Optimal. Leaf size=180 \[ -\frac {\left (2 a^2 A+3 a c C-A c^2\right ) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{e \left (a^2+c^2\right )^{5/2}}-\frac {\left (a^2 (-C)+3 a A c+2 c^2 C\right ) \cosh (d+e x)}{2 e \left (a^2+c^2\right )^2 (a+c \sinh (d+e x))}-\frac {(A c-a C) \cosh (d+e x)}{2 e \left (a^2+c^2\right ) (a+c \sinh (d+e x))^2}-\frac {B}{2 c e (a+c \sinh (d+e x))^2} \]
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Rubi [A] time = 0.27, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {4376, 2754, 12, 2660, 618, 204, 2668, 32} \[ -\frac {\left (2 a^2 A+3 a c C-A c^2\right ) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{e \left (a^2+c^2\right )^{5/2}}-\frac {\left (a^2 (-C)+3 a A c+2 c^2 C\right ) \cosh (d+e x)}{2 e \left (a^2+c^2\right )^2 (a+c \sinh (d+e x))}-\frac {(A c-a C) \cosh (d+e x)}{2 e \left (a^2+c^2\right ) (a+c \sinh (d+e x))^2}-\frac {B}{2 c e (a+c \sinh (d+e x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 32
Rule 204
Rule 618
Rule 2660
Rule 2668
Rule 2754
Rule 4376
Rubi steps
\begin {align*} \int \frac {A+B \cosh (d+e x)+C \sinh (d+e x)}{(a+c \sinh (d+e x))^3} \, dx &=B \int \frac {\cosh (d+e x)}{(a+c \sinh (d+e x))^3} \, dx+\int \frac {A+C \sinh (d+e x)}{(a+c \sinh (d+e x))^3} \, dx\\ &=-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\int \frac {-2 (a A+c C)+(A c-a C) \sinh (d+e x)}{(a+c \sinh (d+e x))^2} \, dx}{2 \left (a^2+c^2\right )}+\frac {B \operatorname {Subst}\left (\int \frac {1}{(a+x)^3} \, dx,x,c \sinh (d+e x)\right )}{c e}\\ &=-\frac {B}{2 c e (a+c \sinh (d+e x))^2}-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\left (3 a A c-a^2 C+2 c^2 C\right ) \cosh (d+e x)}{2 \left (a^2+c^2\right )^2 e (a+c \sinh (d+e x))}+\frac {\int \frac {2 a^2 A-A c^2+3 a c C}{a+c \sinh (d+e x)} \, dx}{2 \left (a^2+c^2\right )^2}\\ &=-\frac {B}{2 c e (a+c \sinh (d+e x))^2}-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\left (3 a A c-a^2 C+2 c^2 C\right ) \cosh (d+e x)}{2 \left (a^2+c^2\right )^2 e (a+c \sinh (d+e x))}+\frac {\left (2 a^2 A-A c^2+3 a c C\right ) \int \frac {1}{a+c \sinh (d+e x)} \, dx}{2 \left (a^2+c^2\right )^2}\\ &=-\frac {B}{2 c e (a+c \sinh (d+e x))^2}-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\left (3 a A c-a^2 C+2 c^2 C\right ) \cosh (d+e x)}{2 \left (a^2+c^2\right )^2 e (a+c \sinh (d+e x))}-\frac {\left (i \left (2 a^2 A-A c^2+3 a c C\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-2 i c x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (i d+i e x)\right )\right )}{\left (a^2+c^2\right )^2 e}\\ &=-\frac {B}{2 c e (a+c \sinh (d+e x))^2}-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\left (3 a A c-a^2 C+2 c^2 C\right ) \cosh (d+e x)}{2 \left (a^2+c^2\right )^2 e (a+c \sinh (d+e x))}+\frac {\left (2 i \left (2 a^2 A-A c^2+3 a c C\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2+c^2\right )-x^2} \, dx,x,-2 i c+2 a \tan \left (\frac {1}{2} (i d+i e x)\right )\right )}{\left (a^2+c^2\right )^2 e}\\ &=-\frac {\left (2 a^2 A-A c^2+3 a c C\right ) \tanh ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2+c^2}}\right )}{\left (a^2+c^2\right )^{5/2} e}-\frac {B}{2 c e (a+c \sinh (d+e x))^2}-\frac {(A c-a C) \cosh (d+e x)}{2 \left (a^2+c^2\right ) e (a+c \sinh (d+e x))^2}-\frac {\left (3 a A c-a^2 C+2 c^2 C\right ) \cosh (d+e x)}{2 \left (a^2+c^2\right )^2 e (a+c \sinh (d+e x))}\\ \end {align*}
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Mathematica [A] time = 0.69, size = 170, normalized size = 0.94 \[ \frac {-\frac {\left (a^2+c^2\right ) \left (B \left (a^2+c^2\right )+c (A c-a C) \cosh (d+e x)\right )}{c (a+c \sinh (d+e x))^2}+\frac {2 \left (2 a^2 A+3 a c C-A c^2\right ) \tan ^{-1}\left (\frac {c-a \tanh \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2-c^2}}\right )}{\sqrt {-a^2-c^2}}+\frac {\left (a^2 C-3 a A c-2 c^2 C\right ) \cosh (d+e x)}{a+c \sinh (d+e x)}}{2 e \left (a^2+c^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.28, size = 1880, normalized size = 10.44 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.35, size = 423, normalized size = 2.35 \[ -\frac {1}{2} \, {\left (\frac {{\left (2 \, A a^{2} + 3 \, C a c - A c^{2}\right )} \log \left (\frac {{\left | -2 \, c e^{\left (x e + d\right )} - 2 \, a - 2 \, \sqrt {a^{2} + c^{2}} \right |}}{{\left | -2 \, c e^{\left (x e + d\right )} - 2 \, a + 2 \, \sqrt {a^{2} + c^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )} \sqrt {a^{2} + c^{2}}} - \frac {2 \, {\left (2 \, A a^{2} c^{2} e^{\left (3 \, x e + 3 \, d\right )} + 3 \, C a c^{3} e^{\left (3 \, x e + 3 \, d\right )} - A c^{4} e^{\left (3 \, x e + 3 \, d\right )} - 2 \, B a^{4} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, C a^{4} e^{\left (2 \, x e + 2 \, d\right )} + 6 \, A a^{3} c e^{\left (2 \, x e + 2 \, d\right )} - 4 \, B a^{2} c^{2} e^{\left (2 \, x e + 2 \, d\right )} + 5 \, C a^{2} c^{2} e^{\left (2 \, x e + 2 \, d\right )} - 3 \, A a c^{3} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, B c^{4} e^{\left (2 \, x e + 2 \, d\right )} - 2 \, C c^{4} e^{\left (2 \, x e + 2 \, d\right )} + 4 \, C a^{3} c e^{\left (x e + d\right )} - 10 \, A a^{2} c^{2} e^{\left (x e + d\right )} - 5 \, C a c^{3} e^{\left (x e + d\right )} - A c^{4} e^{\left (x e + d\right )} - C a^{2} c^{2} + 3 \, A a c^{3} + 2 \, C c^{4}\right )}}{{\left (a^{4} c + 2 \, a^{2} c^{3} + c^{5}\right )} {\left (c e^{\left (2 \, x e + 2 \, d\right )} + 2 \, a e^{\left (x e + d\right )} - c\right )}^{2}}\right )} e^{\left (-1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.21, size = 416, normalized size = 2.31 \[ \frac {-\frac {2 \left (-\frac {\left (5 A \,a^{2} c^{2}+2 A \,c^{4}-2 B \,a^{4}-4 B \,a^{2} c^{2}-2 B \,c^{4}-3 C \,a^{3} c \right ) \left (\tanh ^{3}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 a \left (a^{4}+2 a^{2} c^{2}+c^{4}\right )}-\frac {\left (4 A \,a^{4} c -7 A \,a^{2} c^{3}-2 A \,c^{5}+2 B \,a^{4} c +4 B \,a^{2} c^{3}+2 B \,c^{5}-2 C \,a^{5}+5 C \,a^{3} c^{2}-2 C a \,c^{4}\right ) \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 \left (a^{4}+2 a^{2} c^{2}+c^{4}\right ) a^{2}}+\frac {\left (11 A \,a^{2} c^{2}+2 A \,c^{4}-2 B \,a^{4}-4 B \,a^{2} c^{2}-2 B \,c^{4}-5 C \,a^{3} c +4 C a \,c^{3}\right ) \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )}{2 \left (a^{4}+2 a^{2} c^{2}+c^{4}\right ) a}+\frac {4 A \,a^{2} c +A \,c^{3}-2 C \,a^{3}+C a \,c^{2}}{2 a^{4}+4 a^{2} c^{2}+2 c^{4}}\right )}{\left (a \left (\tanh ^{2}\left (\frac {e x}{2}+\frac {d}{2}\right )\right )-2 c \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-a \right )^{2}}+\frac {\left (2 a^{2} A -A \,c^{2}+3 C a c \right ) \arctanh \left (\frac {2 a \tanh \left (\frac {e x}{2}+\frac {d}{2}\right )-2 c}{2 \sqrt {a^{2}+c^{2}}}\right )}{\left (a^{4}+2 a^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}+c^{2}}}}{e} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.47, size = 726, normalized size = 4.03 \[ \frac {1}{2} \, C {\left (\frac {3 \, a c \log \left (\frac {c e^{\left (-e x - d\right )} - a - \sqrt {a^{2} + c^{2}}}{c e^{\left (-e x - d\right )} - a + \sqrt {a^{2} + c^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )} \sqrt {a^{2} + c^{2}} e} + \frac {2 \, {\left (3 \, a c^{3} e^{\left (-3 \, e x - 3 \, d\right )} + a^{2} c^{2} - 2 \, c^{4} + {\left (4 \, a^{3} c - 5 \, a c^{3}\right )} e^{\left (-e x - d\right )} + {\left (2 \, a^{4} - 5 \, a^{2} c^{2} + 2 \, c^{4}\right )} e^{\left (-2 \, e x - 2 \, d\right )}\right )}}{{\left (a^{4} c^{3} + 2 \, a^{2} c^{5} + c^{7} + 4 \, {\left (a^{5} c^{2} + 2 \, a^{3} c^{4} + a c^{6}\right )} e^{\left (-e x - d\right )} + 2 \, {\left (2 \, a^{6} c + 3 \, a^{4} c^{3} - c^{7}\right )} e^{\left (-2 \, e x - 2 \, d\right )} - 4 \, {\left (a^{5} c^{2} + 2 \, a^{3} c^{4} + a c^{6}\right )} e^{\left (-3 \, e x - 3 \, d\right )} + {\left (a^{4} c^{3} + 2 \, a^{2} c^{5} + c^{7}\right )} e^{\left (-4 \, e x - 4 \, d\right )}\right )} e}\right )} + \frac {1}{2} \, A {\left (\frac {{\left (2 \, a^{2} - c^{2}\right )} \log \left (\frac {c e^{\left (-e x - d\right )} - a - \sqrt {a^{2} + c^{2}}}{c e^{\left (-e x - d\right )} - a + \sqrt {a^{2} + c^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} c^{2} + c^{4}\right )} \sqrt {a^{2} + c^{2}} e} - \frac {2 \, {\left (3 \, a c^{2} + {\left (10 \, a^{2} c + c^{3}\right )} e^{\left (-e x - d\right )} + 3 \, {\left (2 \, a^{3} - a c^{2}\right )} e^{\left (-2 \, e x - 2 \, d\right )} - {\left (2 \, a^{2} c - c^{3}\right )} e^{\left (-3 \, e x - 3 \, d\right )}\right )}}{{\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6} + 4 \, {\left (a^{5} c + 2 \, a^{3} c^{3} + a c^{5}\right )} e^{\left (-e x - d\right )} + 2 \, {\left (2 \, a^{6} + 3 \, a^{4} c^{2} - c^{6}\right )} e^{\left (-2 \, e x - 2 \, d\right )} - 4 \, {\left (a^{5} c + 2 \, a^{3} c^{3} + a c^{5}\right )} e^{\left (-3 \, e x - 3 \, d\right )} + {\left (a^{4} c^{2} + 2 \, a^{2} c^{4} + c^{6}\right )} e^{\left (-4 \, e x - 4 \, d\right )}\right )} e}\right )} - \frac {2 \, B e^{\left (-2 \, e x - 2 \, d\right )}}{{\left (4 \, a c^{2} e^{\left (-e x - d\right )} - 4 \, a c^{2} e^{\left (-3 \, e x - 3 \, d\right )} + c^{3} e^{\left (-4 \, e x - 4 \, d\right )} + c^{3} + 2 \, {\left (2 \, a^{2} c - c^{3}\right )} e^{\left (-2 \, e x - 2 \, d\right )}\right )} e} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\mathrm {cosh}\left (d+e\,x\right )+C\,\mathrm {sinh}\left (d+e\,x\right )}{{\left (a+c\,\mathrm {sinh}\left (d+e\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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