Optimal. Leaf size=277 \[ -\frac {2 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}+\frac {2 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}+\frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {22 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}+\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d} \]
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Rubi [A] time = 0.55, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5767, 5693, 4180, 2531, 2282, 6589, 5717, 8, 5758, 30} \[ -\frac {2 i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}+\frac {2 i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}+\frac {2 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {2 b x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}+\frac {22 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d}+\frac {2 b^2 x^3}{27 c^2 d}-\frac {22 b^2 x}{9 c^4 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 30
Rule 2282
Rule 2531
Rule 4180
Rule 5693
Rule 5717
Rule 5758
Rule 5767
Rule 6589
Rubi steps
\begin {align*} \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx &=\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {\int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^2}-\frac {(2 b) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 c d}\\ &=-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^4}+\frac {(4 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{9 c^3 d}+\frac {(2 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{c^3 d}+\frac {\left (2 b^2\right ) \int x^2 \, dx}{9 c^2 d}\\ &=\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}-\frac {\left (4 b^2\right ) \int 1 \, dx}{9 c^4 d}-\frac {\left (2 b^2\right ) \int 1 \, dx}{c^4 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {(2 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}+\frac {(2 i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {\left (2 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ \end {align*}
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Mathematica [A] time = 1.32, size = 365, normalized size = 1.32 \[ \frac {a^2 c^3 x^3-3 a^2 c x+3 a^2 \tan ^{-1}(c x)-\frac {2}{3} a b \left (-3 c^3 x^3 \sinh ^{-1}(c x)+c^2 x^2 \sqrt {c^2 x^2+1}-11 \sqrt {c^2 x^2+1}+9 i \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )-9 i \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )+9 c x \sinh ^{-1}(c x)-9 i \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+9 i \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )+3 b^2 \left (\frac {5}{2} \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+i \left (-2 \sinh ^{-1}(c x) \left (\text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-\text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )\right )-2 \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )+2 \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )-\left (\sinh ^{-1}(c x)^2 \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )-\frac {5}{4} c x \left (\sinh ^{-1}(c x)^2+2\right )+\frac {1}{108} \left (9 \sinh ^{-1}(c x)^2+2\right ) \sinh \left (3 \sinh ^{-1}(c x)\right )-\frac {1}{18} \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )\right )}{3 c^5 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{4}}{c^{2} d x^{2} + d}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{c^{2} d \,x^{2}+d}\, dx \]
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 \[ \frac {1}{3} \, a^{2} {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4} d} + \frac {3 \, \arctan \left (c x\right )}{c^{5} d}\right )} + \int \frac {b^{2} x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{2} d x^{2} + d} + \frac {2 \, a b x^{4} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{2} d x^{2} + d}\,{d x} \]
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 \[ \int \frac {x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{d\,c^2\,x^2+d} \,d x \]
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 \[ \frac {\int \frac {a^{2} x^{4}}{c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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