosh (6 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}+\frac {3 \text {Subst}\left (\int \frac {\cosh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}+\frac {15 \text {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}\\ &=\frac {5 \log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c}+\frac {\left (15 \cosh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}+\frac {\left (3 \cosh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}-\frac {\left (15 \sinh \left (\frac {2 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}-\frac {\left (3 \sinh \left (\frac {4 a}{b}\right )\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{32 c}\\ &=\frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c}+\frac {3 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c}+\frac {\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c}+\frac {5 \log \left (a+b \sinh ^{-1}(c x)\right )}{16 b c}-\frac {15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \sinh ^{-1}(c x)\right )}{32 b c}-\frac {3 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 a}{b}+4 \sinh ^{-1}(c x)\right )}{16 b c}-\frac {\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (\frac {6 a}{b}+6 \sinh ^{-1}(c x)\right )}{32 b c}\\ \end {align*}
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Mathematica [A]
time = 0.41, size = 153, normalized size = 0.74 \begin {gather*} \frac {15 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+6 \cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+\cosh \left (\frac {6 a}{b}\right ) \text {Chi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+10 \log \left (a+b \sinh ^{-1}(c x)\right )-15 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-6 \sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-\sinh \left (\frac {6 a}{b}\right ) \text {Shi}\left (6 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{32 b c} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 12.64, size = 199, normalized size = 0.97
method | result | size |
default | \(\frac {5 \ln \left (a +b \arcsinh \left (c x \right )\right )}{16 b c}-\frac {{\mathrm e}^{\frac {6 a}{b}} \expIntegral \left (1, 6 \arcsinh \left (c x \right )+\frac {6 a}{b}\right )}{64 c b}-\frac {3 \,{\mathrm e}^{\frac {4 a}{b}} \expIntegral \left (1, 4 \arcsinh \left (c x \right )+\frac {4 a}{b}\right )}{32 c b}-\frac {15 \,{\mathrm e}^{\frac {2 a}{b}} \expIntegral \left (1, 2 \arcsinh \left (c x \right )+\frac {2 a}{b}\right )}{64 c b}-\frac {15 \,{\mathrm e}^{-\frac {2 a}{b}} \expIntegral \left (1, -2 \arcsinh \left (c x \right )-\frac {2 a}{b}\right )}{64 c b}-\frac {3 \,{\mathrm e}^{-\frac {4 a}{b}} \expIntegral \left (1, -4 \arcsinh \left (c x \right )-\frac {4 a}{b}\right )}{32 c b}-\frac {{\mathrm e}^{-\frac {6 a}{b}} \expIntegral \left (1, -6 \arcsinh \left (c x \right )-\frac {6 a}{b}\right )}{64 c b}\) | \(199\) |
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 (c^{2} x^{2} + 1\right )^{\frac {5}{2}}}{a + b \operatorname {asinh}{\left (c x \right )}}\, 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 {{\left (c^2\,x^2+1\right )}^{5/2}}{a+b\,\mathrm {asinh}\left (c\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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