Optimal. Leaf size=156 \[ -\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+\frac {2 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {i b c^3 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {i b c^3 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d} \]
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Rubi [A]
time = 0.18, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps
used = 15, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5809, 5789,
4265, 2317, 2438, 272, 65, 214, 44} \begin {gather*} \frac {2 c^3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}-\frac {i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {b c \sqrt {c^2 x^2+1}}{6 d x^2}+\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{6 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 44
Rule 65
Rule 214
Rule 272
Rule 2317
Rule 2438
Rule 4265
Rule 5789
Rule 5809
Rubi steps
\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^4 \left (d+c^2 d x^2\right )} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}-c^2 \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (d+c^2 d x^2\right )} \, dx+\frac {(b c) \int \frac {1}{x^3 \sqrt {1+c^2 x^2}} \, dx}{3 d}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+c^4 \int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 d}-\frac {\left (b c^3\right ) \int \frac {1}{x \sqrt {1+c^2 x^2}} \, dx}{d}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+\frac {c^3 \text {Subst}\left (\int (a+b x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{12 d}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+\frac {2 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {(b c) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {(b c) \text {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{d}-\frac {\left (i b c^3\right ) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac {\left (i b c^3\right ) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+\frac {2 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {\left (i b c^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac {b c \sqrt {1+c^2 x^2}}{6 d x^2}-\frac {a+b \sinh ^{-1}(c x)}{3 d x^3}+\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d x}+\frac {2 c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {7 b c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{6 d}-\frac {i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac {i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 247, normalized size = 1.58 \begin {gather*} \frac {-2 a+6 a c^2 x^2-b c x \sqrt {1+c^2 x^2}-2 b \sinh ^{-1}(c x)+6 b c^2 x^2 \sinh ^{-1}(c x)+6 a c^3 x^3 \text {ArcTan}(c x)+7 b c^3 x^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )-6 b \left (-c^2\right )^{3/2} x^3 \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+6 b \left (-c^2\right )^{3/2} x^3 \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+6 b \left (-c^2\right )^{3/2} x^3 \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-6 b \left (-c^2\right )^{3/2} x^3 \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{6 d x^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.68, size = 252, normalized size = 1.62
method | result | size |
derivativedivides | \(c^{3} \left (\frac {a \arctan \left (c x \right )}{d}-\frac {a}{3 d \,c^{3} x^{3}}+\frac {a}{d c x}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {b \arcsinh \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {b \arcsinh \left (c x \right )}{d c x}-\frac {b \sqrt {c^{2} x^{2}+1}}{6 d \,c^{2} x^{2}}+\frac {7 b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6 d}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}\right )\) | \(252\) |
default | \(c^{3} \left (\frac {a \arctan \left (c x \right )}{d}-\frac {a}{3 d \,c^{3} x^{3}}+\frac {a}{d c x}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {b \arcsinh \left (c x \right )}{3 d \,c^{3} x^{3}}+\frac {b \arcsinh \left (c x \right )}{d c x}-\frac {b \sqrt {c^{2} x^{2}+1}}{6 d \,c^{2} x^{2}}+\frac {7 b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6 d}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}\right )\) | \(252\) |
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*} \frac {\int \frac {a}{c^{2} x^{6} + x^{4}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{6} + x^{4}}\, dx}{d} \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 {a+b\,\mathrm {asinh}\left (c\,x\right )}{x^4\,\left (d\,c^2\,x^2+d\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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