Optimal. Leaf size=320 \[ -\frac {3 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {3 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^5 d^3}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (c^2 x^2+1\right )}+\frac {3 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {3 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {7 b^2 \tan ^{-1}(c x)}{6 c^5 d^3}-\frac {b^2 x}{12 c^4 d^3 \left (c^2 x^2+1\right )} \]
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Rubi [A] time = 0.55, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5751, 5693, 4180, 2531, 2282, 6589, 5717, 203, 266, 43, 5732, 12, 385} \[ -\frac {3 i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {3 i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {3 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {3 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (c^2 x^2+1\right )}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac {3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^5 d^3}-\frac {b^2 x}{12 c^4 d^3 \left (c^2 x^2+1\right )}+\frac {7 b^2 \tan ^{-1}(c x)}{6 c^5 d^3} \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 203
Rule 266
Rule 385
Rule 2282
Rule 2531
Rule 4180
Rule 5693
Rule 5717
Rule 5732
Rule 5751
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^3} \, dx &=-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}+\frac {b \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 c d^3}+\frac {3 \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{2 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}-\frac {b^2 \int \frac {-2-3 c^2 x^2}{3 c^4 \left (1+c^2 x^2\right )^2} \, dx}{2 d^3}+\frac {(3 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{4 c^3 d^3}+\frac {3 \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{8 c^4 d^2}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 c^5 d^3}-\frac {b^2 \int \frac {-2-3 c^2 x^2}{\left (1+c^2 x^2\right )^2} \, dx}{6 c^4 d^3}+\frac {\left (3 b^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^4 d^3}\\ &=-\frac {b^2 x}{12 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 b^2 \tan ^{-1}(c x)}{4 c^5 d^3}-\frac {(3 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {(3 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {\left (5 b^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{12 c^4 d^3}\\ &=-\frac {b^2 x}{12 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {7 b^2 \tan ^{-1}(c x)}{6 c^5 d^3}-\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^5 d^3}-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 c^5 d^3}\\ &=-\frac {b^2 x}{12 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {7 b^2 \tan ^{-1}(c x)}{6 c^5 d^3}-\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}\\ &=-\frac {b^2 x}{12 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {7 b^2 \tan ^{-1}(c x)}{6 c^5 d^3}-\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {3 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}\\ \end {align*}
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Mathematica [A] time = 3.40, size = 552, normalized size = 1.72 \[ \frac {-\frac {15 a^2 c x}{c^2 x^2+1}+\frac {6 a^2 c x}{\left (c^2 x^2+1\right )^2}+9 a^2 \tan ^{-1}(c x)+\frac {15 a b \left (\sqrt {c^2 x^2+1}-i \sinh ^{-1}(c x)\right )}{-1+i c x}+\frac {15 a b \left (\sqrt {c^2 x^2+1}+i \sinh ^{-1}(c x)\right )}{-1-i c x}-\frac {i a b \left (3 \sinh ^{-1}(c x)+\sqrt {c^2 x^2+1} (c x-2 i)\right )}{(c x-i)^2}+\frac {i a b \left (3 \sinh ^{-1}(c x)+\sqrt {c^2 x^2+1} (c x+2 i)\right )}{(c x+i)^2}+\frac {9}{2} i a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )\right )-\frac {9}{2} i a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )\right )+b^2 \left (-\frac {2 c x}{c^2 x^2+1}-\frac {15 c x \sinh ^{-1}(c x)^2}{c^2 x^2+1}+\frac {6 c x \sinh ^{-1}(c x)^2}{\left (c^2 x^2+1\right )^2}-\frac {30 \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {4 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}-18 i \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )+18 i \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )-18 i \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )+18 i \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )-9 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+9 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+56 \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{24 c^5 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{4}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (c^{2} d \,x^{2}+d \right )^{3}}\, dx \]
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 \[ -\frac {1}{8} \, a^{2} {\left (\frac {5 \, c^{2} x^{3} + 3 \, x}{c^{8} d^{3} x^{4} + 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}} - \frac {3 \, \arctan \left (c x\right )}{c^{5} d^{3}}\right )} + \int \frac {b^{2} x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}} + \frac {2 \, a b x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^3} \,d x \]
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 \[ \frac {\int \frac {a^{2} x^{4}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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