Optimal. Leaf size=87 \[ \frac {b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac {b^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]
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Rubi [A] time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3676, 390, 385, 203} \[ \frac {b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}-\frac {b^3 \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 203
Rule 385
Rule 390
Rule 3676
Rubi steps
\begin {align*} \int \cosh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+(a+b) x^2\right )^3}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left ((a-2 b) (a+b)^2+(a+b)^3 x^2+\frac {b^2 (3 a+2 b)+3 b^2 (a+b) x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}+\frac {\operatorname {Subst}\left (\int \frac {b^2 (3 a+2 b)+3 b^2 (a+b) x^2}{\left (1+x^2\right )^2} \, dx,x,\sinh (c+d x)\right )}{d}\\ &=\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}-\frac {b^3 \text {sech}(c+d x) \tanh (c+d x)}{2 d}+\frac {\left (b^2 (6 a+5 b)\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (c+d x)\right )}{2 d}\\ &=\frac {b^2 (6 a+5 b) \tan ^{-1}(\sinh (c+d x))}{2 d}+\frac {(a-2 b) (a+b)^2 \sinh (c+d x)}{d}+\frac {(a+b)^3 \sinh ^3(c+d x)}{3 d}-\frac {b^3 \text {sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end {align*}
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Mathematica [C] time = 6.89, size = 494, normalized size = 5.68 \[ \frac {\text {csch}^5(c+d x) \left (-256 \sinh ^8(c+d x) \left (a \sinh ^2(c+d x)+a+b \sinh ^2(c+d x)\right )^3 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};-\sinh ^2(c+d x)\right )+21 \left (a^3 \left (753 \sinh ^{10}(c+d x)+19579 \sinh ^8(c+d x)+89514 \sinh ^6(c+d x)+157878 \sinh ^4(c+d x)+124165 \sinh ^2(c+d x)+36015\right )+3 a^2 b \left (753 \sinh ^8(c+d x)+18826 \sinh ^6(c+d x)+69728 \sinh ^4(c+d x)+88150 \sinh ^2(c+d x)+36015\right ) \sinh ^2(c+d x)+3 a b^2 \left (753 \sinh ^6(c+d x)+18073 \sinh ^4(c+d x)+50695 \sinh ^2(c+d x)+36015\right ) \sinh ^4(c+d x)+b^3 \left (753 \sinh ^4(c+d x)+17320 \sinh ^2(c+d x)+32415\right ) \sinh ^6(c+d x)\right )-\frac {315 \tanh ^{-1}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (a^3 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right ) \cosh ^6(c+d x)+3 a^2 b \left (\sinh ^3(c+d x)+\sinh (c+d x)\right )^2 \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2401\right )+3 a b^2 \sinh ^4(c+d x) \left (\sinh ^8(c+d x)+244 \sinh ^6(c+d x)+2118 \sinh ^4(c+d x)+4180 \sinh ^2(c+d x)+2401\right )+b^3 \sinh ^6(c+d x) \left (\sinh ^6(c+d x)+243 \sinh ^4(c+d x)+1875 \sinh ^2(c+d x)+2161\right )\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{30240 d} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.45, size = 1840, normalized size = 21.15 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.66, size = 279, normalized size = 3.21 \[ \frac {24 \, {\left (6 \, a b^{2} e^{c} + 5 \, b^{3} e^{c}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) e^{\left (-c\right )} - {\left (9 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 9 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 45 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 27 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} + {\left (a^{3} e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a^{2} b e^{\left (3 \, d x + 30 \, c\right )} + 3 \, a b^{2} e^{\left (3 \, d x + 30 \, c\right )} + b^{3} e^{\left (3 \, d x + 30 \, c\right )} + 9 \, a^{3} e^{\left (d x + 28 \, c\right )} - 9 \, a^{2} b e^{\left (d x + 28 \, c\right )} - 45 \, a b^{2} e^{\left (d x + 28 \, c\right )} - 27 \, b^{3} e^{\left (d x + 28 \, c\right )}\right )} e^{\left (-27 \, c\right )} - \frac {24 \, {\left (b^{3} e^{\left (3 \, d x + 3 \, c\right )} - b^{3} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.38, size = 206, normalized size = 2.37 \[ \frac {2 a^{3} \sinh \left (d x +c \right )}{3 d}+\frac {a^{3} \sinh \left (d x +c \right ) \left (\cosh ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{2} b \left (\sinh ^{3}\left (d x +c \right )\right )}{d}+\frac {a \,b^{2} \left (\sinh ^{3}\left (d x +c \right )\right )}{d}-\frac {3 a \,b^{2} \sinh \left (d x +c \right )}{d}+\frac {6 a \,b^{2} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {b^{3} \left (\sinh ^{5}\left (d x +c \right )\right )}{3 d \cosh \left (d x +c \right )^{2}}-\frac {5 b^{3} \left (\sinh ^{3}\left (d x +c \right )\right )}{3 d \cosh \left (d x +c \right )^{2}}-\frac {5 b^{3} \sinh \left (d x +c \right )}{d \cosh \left (d x +c \right )^{2}}+\frac {5 b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2 d}+\frac {5 b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 284, normalized size = 3.26 \[ \frac {a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3}}{8 \, d} - \frac {1}{8} \, a b^{2} {\left (\frac {{\left (15 \, e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )} e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {15 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {48 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d}\right )} + \frac {1}{24} \, b^{3} {\left (\frac {27 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac {120 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {25 \, e^{\left (-2 \, d x - 2 \, c\right )} + 77 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d {\left (e^{\left (-3 \, d x - 3 \, c\right )} + 2 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac {1}{24} \, a^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.36, size = 232, normalized size = 2.67 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (5\,b^3\,\sqrt {d^2}+6\,a\,b^2\,\sqrt {d^2}\right )}{d\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}\right )\,\sqrt {36\,a^2\,b^4+60\,a\,b^5+25\,b^6}}{\sqrt {d^2}}-\frac {{\mathrm {e}}^{-3\,c-3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {{\mathrm {e}}^{3\,c+3\,d\,x}\,{\left (a+b\right )}^3}{24\,d}+\frac {3\,{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}-\frac {b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{-c-d\,x}\,{\left (a+b\right )}^2\,\left (a-3\,b\right )}{8\,d}+\frac {2\,b^3\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]
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 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \cosh ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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