Optimal. Leaf size=165 \[ \frac {\left (3-4 a^2\right ) a e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}+\frac {\left (3-4 a^2\right ) a \sinh ^{-1}(a+b x)}{8 b^4}-\frac {\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}-\frac {\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}+\frac {3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac {3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac {e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac {e^{5 \sinh ^{-1}(a+b x)}}{80 b^4} \]
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Rubi [A] time = 0.17, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5898, 2282, 12, 1628} \[ \frac {\left (3-4 a^2\right ) a e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}+\frac {\left (3-4 a^2\right ) a \sinh ^{-1}(a+b x)}{8 b^4}-\frac {\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}-\frac {\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}+\frac {3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac {3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac {e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac {e^{5 \sinh ^{-1}(a+b x)}}{80 b^4} \]
Antiderivative was successfully verified.
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Rule 12
Rule 1628
Rule 2282
Rule 5898
Rubi steps
\begin {align*} \int e^{\sinh ^{-1}(a+b x)} x^3 \, dx &=\frac {\operatorname {Subst}\left (\int e^x \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^3 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1-x^2\right ) \left (1+2 a x-x^2\right )^3}{16 b^3 x^4} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1-x^2\right ) \left (1+2 a x-x^2\right )^3}{x^4} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{16 b^4}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{x^4}-\frac {6 a}{x^3}-\frac {2 \left (-1+6 a^2\right )}{x^2}+\frac {2 a \left (3-4 a^2\right )}{x}+2 a \left (3-4 a^2\right ) x+2 \left (-1+6 a^2\right ) x^2-6 a x^3+x^4\right ) \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{16 b^4}\\ &=\frac {e^{-3 \sinh ^{-1}(a+b x)}}{48 b^4}+\frac {3 a e^{-2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac {\left (1-6 a^2\right ) e^{-\sinh ^{-1}(a+b x)}}{8 b^4}+\frac {a \left (3-4 a^2\right ) e^{2 \sinh ^{-1}(a+b x)}}{16 b^4}-\frac {\left (1-6 a^2\right ) e^{3 \sinh ^{-1}(a+b x)}}{24 b^4}-\frac {3 a e^{4 \sinh ^{-1}(a+b x)}}{32 b^4}+\frac {e^{5 \sinh ^{-1}(a+b x)}}{80 b^4}+\frac {a \left (3-4 a^2\right ) \sinh ^{-1}(a+b x)}{8 b^4}\\ \end {align*}
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Mathematica [A] time = 0.09, size = 119, normalized size = 0.72 \[ \frac {15 a \left (3-4 a^2\right ) \sinh ^{-1}(a+b x)-\sqrt {a^2+2 a b x+b^2 x^2+1} \left (6 a^4+2 \left (3 a^2-4\right ) b^2 x^2+\left (29-6 a^2\right ) a b x-83 a^2-6 a b^3 x^3-24 b^4 x^4+16\right )+30 a b^4 x^4+24 b^5 x^5}{120 b^4} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.56, size = 138, normalized size = 0.84 \[ \frac {24 \, b^{5} x^{5} + 30 \, a b^{4} x^{4} + 15 \, {\left (4 \, a^{3} - 3 \, a\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (24 \, b^{4} x^{4} + 6 \, a b^{3} x^{3} - 2 \, {\left (3 \, a^{2} - 4\right )} b^{2} x^{2} - 6 \, a^{4} + {\left (6 \, a^{3} - 29 \, a\right )} b x + 83 \, a^{2} - 16\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{120 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.41, size = 173, normalized size = 1.05 \[ \frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} + \frac {1}{120} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left ({\left (2 \, {\left (3 \, {\left (4 \, x + \frac {a}{b}\right )} x - \frac {3 \, a^{2} b^{5} - 4 \, b^{5}}{b^{7}}\right )} x + \frac {6 \, a^{3} b^{4} - 29 \, a b^{4}}{b^{7}}\right )} x - \frac {6 \, a^{4} b^{3} - 83 \, a^{2} b^{3} + 16 \, b^{3}}{b^{7}}\right )} + \frac {{\left (4 \, a^{3} - 3 \, a\right )} \log \left (-a b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} {\left | b \right |}\right )}{8 \, b^{3} {\left | b \right |}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 322, normalized size = 1.95 \[ \frac {x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{5 b^{2}}-\frac {7 a x \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{20 b^{3}}+\frac {9 a^{2} \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{20 b^{4}}-\frac {a^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{2 b^{3}}-\frac {a^{4} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 b^{4}}-\frac {a^{3} \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{2 b^{3} \sqrt {b^{2}}}+\frac {3 a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\, x}{8 b^{3}}+\frac {3 a^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{8 b^{4}}+\frac {3 a \ln \left (\frac {b^{2} x +a b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}\right )}{8 b^{3} \sqrt {b^{2}}}-\frac {2 \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )^{\frac {3}{2}}}{15 b^{4}}+\frac {b \,x^{5}}{5}+\frac {x^{4} a}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.70, size = 491, normalized size = 2.98 \[ \frac {1}{5} \, b x^{5} + \frac {1}{4} \, a x^{4} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{5 \, b^{2}} - \frac {{\left (a^{2} + 1\right )} a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a x}{20 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a x}{5 \, b^{3}} + \frac {{\left (a^{2} + 1\right )}^{2} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{5 \, b^{4}} + \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} a^{2}}{12 \, b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + 1\right )} a^{2}}{5 \, b^{4}} + \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} a^{3} \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{40 \, b^{6}} - \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac {3}{2}} {\left (a^{2} + 1\right )}}{15 \, b^{4}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a x}{40 \, b^{5}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} {\left (a^{2} + 1\right )} a \operatorname {arsinh}\left (\frac {2 \, {\left (b^{2} x + a b\right )}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}}}\right )}{40 \, b^{6}} - \frac {7 \, {\left (5 \, a^{2} b^{2} - {\left (a^{2} + 1\right )} b^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2}}{40 \, b^{6}} \]
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 \[ \int x^3\,\left (a+\sqrt {{\left (a+b\,x\right )}^2+1}+b\,x\right ) \,d x \]
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 \[ \int x^{3} \left (a + b x + \sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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