3.172 \(\int \sinh ^3(a+b x) \sinh ^3(c+d x) \, dx\)

Optimal. Leaf size=195 \[ \frac {3 \sinh (a+x (b-3 d)-3 c)}{32 (b-3 d)}-\frac {9 \sinh (a+x (b-d)-c)}{32 (b-d)}-\frac {\sinh (3 (a-c)+3 x (b-d))}{96 (b-d)}+\frac {3 \sinh (3 a+x (3 b-d)-c)}{32 (3 b-d)}+\frac {9 \sinh (a+x (b+d)+c)}{32 (b+d)}+\frac {\sinh (3 (a+c)+3 x (b+d))}{96 (b+d)}-\frac {3 \sinh (3 a+x (3 b+d)+c)}{32 (3 b+d)}-\frac {3 \sinh (a+x (b+3 d)+3 c)}{32 (b+3 d)} \]

[Out]

3/32*sinh(a-3*c+(b-3*d)*x)/(b-3*d)-9/32*sinh(a-c+(b-d)*x)/(b-d)-1/96*sinh(3*a-3*c+3*(b-d)*x)/(b-d)+3/32*sinh(3
*a-c+(3*b-d)*x)/(3*b-d)+9/32*sinh(a+c+(b+d)*x)/(b+d)+1/96*sinh(3*a+3*c+3*(b+d)*x)/(b+d)-3/32*sinh(3*a+c+(3*b+d
)*x)/(3*b+d)-3/32*sinh(a+3*c+(b+3*d)*x)/(b+3*d)

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Rubi [A]  time = 0.14, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {5613, 2637} \[ \frac {3 \sinh (a+x (b-3 d)-3 c)}{32 (b-3 d)}-\frac {9 \sinh (a+x (b-d)-c)}{32 (b-d)}-\frac {\sinh (3 (a-c)+3 x (b-d))}{96 (b-d)}+\frac {3 \sinh (3 a+x (3 b-d)-c)}{32 (3 b-d)}+\frac {9 \sinh (a+x (b+d)+c)}{32 (b+d)}+\frac {\sinh (3 (a+c)+3 x (b+d))}{96 (b+d)}-\frac {3 \sinh (3 a+x (3 b+d)+c)}{32 (3 b+d)}-\frac {3 \sinh (a+x (b+3 d)+3 c)}{32 (b+3 d)} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^3*Sinh[c + d*x]^3,x]

[Out]

(3*Sinh[a - 3*c + (b - 3*d)*x])/(32*(b - 3*d)) - (9*Sinh[a - c + (b - d)*x])/(32*(b - d)) - Sinh[3*(a - c) + 3
*(b - d)*x]/(96*(b - d)) + (3*Sinh[3*a - c + (3*b - d)*x])/(32*(3*b - d)) + (9*Sinh[a + c + (b + d)*x])/(32*(b
+ d)) + Sinh[3*(a + c) + 3*(b + d)*x]/(96*(b + d)) - (3*Sinh[3*a + c + (3*b + d)*x])/(32*(3*b + d)) - (3*Sinh
[a + 3*c + (b + 3*d)*x])/(32*(b + 3*d))

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 5613

Int[Sinh[v_]^(p_.)*Sinh[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Sinh[v]^p*Sinh[w]^q, x], x] /; IGtQ[p, 0]
&& IGtQ[q, 0] && ((PolynomialQ[v, x] && PolynomialQ[w, x]) || (BinomialQ[{v, w}, x] && IndependentQ[Cancel[v/
w], x]))

Rubi steps

\begin {align*} \int \sinh ^3(a+b x) \sinh ^3(c+d x) \, dx &=\int \left (\frac {3}{32} \cosh (a-3 c+(b-3 d) x)-\frac {9}{32} \cosh (a-c+(b-d) x)-\frac {1}{32} \cosh (3 (a-c)+3 (b-d) x)+\frac {3}{32} \cosh (3 a-c+(3 b-d) x)+\frac {9}{32} \cosh (a+c+(b+d) x)+\frac {1}{32} \cosh (3 (a+c)+3 (b+d) x)-\frac {3}{32} \cosh (3 a+c+(3 b+d) x)-\frac {3}{32} \cosh (a+3 c+(b+3 d) x)\right ) \, dx\\ &=-\left (\frac {1}{32} \int \cosh (3 (a-c)+3 (b-d) x) \, dx\right )+\frac {1}{32} \int \cosh (3 (a+c)+3 (b+d) x) \, dx+\frac {3}{32} \int \cosh (a-3 c+(b-3 d) x) \, dx+\frac {3}{32} \int \cosh (3 a-c+(3 b-d) x) \, dx-\frac {3}{32} \int \cosh (3 a+c+(3 b+d) x) \, dx-\frac {3}{32} \int \cosh (a+3 c+(b+3 d) x) \, dx-\frac {9}{32} \int \cosh (a-c+(b-d) x) \, dx+\frac {9}{32} \int \cosh (a+c+(b+d) x) \, dx\\ &=\frac {3 \sinh (a-3 c+(b-3 d) x)}{32 (b-3 d)}-\frac {9 \sinh (a-c+(b-d) x)}{32 (b-d)}-\frac {\sinh (3 (a-c)+3 (b-d) x)}{96 (b-d)}+\frac {3 \sinh (3 a-c+(3 b-d) x)}{32 (3 b-d)}+\frac {9 \sinh (a+c+(b+d) x)}{32 (b+d)}+\frac {\sinh (3 (a+c)+3 (b+d) x)}{96 (b+d)}-\frac {3 \sinh (3 a+c+(3 b+d) x)}{32 (3 b+d)}-\frac {3 \sinh (a+3 c+(b+3 d) x)}{32 (b+3 d)}\\ \end {align*}

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Mathematica [A]  time = 1.59, size = 177, normalized size = 0.91 \[ \frac {1}{96} \left (\frac {9 \sinh (a+b x-3 c-3 d x)}{b-3 d}-\frac {27 \sinh (a+b x-c-d x)}{b-d}-\frac {\sinh (3 (a+b x-c-d x))}{b-d}+\frac {9 \sinh (3 a+3 b x-c-d x)}{3 b-d}-\frac {9 \sinh (3 a+3 b x+c+d x)}{3 b+d}-\frac {9 \sinh (a+b x+3 c+3 d x)}{b+3 d}+\frac {27 \sinh (a+x (b+d)+c)}{b+d}+\frac {\sinh (3 (a+x (b+d)+c))}{b+d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^3*Sinh[c + d*x]^3,x]

[Out]

((9*Sinh[a - 3*c + b*x - 3*d*x])/(b - 3*d) - (27*Sinh[a - c + b*x - d*x])/(b - d) - Sinh[3*(a - c + b*x - d*x)
]/(b - d) + (9*Sinh[3*a - c + 3*b*x - d*x])/(3*b - d) - (9*Sinh[3*a + c + 3*b*x + d*x])/(3*b + d) - (9*Sinh[a
+ 3*c + b*x + 3*d*x])/(b + 3*d) + (27*Sinh[a + c + (b + d)*x])/(b + d) + Sinh[3*(a + c + (b + d)*x)]/(b + d))/
96

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fricas [B]  time = 0.46, size = 731, normalized size = 3.75 \[ -\frac {{\left ({\left (9 \, b^{4} d - 82 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (d x + c\right )^{3} - 9 \, {\left (b^{4} d - 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right )^{3} - {\left ({\left (9 \, b^{5} - 82 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right )^{3} + 3 \, {\left (9 \, b^{5} - 82 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 9 \, {\left (9 \, b^{5} - 10 \, b^{3} d^{2} + b d^{4}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left ({\left (9 \, b^{4} d - 82 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{3} - 3 \, {\left (81 \, b^{4} d - 90 \, b^{2} d^{3} + 9 \, d^{5} - {\left (9 \, b^{4} d - 82 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \, {\left ({\left (81 \, b^{4} d - 90 \, b^{2} d^{3} + 9 \, d^{5} - {\left (9 \, b^{4} d - 82 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right )^{3} - 9 \, {\left (9 \, b^{4} d - 82 \, b^{2} d^{3} + 9 \, d^{5} - {\left (b^{4} d - 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right ) + 3 \, {\left (9 \, {\left (b^{5} - 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right )^{3} - {\left ({\left (9 \, b^{5} - 82 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right )^{3} - 9 \, {\left (9 \, b^{5} - 10 \, b^{3} d^{2} + b d^{4}\right )} \cosh \left (b x + a\right )\right )} \cosh \left (d x + c\right )^{2} - 3 \, {\left ({\left (9 \, b^{5} - 82 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} - 9 \, {\left (b^{5} - 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 9 \, {\left (9 \, b^{5} - 82 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{48 \, {\left ({\left (9 \, b^{6} - 91 \, b^{4} d^{2} + 91 \, b^{2} d^{4} - 9 \, d^{6}\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (9 \, b^{6} - 91 \, b^{4} d^{2} + 91 \, b^{2} d^{4} - 9 \, d^{6}\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (9 \, b^{6} - 91 \, b^{4} d^{2} + 91 \, b^{2} d^{4} - 9 \, d^{6}\right )} \sinh \left (b x + a\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^3*sinh(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/48*(((9*b^4*d - 82*b^2*d^3 + 9*d^5)*cosh(d*x + c)^3 - 9*(b^4*d - 10*b^2*d^3 + 9*d^5)*cosh(d*x + c))*sinh(b*
x + a)^3 - ((9*b^5 - 82*b^3*d^2 + 9*b*d^4)*cosh(b*x + a)^3 + 3*(9*b^5 - 82*b^3*d^2 + 9*b*d^4)*cosh(b*x + a)*si
nh(b*x + a)^2 - 9*(9*b^5 - 10*b^3*d^2 + b*d^4)*cosh(b*x + a))*sinh(d*x + c)^3 + 3*((9*b^4*d - 82*b^2*d^3 + 9*d
^5)*cosh(d*x + c)*sinh(b*x + a)^3 - 3*(81*b^4*d - 90*b^2*d^3 + 9*d^5 - (9*b^4*d - 82*b^2*d^3 + 9*d^5)*cosh(b*x
+ a)^2)*cosh(d*x + c)*sinh(b*x + a))*sinh(d*x + c)^2 - 3*((81*b^4*d - 90*b^2*d^3 + 9*d^5 - (9*b^4*d - 82*b^2*
d^3 + 9*d^5)*cosh(b*x + a)^2)*cosh(d*x + c)^3 - 9*(9*b^4*d - 82*b^2*d^3 + 9*d^5 - (b^4*d - 10*b^2*d^3 + 9*d^5)
*cosh(b*x + a)^2)*cosh(d*x + c))*sinh(b*x + a) + 3*(9*(b^5 - 10*b^3*d^2 + 9*b*d^4)*cosh(b*x + a)^3 - ((9*b^5 -
82*b^3*d^2 + 9*b*d^4)*cosh(b*x + a)^3 - 9*(9*b^5 - 10*b^3*d^2 + b*d^4)*cosh(b*x + a))*cosh(d*x + c)^2 - 3*((9
*b^5 - 82*b^3*d^2 + 9*b*d^4)*cosh(b*x + a)*cosh(d*x + c)^2 - 9*(b^5 - 10*b^3*d^2 + 9*b*d^4)*cosh(b*x + a))*sin
h(b*x + a)^2 - 9*(9*b^5 - 82*b^3*d^2 + 9*b*d^4)*cosh(b*x + a))*sinh(d*x + c))/((9*b^6 - 91*b^4*d^2 + 91*b^2*d^
4 - 9*d^6)*cosh(b*x + a)^4 - 2*(9*b^6 - 91*b^4*d^2 + 91*b^2*d^4 - 9*d^6)*cosh(b*x + a)^2*sinh(b*x + a)^2 + (9*
b^6 - 91*b^4*d^2 + 91*b^2*d^4 - 9*d^6)*sinh(b*x + a)^4)

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giac [B]  time = 0.15, size = 373, normalized size = 1.91 \[ \frac {e^{\left (3 \, b x + 3 \, d x + 3 \, a + 3 \, c\right )}}{192 \, {\left (b + d\right )}} - \frac {3 \, e^{\left (3 \, b x + d x + 3 \, a + c\right )}}{64 \, {\left (3 \, b + d\right )}} + \frac {3 \, e^{\left (3 \, b x - d x + 3 \, a - c\right )}}{64 \, {\left (3 \, b - d\right )}} - \frac {e^{\left (3 \, b x - 3 \, d x + 3 \, a - 3 \, c\right )}}{192 \, {\left (b - d\right )}} - \frac {3 \, e^{\left (b x + 3 \, d x + a + 3 \, c\right )}}{64 \, {\left (b + 3 \, d\right )}} + \frac {9 \, e^{\left (b x + d x + a + c\right )}}{64 \, {\left (b + d\right )}} - \frac {9 \, e^{\left (b x - d x + a - c\right )}}{64 \, {\left (b - d\right )}} + \frac {3 \, e^{\left (b x - 3 \, d x + a - 3 \, c\right )}}{64 \, {\left (b - 3 \, d\right )}} - \frac {3 \, e^{\left (-b x + 3 \, d x - a + 3 \, c\right )}}{64 \, {\left (b - 3 \, d\right )}} + \frac {9 \, e^{\left (-b x + d x - a + c\right )}}{64 \, {\left (b - d\right )}} - \frac {9 \, e^{\left (-b x - d x - a - c\right )}}{64 \, {\left (b + d\right )}} + \frac {3 \, e^{\left (-b x - 3 \, d x - a - 3 \, c\right )}}{64 \, {\left (b + 3 \, d\right )}} + \frac {e^{\left (-3 \, b x + 3 \, d x - 3 \, a + 3 \, c\right )}}{192 \, {\left (b - d\right )}} - \frac {3 \, e^{\left (-3 \, b x + d x - 3 \, a + c\right )}}{64 \, {\left (3 \, b - d\right )}} + \frac {3 \, e^{\left (-3 \, b x - d x - 3 \, a - c\right )}}{64 \, {\left (3 \, b + d\right )}} - \frac {e^{\left (-3 \, b x - 3 \, d x - 3 \, a - 3 \, c\right )}}{192 \, {\left (b + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^3*sinh(d*x+c)^3,x, algorithm="giac")

[Out]

1/192*e^(3*b*x + 3*d*x + 3*a + 3*c)/(b + d) - 3/64*e^(3*b*x + d*x + 3*a + c)/(3*b + d) + 3/64*e^(3*b*x - d*x +
3*a - c)/(3*b - d) - 1/192*e^(3*b*x - 3*d*x + 3*a - 3*c)/(b - d) - 3/64*e^(b*x + 3*d*x + a + 3*c)/(b + 3*d) +
9/64*e^(b*x + d*x + a + c)/(b + d) - 9/64*e^(b*x - d*x + a - c)/(b - d) + 3/64*e^(b*x - 3*d*x + a - 3*c)/(b -
3*d) - 3/64*e^(-b*x + 3*d*x - a + 3*c)/(b - 3*d) + 9/64*e^(-b*x + d*x - a + c)/(b - d) - 9/64*e^(-b*x - d*x -
a - c)/(b + d) + 3/64*e^(-b*x - 3*d*x - a - 3*c)/(b + 3*d) + 1/192*e^(-3*b*x + 3*d*x - 3*a + 3*c)/(b - d) - 3
/64*e^(-3*b*x + d*x - 3*a + c)/(3*b - d) + 3/64*e^(-3*b*x - d*x - 3*a - c)/(3*b + d) - 1/192*e^(-3*b*x - 3*d*x
- 3*a - 3*c)/(b + d)

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maple [A]  time = 0.54, size = 184, normalized size = 0.94 \[ \frac {3 \sinh \left (a -3 c +\left (b -3 d \right ) x \right )}{32 \left (b -3 d \right )}-\frac {9 \sinh \left (a -c +\left (b -d \right ) x \right )}{32 \left (b -d \right )}+\frac {9 \sinh \left (a +c +\left (b +d \right ) x \right )}{32 \left (b +d \right )}-\frac {3 \sinh \left (a +3 c +\left (b +3 d \right ) x \right )}{32 \left (b +3 d \right )}-\frac {\sinh \left (\left (3 b -3 d \right ) x +3 a -3 c \right )}{96 \left (b -d \right )}+\frac {3 \sinh \left (3 a -c +\left (3 b -d \right ) x \right )}{32 \left (3 b -d \right )}-\frac {3 \sinh \left (3 a +c +\left (3 b +d \right ) x \right )}{32 \left (3 b +d \right )}+\frac {\sinh \left (\left (3 b +3 d \right ) x +3 a +3 c \right )}{96 b +96 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^3*sinh(d*x+c)^3,x)

[Out]

3/32*sinh(a-3*c+(b-3*d)*x)/(b-3*d)-9/32*sinh(a-c+(b-d)*x)/(b-d)+9/32*sinh(a+c+(b+d)*x)/(b+d)-3/32*sinh(a+3*c+(
b+3*d)*x)/(b+3*d)-1/96/(b-d)*sinh((3*b-3*d)*x+3*a-3*c)+3/32*sinh(3*a-c+(3*b-d)*x)/(3*b-d)-3/32*sinh(3*a+c+(3*b
+d)*x)/(3*b+d)+1/96/(b+d)*sinh((3*b+3*d)*x+3*a+3*c)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^3*sinh(d*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-(3*d)/b>0)', see `assume?` fo
r more details)Is -(3*d)/b equal to -1?

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mupad [B]  time = 2.08, size = 906, normalized size = 4.65 \[ {\mathrm {e}}^{3\,a+c+3\,b\,x+d\,x}\,\left (\frac {-9\,b^3+3\,b^2\,d+9\,b\,d^2-3\,d^3}{576\,b^4-640\,b^2\,d^2+64\,d^4}+\frac {{\mathrm {e}}^{-6\,a-6\,b\,x}\,\left (-9\,b^3-3\,b^2\,d+9\,b\,d^2+3\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left (-81\,b^3+81\,b^2\,d+9\,b\,d^2-9\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}-\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}\,\left (-81\,b^3-81\,b^2\,d+9\,b\,d^2+9\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}\right )-{\mathrm {e}}^{3\,a-c+3\,b\,x-d\,x}\,\left (\frac {-9\,b^3-3\,b^2\,d+9\,b\,d^2+3\,d^3}{576\,b^4-640\,b^2\,d^2+64\,d^4}+\frac {{\mathrm {e}}^{-6\,a-6\,b\,x}\,\left (-9\,b^3+3\,b^2\,d+9\,b\,d^2-3\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left (-81\,b^3-81\,b^2\,d+9\,b\,d^2+9\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}-\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}\,\left (-81\,b^3+81\,b^2\,d+9\,b\,d^2-9\,d^3\right )}{576\,b^4-640\,b^2\,d^2+64\,d^4}\right )+{\mathrm {e}}^{3\,a-3\,c+3\,b\,x-3\,d\,x}\,\left (\frac {-b^3-b^2\,d+9\,b\,d^2+9\,d^3}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}+\frac {{\mathrm {e}}^{-6\,a-6\,b\,x}\,\left (-b^3+b^2\,d+9\,b\,d^2-9\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left (-9\,b^3-27\,b^2\,d+9\,b\,d^2+27\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}-\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}\,\left (-9\,b^3+27\,b^2\,d+9\,b\,d^2-27\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}\right )-{\mathrm {e}}^{3\,a+3\,c+3\,b\,x+3\,d\,x}\,\left (\frac {-b^3+b^2\,d+9\,b\,d^2-9\,d^3}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}+\frac {{\mathrm {e}}^{-6\,a-6\,b\,x}\,\left (-b^3-b^2\,d+9\,b\,d^2+9\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}\,\left (-9\,b^3+27\,b^2\,d+9\,b\,d^2-27\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}-\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}\,\left (-9\,b^3-27\,b^2\,d+9\,b\,d^2+27\,d^3\right )}{192\,b^4-1920\,b^2\,d^2+1728\,d^4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x)^3*sinh(c + d*x)^3,x)

[Out]

exp(3*a + c + 3*b*x + d*x)*((9*b*d^2 + 3*b^2*d - 9*b^3 - 3*d^3)/(576*b^4 + 64*d^4 - 640*b^2*d^2) + (exp(- 6*a
- 6*b*x)*(9*b*d^2 - 3*b^2*d - 9*b^3 + 3*d^3))/(576*b^4 + 64*d^4 - 640*b^2*d^2) - (exp(- 2*a - 2*b*x)*(9*b*d^2
+ 81*b^2*d - 81*b^3 - 9*d^3))/(576*b^4 + 64*d^4 - 640*b^2*d^2) - (exp(- 4*a - 4*b*x)*(9*b*d^2 - 81*b^2*d - 81*
b^3 + 9*d^3))/(576*b^4 + 64*d^4 - 640*b^2*d^2)) - exp(3*a - c + 3*b*x - d*x)*((9*b*d^2 - 3*b^2*d - 9*b^3 + 3*d
^3)/(576*b^4 + 64*d^4 - 640*b^2*d^2) + (exp(- 6*a - 6*b*x)*(9*b*d^2 + 3*b^2*d - 9*b^3 - 3*d^3))/(576*b^4 + 64*
d^4 - 640*b^2*d^2) - (exp(- 2*a - 2*b*x)*(9*b*d^2 - 81*b^2*d - 81*b^3 + 9*d^3))/(576*b^4 + 64*d^4 - 640*b^2*d^
2) - (exp(- 4*a - 4*b*x)*(9*b*d^2 + 81*b^2*d - 81*b^3 - 9*d^3))/(576*b^4 + 64*d^4 - 640*b^2*d^2)) + exp(3*a -
3*c + 3*b*x - 3*d*x)*((9*b*d^2 - b^2*d - b^3 + 9*d^3)/(192*b^4 + 1728*d^4 - 1920*b^2*d^2) + (exp(- 6*a - 6*b*x
)*(9*b*d^2 + b^2*d - b^3 - 9*d^3))/(192*b^4 + 1728*d^4 - 1920*b^2*d^2) - (exp(- 2*a - 2*b*x)*(9*b*d^2 - 27*b^2
*d - 9*b^3 + 27*d^3))/(192*b^4 + 1728*d^4 - 1920*b^2*d^2) - (exp(- 4*a - 4*b*x)*(9*b*d^2 + 27*b^2*d - 9*b^3 -
27*d^3))/(192*b^4 + 1728*d^4 - 1920*b^2*d^2)) - exp(3*a + 3*c + 3*b*x + 3*d*x)*((9*b*d^2 + b^2*d - b^3 - 9*d^3
)/(192*b^4 + 1728*d^4 - 1920*b^2*d^2) + (exp(- 6*a - 6*b*x)*(9*b*d^2 - b^2*d - b^3 + 9*d^3))/(192*b^4 + 1728*d
^4 - 1920*b^2*d^2) - (exp(- 2*a - 2*b*x)*(9*b*d^2 + 27*b^2*d - 9*b^3 - 27*d^3))/(192*b^4 + 1728*d^4 - 1920*b^2
*d^2) - (exp(- 4*a - 4*b*x)*(9*b*d^2 - 27*b^2*d - 9*b^3 + 27*d^3))/(192*b^4 + 1728*d^4 - 1920*b^2*d^2))

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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.

[In]

integrate(sinh(b*x+a)**3*sinh(d*x+c)**3,x)

[Out]

Timed out

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