Optimal. Leaf size=107 \[ a^4 x+\frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {2 a b \left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{d}+\frac {4 a b^3 \tanh (c+d x) \text {sech}(c+d x)}{3 d}+\frac {b^2 \tanh (c+d x) (a+b \text {sech}(c+d x))^2}{3 d} \]
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Rubi [A] time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3782, 4048, 3770, 3767, 8} \[ \frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {2 a b \left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{d}+a^4 x+\frac {4 a b^3 \tanh (c+d x) \text {sech}(c+d x)}{3 d}+\frac {b^2 \tanh (c+d x) (a+b \text {sech}(c+d x))^2}{3 d} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3767
Rule 3770
Rule 3782
Rule 4048
Rubi steps
\begin {align*} \int (a+b \text {sech}(c+d x))^4 \, dx &=\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {1}{3} \int (a+b \text {sech}(c+d x)) \left (3 a^3+b \left (9 a^2+2 b^2\right ) \text {sech}(c+d x)+8 a b^2 \text {sech}^2(c+d x)\right ) \, dx\\ &=\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^4+12 a b \left (2 a^2+b^2\right ) \text {sech}(c+d x)+2 b^2 \left (17 a^2+2 b^2\right ) \text {sech}^2(c+d x)\right ) \, dx\\ &=a^4 x+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\left (2 a b \left (2 a^2+b^2\right )\right ) \int \text {sech}(c+d x) \, dx+\frac {1}{3} \left (b^2 \left (17 a^2+2 b^2\right )\right ) \int \text {sech}^2(c+d x) \, dx\\ &=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{d}+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}+\frac {\left (i b^2 \left (17 a^2+2 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{3 d}\\ &=a^4 x+\frac {2 a b \left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))}{d}+\frac {b^2 \left (17 a^2+2 b^2\right ) \tanh (c+d x)}{3 d}+\frac {4 a b^3 \text {sech}(c+d x) \tanh (c+d x)}{3 d}+\frac {b^2 (a+b \text {sech}(c+d x))^2 \tanh (c+d x)}{3 d}\\ \end {align*}
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Mathematica [A] time = 0.25, size = 78, normalized size = 0.73 \[ \frac {3 a^4 d x+6 a b \left (2 a^2+b^2\right ) \tan ^{-1}(\sinh (c+d x))+3 b^2 \tanh (c+d x) \left (6 a^2+2 a b \text {sech}(c+d x)+b^2\right )-b^4 \tanh ^3(c+d x)}{3 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.41, size = 1028, normalized size = 9.61 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.13, size = 141, normalized size = 1.32 \[ \frac {3 \, {\left (d x + c\right )} a^{4} + 12 \, {\left (2 \, a^{3} b + a b^{3}\right )} \arctan \left (e^{\left (d x + c\right )}\right ) + \frac {4 \, {\left (3 \, a b^{3} e^{\left (5 \, d x + 5 \, c\right )} - 9 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{3} e^{\left (d x + c\right )} - 9 \, a^{2} b^{2} - b^{4}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.43, size = 121, normalized size = 1.13 \[ a^{4} x +\frac {a^{4} c}{d}+\frac {8 a^{3} b \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {6 a^{2} b^{2} \tanh \left (d x +c \right )}{d}+\frac {2 a \,b^{3} \mathrm {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{d}+\frac {4 a \,b^{3} \arctan \left ({\mathrm e}^{d x +c}\right )}{d}+\frac {2 b^{4} \tanh \left (d x +c \right )}{3 d}+\frac {b^{4} \tanh \left (d x +c \right ) \mathrm {sech}\left (d x +c \right )^{2}}{3 d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.92, size = 211, normalized size = 1.97 \[ a^{4} x - 4 \, a b^{3} {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} - \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {4}{3} \, b^{4} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {4 \, a^{3} b \arctan \left (\sinh \left (d x + c\right )\right )}{d} + \frac {12 \, a^{2} b^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.41, size = 233, normalized size = 2.18 \[ a^4\,x-\frac {\frac {12\,a^2\,b^2}{d}-\frac {4\,a\,b^3\,{\mathrm {e}}^{c+d\,x}}{d}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {\frac {4\,b^4}{d}+\frac {8\,a\,b^3\,{\mathrm {e}}^{c+d\,x}}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {8\,b^4}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {4\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a\,b^3\,\sqrt {d^2}+2\,a^3\,b\,\sqrt {d^2}\right )}{d\,\sqrt {4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6}}\right )\,\sqrt {4\,a^6\,b^2+4\,a^4\,b^4+a^2\,b^6}}{\sqrt {d^2}} \]
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 \[ \int \left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{4}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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