Optimal. Leaf size=279 \[ \frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}-\frac {9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac {3 b e^3 \sqrt {(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac {9 b e^3 \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}-\frac {3 b^3 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{128 d}+\frac {45 b^3 e^3 \sqrt {(c+d x)^2+1} (c+d x)}{256 d}-\frac {45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d} \]
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Rubi [A] time = 0.39, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {5865, 12, 5661, 5758, 5675, 321, 215} \[ \frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}-\frac {9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac {3 b e^3 \sqrt {(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac {9 b e^3 \sqrt {(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}-\frac {3 b^3 e^3 \sqrt {(c+d x)^2+1} (c+d x)^3}{128 d}+\frac {45 b^3 e^3 \sqrt {(c+d x)^2+1} (c+d x)}{256 d}-\frac {45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 215
Rule 321
Rule 5661
Rule 5675
Rule 5758
Rule 5865
Rubi steps
\begin {align*} \int (c e+d e x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int e^3 x^3 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 \operatorname {Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac {\left (3 b e^3\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{4 d}\\ &=-\frac {3 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac {\left (9 b e^3\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{16 d}+\frac {\left (3 b^2 e^3\right ) \operatorname {Subst}\left (\int x^3 \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{8 d}\\ &=\frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {9 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac {\left (9 b e^3\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{32 d}-\frac {\left (9 b^2 e^3\right ) \operatorname {Subst}\left (\int x \left (a+b \sinh ^{-1}(x)\right ) \, dx,x,c+d x\right )}{16 d}-\frac {\left (3 b^3 e^3\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{32 d}\\ &=-\frac {3 b^3 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{128 d}-\frac {9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {9 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}+\frac {\left (9 b^3 e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{128 d}+\frac {\left (9 b^3 e^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{32 d}\\ &=\frac {45 b^3 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{256 d}-\frac {3 b^3 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{128 d}-\frac {9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {9 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}-\frac {\left (9 b^3 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{256 d}-\frac {\left (9 b^3 e^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{64 d}\\ &=\frac {45 b^3 e^3 (c+d x) \sqrt {1+(c+d x)^2}}{256 d}-\frac {3 b^3 e^3 (c+d x)^3 \sqrt {1+(c+d x)^2}}{128 d}-\frac {45 b^3 e^3 \sinh ^{-1}(c+d x)}{256 d}-\frac {9 b^2 e^3 (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {3 b^2 e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )}{32 d}+\frac {9 b e^3 (c+d x) \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{32 d}-\frac {3 b e^3 (c+d x)^3 \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{16 d}-\frac {3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{32 d}+\frac {e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^3}{4 d}\\ \end {align*}
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Mathematica [A] time = 0.40, size = 303, normalized size = 1.09 \[ \frac {e^3 \left (8 a \left (8 a^2+3 b^2\right ) (c+d x)^4+3 b \sqrt {(c+d x)^2+1} (c+d x) \left (3 \left (8 a^2+5 b^2\right )-2 \left (8 a^2+b^2\right ) (c+d x)^2\right )-24 b (c+d x) \sinh ^{-1}(c+d x) \left (-8 a^2 (c+d x)^3+4 a b \sqrt {(c+d x)^2+1} (c+d x)^2-6 a b \sqrt {(c+d x)^2+1}-b^2 (c+d x)^3+3 b^2 (c+d x)\right )-9 b \left (8 a^2+5 b^2\right ) \sinh ^{-1}(c+d x)-72 a b^2 (c+d x)^2+24 b^2 \sinh ^{-1}(c+d x)^2 \left (8 a (c+d x)^4-3 a-2 b \sqrt {(c+d x)^2+1} (c+d x)^3+3 b \sqrt {(c+d x)^2+1} (c+d x)\right )+8 b^3 \left (8 (c+d x)^4-3\right ) \sinh ^{-1}(c+d x)^3\right )}{256 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.63, size = 832, normalized size = 2.98 \[ \frac {8 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} d^{4} e^{3} x^{4} + 32 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} c d^{3} e^{3} x^{3} - 24 \, {\left (3 \, a b^{2} - 2 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} c^{2}\right )} d^{2} e^{3} x^{2} - 16 \, {\left (9 \, a b^{2} c - 2 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} c^{3}\right )} d e^{3} x + 8 \, {\left (8 \, b^{3} d^{4} e^{3} x^{4} + 32 \, b^{3} c d^{3} e^{3} x^{3} + 48 \, b^{3} c^{2} d^{2} e^{3} x^{2} + 32 \, b^{3} c^{3} d e^{3} x + {\left (8 \, b^{3} c^{4} - 3 \, b^{3}\right )} e^{3}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{3} + 24 \, {\left (8 \, a b^{2} d^{4} e^{3} x^{4} + 32 \, a b^{2} c d^{3} e^{3} x^{3} + 48 \, a b^{2} c^{2} d^{2} e^{3} x^{2} + 32 \, a b^{2} c^{3} d e^{3} x + {\left (8 \, a b^{2} c^{4} - 3 \, a b^{2}\right )} e^{3} - {\left (2 \, b^{3} d^{3} e^{3} x^{3} + 6 \, b^{3} c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b^{3} c^{2} - b^{3}\right )} d e^{3} x + {\left (2 \, b^{3} c^{3} - 3 \, b^{3} c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2} + 3 \, {\left (8 \, {\left (8 \, a^{2} b + b^{3}\right )} d^{4} e^{3} x^{4} + 32 \, {\left (8 \, a^{2} b + b^{3}\right )} c d^{3} e^{3} x^{3} - 24 \, {\left (b^{3} - 2 \, {\left (8 \, a^{2} b + b^{3}\right )} c^{2}\right )} d^{2} e^{3} x^{2} - 16 \, {\left (3 \, b^{3} c - 2 \, {\left (8 \, a^{2} b + b^{3}\right )} c^{3}\right )} d e^{3} x - {\left (24 \, b^{3} c^{2} - 8 \, {\left (8 \, a^{2} b + b^{3}\right )} c^{4} + 24 \, a^{2} b + 15 \, b^{3}\right )} e^{3} - 16 \, {\left (2 \, a b^{2} d^{3} e^{3} x^{3} + 6 \, a b^{2} c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, a b^{2} c^{2} - a b^{2}\right )} d e^{3} x + {\left (2 \, a b^{2} c^{3} - 3 \, a b^{2} c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) - 3 \, {\left (2 \, {\left (8 \, a^{2} b + b^{3}\right )} d^{3} e^{3} x^{3} + 6 \, {\left (8 \, a^{2} b + b^{3}\right )} c d^{2} e^{3} x^{2} - 3 \, {\left (8 \, a^{2} b + 5 \, b^{3} - 2 \, {\left (8 \, a^{2} b + b^{3}\right )} c^{2}\right )} d e^{3} x + {\left (2 \, {\left (8 \, a^{2} b + b^{3}\right )} c^{3} - 3 \, {\left (8 \, a^{2} b + 5 \, b^{3}\right )} c\right )} e^{3}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}{256 \, d} \]
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 \[ \int {\left (d e x + c e\right )}^{3} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 365, normalized size = 1.31 \[ \frac {\frac {\left (d x +c \right )^{4} e^{3} a^{3}}{4}+e^{3} b^{3} \left (\frac {\left (d x +c \right )^{4} \arcsinh \left (d x +c \right )^{3}}{4}-\frac {3 \left (d x +c \right )^{3} \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {9 \arcsinh \left (d x +c \right )^{2} \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right )}{32}-\frac {3 \arcsinh \left (d x +c \right )^{3}}{32}+\frac {3 \left (d x +c \right )^{4} \arcsinh \left (d x +c \right )}{32}-\frac {3 \left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{128}+\frac {45 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{256}+\frac {27 \arcsinh \left (d x +c \right )}{256}-\frac {9 \arcsinh \left (d x +c \right ) \left (1+\left (d x +c \right )^{2}\right )}{32}\right )+3 e^{3} a \,b^{2} \left (\frac {\left (d x +c \right )^{4} \arcsinh \left (d x +c \right )^{2}}{4}-\frac {\left (d x +c \right )^{3} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{8}+\frac {3 \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}\, \left (d x +c \right )}{16}-\frac {3 \arcsinh \left (d x +c \right )^{2}}{32}+\frac {\left (d x +c \right )^{4}}{32}-\frac {3 \left (d x +c \right )^{2}}{32}-\frac {3}{32}\right )+3 e^{3} a^{2} b \left (\frac {\left (d x +c \right )^{4} \arcsinh \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3} \sqrt {1+\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsinh \left (d x +c \right )}{32}\right )}{d} \]
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 \[ \text {result too large to display} \]
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 \[ \int {\left (c\,e+d\,e\,x\right )}^3\,{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 10.07, size = 1828, normalized size = 6.55 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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