### 3.171 $$\int \sinh ^2(a+b x) \sinh ^3(c+d x) \, dx$$

Optimal. Leaf size=144 $-\frac{\cosh (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac{3 \cosh (2 a+x (2 b-d)-c)}{16 (2 b-d)}-\frac{3 \cosh (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac{\cosh (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac{3 \cosh (c+d x)}{8 d}-\frac{\cosh (3 c+3 d x)}{24 d}$

[Out]

-Cosh[2*a - 3*c + (2*b - 3*d)*x]/(16*(2*b - 3*d)) + (3*Cosh[2*a - c + (2*b - d)*x])/(16*(2*b - d)) + (3*Cosh[c
+ d*x])/(8*d) - Cosh[3*c + 3*d*x]/(24*d) - (3*Cosh[2*a + c + (2*b + d)*x])/(16*(2*b + d)) + Cosh[2*a + 3*c +
(2*b + 3*d)*x]/(16*(2*b + 3*d))

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Rubi [A]  time = 0.121526, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {5613, 2638} $-\frac{\cosh (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac{3 \cosh (2 a+x (2 b-d)-c)}{16 (2 b-d)}-\frac{3 \cosh (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac{\cosh (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac{3 \cosh (c+d x)}{8 d}-\frac{\cosh (3 c+3 d x)}{24 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[a + b*x]^2*Sinh[c + d*x]^3,x]

[Out]

-Cosh[2*a - 3*c + (2*b - 3*d)*x]/(16*(2*b - 3*d)) + (3*Cosh[2*a - c + (2*b - d)*x])/(16*(2*b - d)) + (3*Cosh[c
+ d*x])/(8*d) - Cosh[3*c + 3*d*x]/(24*d) - (3*Cosh[2*a + c + (2*b + d)*x])/(16*(2*b + d)) + Cosh[2*a + 3*c +
(2*b + 3*d)*x]/(16*(2*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 2638

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

Rubi steps

\begin{align*} \int \sinh ^2(a+b x) \sinh ^3(c+d x) \, dx &=\int \left (-\frac{1}{16} \sinh (2 a-3 c+(2 b-3 d) x)+\frac{3}{16} \sinh (2 a-c+(2 b-d) x)+\frac{3}{8} \sinh (c+d x)-\frac{1}{8} \sinh (3 c+3 d x)-\frac{3}{16} \sinh (2 a+c+(2 b+d) x)+\frac{1}{16} \sinh (2 a+3 c+(2 b+3 d) x)\right ) \, dx\\ &=-\left (\frac{1}{16} \int \sinh (2 a-3 c+(2 b-3 d) x) \, dx\right )+\frac{1}{16} \int \sinh (2 a+3 c+(2 b+3 d) x) \, dx-\frac{1}{8} \int \sinh (3 c+3 d x) \, dx+\frac{3}{16} \int \sinh (2 a-c+(2 b-d) x) \, dx-\frac{3}{16} \int \sinh (2 a+c+(2 b+d) x) \, dx+\frac{3}{8} \int \sinh (c+d x) \, dx\\ &=-\frac{\cosh (2 a-3 c+(2 b-3 d) x)}{16 (2 b-3 d)}+\frac{3 \cosh (2 a-c+(2 b-d) x)}{16 (2 b-d)}+\frac{3 \cosh (c+d x)}{8 d}-\frac{\cosh (3 c+3 d x)}{24 d}-\frac{3 \cosh (2 a+c+(2 b+d) x)}{16 (2 b+d)}+\frac{\cosh (2 a+3 c+(2 b+3 d) x)}{16 (2 b+3 d)}\\ \end{align*}

Mathematica [A]  time = 1.62358, size = 158, normalized size = 1.1 $\frac{1}{48} \left (-\frac{3 \cosh (2 a+2 b x-3 c-3 d x)}{2 b-3 d}+\frac{9 \cosh (2 a+2 b x-c-d x)}{2 b-d}-\frac{9 \cosh (2 a+2 b x+c+d x)}{2 b+d}+\frac{3 \cosh (2 a+2 b x+3 c+3 d x)}{2 b+3 d}+\frac{18 \sinh (c) \sinh (d x)}{d}-\frac{2 \sinh (3 c) \sinh (3 d x)}{d}+\frac{18 \cosh (c) \cosh (d x)}{d}-\frac{2 \cosh (3 c) \cosh (3 d x)}{d}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[a + b*x]^2*Sinh[c + d*x]^3,x]

[Out]

((18*Cosh[c]*Cosh[d*x])/d - (2*Cosh[3*c]*Cosh[3*d*x])/d - (3*Cosh[2*a - 3*c + 2*b*x - 3*d*x])/(2*b - 3*d) + (9
*Cosh[2*a - c + 2*b*x - d*x])/(2*b - d) - (9*Cosh[2*a + c + 2*b*x + d*x])/(2*b + d) + (3*Cosh[2*a + 3*c + 2*b*
x + 3*d*x])/(2*b + 3*d) + (18*Sinh[c]*Sinh[d*x])/d - (2*Sinh[3*c]*Sinh[3*d*x])/d)/48

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Maple [A]  time = 0.017, size = 133, normalized size = 0.9 \begin{align*} -{\frac{\cosh \left ( 2\,a-3\,c+ \left ( 2\,b-3\,d \right ) x \right ) }{32\,b-48\,d}}+{\frac{3\,\cosh \left ( 2\,a-c+ \left ( 2\,b-d \right ) x \right ) }{32\,b-16\,d}}+{\frac{3\,\cosh \left ( dx+c \right ) }{8\,d}}-{\frac{\cosh \left ( 3\,dx+3\,c \right ) }{24\,d}}-{\frac{3\,\cosh \left ( 2\,a+c+ \left ( 2\,b+d \right ) x \right ) }{32\,b+16\,d}}+{\frac{\cosh \left ( 2\,a+3\,c+ \left ( 2\,b+3\,d \right ) x \right ) }{32\,b+48\,d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x+a)^2*sinh(d*x+c)^3,x)

[Out]

-1/16*cosh(2*a-3*c+(2*b-3*d)*x)/(2*b-3*d)+3/16*cosh(2*a-c+(2*b-d)*x)/(2*b-d)+3/8*cosh(d*x+c)/d-1/24*cosh(3*d*x
+3*c)/d-3/16*cosh(2*a+c+(2*b+d)*x)/(2*b+d)+1/16*cosh(2*a+3*c+(2*b+3*d)*x)/(2*b+3*d)

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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)^2*sinh(d*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.89329, size = 961, normalized size = 6.67 \begin{align*} \frac{12 \,{\left (4 \, b^{3} d - b d^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) \sinh \left (d x + c\right )^{3} -{\left (16 \, b^{4} - 40 \, b^{2} d^{2} + 9 \, d^{4} + 9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right )^{3} - 9 \,{\left ({\left (4 \, b^{2} d^{2} - d^{4}\right )} \cosh \left (d x + c\right )^{3} -{\left (4 \, b^{2} d^{2} - 9 \, d^{4}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right )^{2} + 36 \,{\left ({\left (4 \, b^{3} d - b d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} -{\left (4 \, b^{3} d - 9 \, b d^{3}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) \sinh \left (d x + c\right ) - 3 \,{\left (9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{2} +{\left (16 \, b^{4} - 40 \, b^{2} d^{2} + 9 \, d^{4} + 9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 9 \,{\left (16 \, b^{4} - 40 \, b^{2} d^{2} + 9 \, d^{4} +{\left (4 \, b^{2} d^{2} - 9 \, d^{4}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right )}{24 \,{\left ({\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cosh \left (b x + a\right )^{2} -{\left (16 \, b^{4} d - 40 \, b^{2} d^{3} + 9 \, d^{5}\right )} \sinh \left (b x + a\right )^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2*sinh(d*x+c)^3,x, algorithm="fricas")

[Out]

1/24*(12*(4*b^3*d - b*d^3)*cosh(b*x + a)*sinh(b*x + a)*sinh(d*x + c)^3 - (16*b^4 - 40*b^2*d^2 + 9*d^4 + 9*(4*b
^2*d^2 - d^4)*cosh(b*x + a)^2)*cosh(d*x + c)^3 - 9*((4*b^2*d^2 - d^4)*cosh(d*x + c)^3 - (4*b^2*d^2 - 9*d^4)*co
sh(d*x + c))*sinh(b*x + a)^2 + 36*((4*b^3*d - b*d^3)*cosh(b*x + a)*cosh(d*x + c)^2 - (4*b^3*d - 9*b*d^3)*cosh(
b*x + a))*sinh(b*x + a)*sinh(d*x + c) - 3*(9*(4*b^2*d^2 - d^4)*cosh(d*x + c)*sinh(b*x + a)^2 + (16*b^4 - 40*b^
2*d^2 + 9*d^4 + 9*(4*b^2*d^2 - d^4)*cosh(b*x + a)^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 9*(16*b^4 - 40*b^2*d^2 +
9*d^4 + (4*b^2*d^2 - 9*d^4)*cosh(b*x + a)^2)*cosh(d*x + c))/((16*b^4*d - 40*b^2*d^3 + 9*d^5)*cosh(b*x + a)^2
- (16*b^4*d - 40*b^2*d^3 + 9*d^5)*sinh(b*x + a)^2)

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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)**2*sinh(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.20852, size = 351, normalized size = 2.44 \begin{align*} \frac{e^{\left (2 \, b x + 3 \, d x + 2 \, a + 3 \, c\right )}}{32 \,{\left (2 \, b + 3 \, d\right )}} - \frac{3 \, e^{\left (2 \, b x + d x + 2 \, a + c\right )}}{32 \,{\left (2 \, b + d\right )}} + \frac{3 \, e^{\left (2 \, b x - d x + 2 \, a - c\right )}}{32 \,{\left (2 \, b - d\right )}} - \frac{e^{\left (2 \, b x - 3 \, d x + 2 \, a - 3 \, c\right )}}{32 \,{\left (2 \, b - 3 \, d\right )}} - \frac{e^{\left (-2 \, b x + 3 \, d x - 2 \, a + 3 \, c\right )}}{32 \,{\left (2 \, b - 3 \, d\right )}} + \frac{3 \, e^{\left (-2 \, b x + d x - 2 \, a + c\right )}}{32 \,{\left (2 \, b - d\right )}} - \frac{3 \, e^{\left (-2 \, b x - d x - 2 \, a - c\right )}}{32 \,{\left (2 \, b + d\right )}} + \frac{e^{\left (-2 \, b x - 3 \, d x - 2 \, a - 3 \, c\right )}}{32 \,{\left (2 \, b + 3 \, d\right )}} - \frac{e^{\left (3 \, d x + 3 \, c\right )}}{48 \, d} + \frac{3 \, e^{\left (d x + c\right )}}{16 \, d} + \frac{3 \, e^{\left (-d x - c\right )}}{16 \, d} - \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{48 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(b*x+a)^2*sinh(d*x+c)^3,x, algorithm="giac")

[Out]

1/32*e^(2*b*x + 3*d*x + 2*a + 3*c)/(2*b + 3*d) - 3/32*e^(2*b*x + d*x + 2*a + c)/(2*b + d) + 3/32*e^(2*b*x - d*
x + 2*a - c)/(2*b - d) - 1/32*e^(2*b*x - 3*d*x + 2*a - 3*c)/(2*b - 3*d) - 1/32*e^(-2*b*x + 3*d*x - 2*a + 3*c)/
(2*b - 3*d) + 3/32*e^(-2*b*x + d*x - 2*a + c)/(2*b - d) - 3/32*e^(-2*b*x - d*x - 2*a - c)/(2*b + d) + 1/32*e^(
-2*b*x - 3*d*x - 2*a - 3*c)/(2*b + 3*d) - 1/48*e^(3*d*x + 3*c)/d + 3/16*e^(d*x + c)/d + 3/16*e^(-d*x - c)/d -
1/48*e^(-3*d*x - 3*c)/d