Optimal. Leaf size=122 \[ \frac {b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac {3 b (a+b)^2 \tanh (c+d x)}{d}+\frac {(a+b)^3}{4 d (1-\tanh (c+d x))}-\frac {(a+b)^3}{4 d (\tanh (c+d x)+1)}-\frac {1}{2} x (a+b)^2 (a+7 b)+\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.19, antiderivative size = 139, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3663, 467, 528, 388, 206} \[ \frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {\sinh (c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {1}{2} x (a+b)^2 (a+7 b) \]
Antiderivative was successfully verified.
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Rule 206
Rule 388
Rule 467
Rule 528
Rule 3663
Rubi steps
\begin {align*} \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^2 \left (a+7 b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}+\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right ) \left (-a (5 a+7 b)-b (33 a+35 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {a \left (15 a^2+54 a b+35 b^2\right )+b \left (81 a^2+190 a b+105 b^2\right ) x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{30 d}\\ &=\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac {\left ((a+b)^2 (a+7 b)\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {1}{2} (a+b)^2 (a+7 b) x+\frac {b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac {b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac {7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac {\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}\\ \end {align*}
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Mathematica [A] time = 2.20, size = 95, normalized size = 0.78 \[ \frac {4 b \tanh (c+d x) \left (45 a^2-b (15 a+16 b) \text {sech}^2(c+d x)+105 a b+3 b^2 \text {sech}^4(c+d x)+58 b^2\right )-30 (a+7 b) (a+b)^2 (c+d x)+15 (a+b)^3 \sinh (2 (c+d x))}{60 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.87, size = 725, normalized size = 5.94 \[ \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{7} - 4 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{5} - 20 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3} + 315 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{5} - 20 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (105 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 27 \, a^{3} + 297 \, a^{2} b + 489 \, a b^{2} + 203 \, b^{3} + 2 \, {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} - 20 \, {\left (2 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 40 \, {\left (90 \, a^{2} b + 210 \, a b^{2} + 116 \, b^{3} + 15 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 5 \, {\left (21 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{6} + {\left (75 \, a^{3} + 585 \, a^{2} b + 1065 \, a b^{2} + 539 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} + 189 \, a^{2} b + 285 \, a b^{2} + 175 \, b^{3} + 3 \, {\left (27 \, a^{3} + 297 \, a^{2} b + 489 \, a b^{2} + 203 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{120 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.64, size = 395, normalized size = 3.24 \[ -\frac {60 \, {\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x - 15 \, {\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 15 \, {\left (a^{3} e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 16 \, c\right )} + b^{3} e^{\left (2 \, d x + 16 \, c\right )}\right )} e^{\left (-14 \, c\right )} + \frac {16 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 135 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 450 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 240 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 600 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 340 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 390 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 200 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 105 \, a b^{2} + 58 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 180, normalized size = 1.48 \[ \frac {a^{3} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+3 a^{2} b \left (\frac {\sinh ^{3}\left (d x +c \right )}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+3 a \,b^{2} \left (\frac {\sinh ^{5}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}\right )+b^{3} \left (\frac {\sinh ^{7}\left (d x +c \right )}{2 \cosh \left (d x +c \right )^{5}}-\frac {7 d x}{2}-\frac {7 c}{2}+\frac {7 \tanh \left (d x +c \right )}{2}+\frac {7 \left (\tanh ^{3}\left (d x +c \right )\right )}{6}+\frac {7 \left (\tanh ^{5}\left (d x +c \right )\right )}{10}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.33, size = 377, normalized size = 3.09 \[ -\frac {1}{8} \, a^{3} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{120} \, b^{3} {\left (\frac {420 \, {\left (d x + c\right )}}{d} + \frac {15 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {1003 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3350 \, e^{\left (-4 \, d x - 4 \, c\right )} + 5590 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3915 \, e^{\left (-8 \, d x - 8 \, c\right )} + 1455 \, e^{\left (-10 \, d x - 10 \, c\right )} + 15}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )}\right )}}\right )} - \frac {1}{8} \, a b^{2} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac {3}{8} \, a^{2} b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.32, size = 668, normalized size = 5.48 \[ \frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{15\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {6\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^3}{8\,d}-\frac {\frac {6\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^2\,b+3\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+6\,a\,b^2+2\,b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (9\,a^2\,b+15\,a\,b^2+10\,b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {x\,{\left (a+b\right )}^2\,\left (a+7\,b\right )}{2} \]
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 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \sinh ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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