Optimal. Leaf size=275 \[ -\frac{b \left (3 a^2+10 b^2\right ) \tanh ^2(c+d x)}{2 d}+\frac{3 b \left (3 a^2+5 b^2\right ) \log (\cosh (c+d x))}{d}+\frac{\sinh (c+d x) \cosh ^3(c+d x) \left (b \left (3 a^2+b^2\right ) \tanh (c+d x)+a \left (a^2+3 b^2\right )\right )}{4 d}-\frac{\sinh (c+d x) \cosh (c+d x) \left (2 b \left (15 a^2+11 b^2\right ) \tanh (c+d x)+a \left (5 a^2+51 b^2\right )\right )}{8 d}+\frac{3}{8} a x \left (a^2+63 b^2\right )-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{18 a b^2 \tanh (c+d x)}{d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d} \]
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Rubi [A] time = 0.493673, antiderivative size = 306, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3663, 1804, 1802, 633, 31} \[ -\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+23 a b+40 b^2\right ) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-23 a b+40 b^2\right ) \log (\tanh (c+d x)+1)}{16 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)+a \left (a^2+39 b^2\right )\right )}{8 d}+\frac{\sinh ^4(c+d x) \left (a \left (a^2+3 b^2\right ) \tanh (c+d x)+b \left (3 a^2+b^2\right )\right )}{4 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 1804
Rule 1802
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^3\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (-4 b \left (3 a^2+b^2\right )-a \left (a^2+15 b^2\right ) x-4 b \left (3 a^2+b^2\right ) x^2-12 a b^2 x^3-4 b^3 x^4-12 a b^2 x^5-4 b^3 x^6-4 b^3 x^8\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a \left (a^2+39 b^2\right )+72 b \left (a^2+b^2\right ) x+48 a b^2 x^2+24 b^3 x^3+24 a b^2 x^4+16 b^3 x^5+8 b^3 x^7\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{\operatorname{Subst}\left (\int \left (-3 a \left (a^2+63 b^2\right )-24 b \left (3 a^2+5 b^2\right ) x-72 a b^2 x^2-48 b^3 x^3-24 a b^2 x^4-24 b^3 x^5-8 b^3 x^7+\frac{3 \left (a^3+63 a b^2+8 b \left (3 a^2+5 b^2\right ) x\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}+\frac{3 \operatorname{Subst}\left (\int \frac{a^3+63 a b^2+8 b \left (3 a^2+5 b^2\right ) x}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}-\frac{\left (3 (a-b) \left (a^2-23 a b+40 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}+\frac{\left (3 (a+b) \left (a^2+23 a b+40 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{3 (a+b) \left (a^2+23 a b+40 b^2\right ) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-b) \left (a^2-23 a b+40 b^2\right ) \log (1+\tanh (c+d x))}{16 d}-\frac{3 a \left (a^2+63 b^2\right ) \tanh (c+d x)}{8 d}-\frac{3 b \left (3 a^2+5 b^2\right ) \tanh ^2(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^3(c+d x)}{d}-\frac{3 b^3 \tanh ^4(c+d x)}{2 d}-\frac{3 a b^2 \tanh ^5(c+d x)}{5 d}-\frac{b^3 \tanh ^6(c+d x)}{2 d}-\frac{b^3 \tanh ^8(c+d x)}{8 d}+\frac{\sinh ^4(c+d x) \left (b \left (3 a^2+b^2\right )+a \left (a^2+3 b^2\right ) \tanh (c+d x)\right )}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) \left (a \left (a^2+39 b^2\right )+4 b \left (6 a^2+5 b^2\right ) \tanh (c+d x)\right )}{8 d}\\ \end{align*}
Mathematica [A] time = 6.23935, size = 294, normalized size = 1.07 \[ \frac{3 a \left (a^2+63 b^2\right ) (c+d x)}{8 d}-\frac{a \left (a^2+12 b^2\right ) \sinh (2 (c+d x))}{4 d}+\frac{a \left (a^2+3 b^2\right ) \sinh (4 (c+d x))}{32 d}-\frac{b \left (15 a^2+11 b^2\right ) \cosh (2 (c+d x))}{8 d}+\frac{b \left (3 a^2+b^2\right ) \cosh (4 (c+d x))}{32 d}+\frac{b \left (3 a^2+20 b^2\right ) \text{sech}^2(c+d x)}{2 d}+\frac{3 \left (3 a^2 b+5 b^3\right ) \log (\cosh (c+d x))}{d}-\frac{108 a b^2 \tanh (c+d x)}{5 d}-\frac{3 a b^2 \tanh (c+d x) \text{sech}^4(c+d x)}{5 d}+\frac{21 a b^2 \tanh (c+d x) \text{sech}^2(c+d x)}{5 d}-\frac{b^3 \text{sech}^8(c+d x)}{8 d}+\frac{b^3 \text{sech}^6(c+d x)}{d}-\frac{15 b^3 \text{sech}^4(c+d x)}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.076, size = 385, normalized size = 1.4 \begin{align*}{\frac{{a}^{3}\cosh \left ( dx+c \right ) \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,{a}^{3}\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}x}{8}}+{\frac{3\,{a}^{3}c}{8\,d}}+{\frac{3\,{a}^{2}b \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{9\,{a}^{2}b \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+9\,{\frac{{a}^{2}b\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{9\,{a}^{2}b \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}}+{\frac{3\,a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{9}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{27\,a{b}^{2} \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{189\,a{b}^{2}x}{8}}+{\frac{189\,a{b}^{2}c}{8\,d}}-{\frac{189\,a{b}^{2}\tanh \left ( dx+c \right ) }{8\,d}}-{\frac{63\,a{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{189\,a{b}^{2} \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{40\,d}}+{\frac{{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{12}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{8}}}-{\frac{3\,{b}^{3} \left ( \sinh \left ( dx+c \right ) \right ) ^{10}}{2\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{8}}}+15\,{\frac{{b}^{3}\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{5\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{6}}{2\,d}}-{\frac{15\,{b}^{3} \left ( \tanh \left ( dx+c \right ) \right ) ^{8}}{8\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.80729, size = 873, normalized size = 3.17 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 4.87444, size = 956, normalized size = 3.48 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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