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*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))

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Rubi [A]  time = 0.144429, antiderivative size = 195, normalized size of antiderivative = 1., 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 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]))

Rule 2637

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

Mathematica [A]  time = 1.57155, 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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Maple [A]  time = 0.023, size = 184, normalized size = 0.9 \begin{align*}{\frac{3\,\sinh \left ( a-3\,c+ \left ( b-3\,d \right ) x \right ) }{32\,b-96\,d}}-{\frac{9\,\sinh \left ( a-c+ \left ( b-d \right ) x \right ) }{32\,b-32\,d}}+{\frac{9\,\sinh \left ( a+c+ \left ( b+d \right ) x \right ) }{32\,b+32\,d}}-{\frac{3\,\sinh \left ( a+3\,c+ \left ( b+3\,d \right ) x \right ) }{32\,b+96\,d}}-{\frac{\sinh \left ( \left ( 3\,b-3\,d \right ) x+3\,a-3\,c \right ) }{96\,b-96\,d}}+{\frac{3\,\sinh \left ( 3\,a-c+ \left ( 3\,b-d \right ) x \right ) }{96\,b-32\,d}}-{\frac{3\,\sinh \left ( 3\,a+c+ \left ( 3\,b+d \right ) x \right ) }{96\,b+32\,d}}+{\frac{\sinh \left ( \left ( 3\,b+3\,d \right ) x+3\,a+3\,c \right ) }{96\,b+96\,d}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.08868, size = 1710, normalized size = 8.77 \begin{align*} \text{result too large to display} \end{align*}

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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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.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.24565, size = 504, normalized size = 2.58 \begin{align*} \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 )}} \end{align*}

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)