### 3.177 $$\int \cosh ^2(a+b x) \cosh ^3(c+d x) \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0998268, 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 = {5614, 2637} $\frac{\sinh (2 a+x (2 b-3 d)-3 c)}{16 (2 b-3 d)}+\frac{3 \sinh (2 a+x (2 b-d)-c)}{16 (2 b-d)}+\frac{3 \sinh (2 a+x (2 b+d)+c)}{16 (2 b+d)}+\frac{\sinh (2 a+x (2 b+3 d)+3 c)}{16 (2 b+3 d)}+\frac{3 \sinh (c+d x)}{8 d}+\frac{\sinh (3 c+3 d x)}{24 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 5614

Int[Cosh[v_]^(p_.)*Cosh[w_]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[Cosh[v]^p*Cosh[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 \cosh ^2(a+b x) \cosh ^3(c+d x) \, dx &=\int \left (\frac{1}{16} \cosh (2 a-3 c+(2 b-3 d) x)+\frac{3}{16} \cosh (2 a-c+(2 b-d) x)+\frac{3}{8} \cosh (c+d x)+\frac{1}{8} \cosh (3 c+3 d x)+\frac{3}{16} \cosh (2 a+c+(2 b+d) x)+\frac{1}{16} \cosh (2 a+3 c+(2 b+3 d) x)\right ) \, dx\\ &=\frac{1}{16} \int \cosh (2 a-3 c+(2 b-3 d) x) \, dx+\frac{1}{16} \int \cosh (2 a+3 c+(2 b+3 d) x) \, dx+\frac{1}{8} \int \cosh (3 c+3 d x) \, dx+\frac{3}{16} \int \cosh (2 a-c+(2 b-d) x) \, dx+\frac{3}{16} \int \cosh (2 a+c+(2 b+d) x) \, dx+\frac{3}{8} \int \cosh (c+d x) \, dx\\ &=\frac{\sinh (2 a-3 c+(2 b-3 d) x)}{16 (2 b-3 d)}+\frac{3 \sinh (2 a-c+(2 b-d) x)}{16 (2 b-d)}+\frac{3 \sinh (c+d x)}{8 d}+\frac{\sinh (3 c+3 d x)}{24 d}+\frac{3 \sinh (2 a+c+(2 b+d) x)}{16 (2 b+d)}+\frac{\sinh (2 a+3 c+(2 b+3 d) x)}{16 (2 b+3 d)}\\ \end{align*}

Mathematica [A]  time = 1.61184, size = 158, normalized size = 1.1 $\frac{1}{48} \left (\frac{3 \sinh (2 a+2 b x-3 c-3 d x)}{2 b-3 d}+\frac{9 \sinh (2 a+2 b x-c-d x)}{2 b-d}+\frac{9 \sinh (2 a+2 b x+c+d x)}{2 b+d}+\frac{3 \sinh (2 a+2 b x+3 c+3 d x)}{2 b+3 d}+\frac{18 \sinh (c) \cosh (d x)}{d}+\frac{2 \sinh (3 c) \cosh (3 d x)}{d}+\frac{18 \cosh (c) \sinh (d x)}{d}+\frac{2 \cosh (3 c) \sinh (3 d x)}{d}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 133, normalized size = 0.9 \begin{align*}{\frac{\sinh \left ( 2\,a-3\,c+ \left ( 2\,b-3\,d \right ) x \right ) }{32\,b-48\,d}}+{\frac{3\,\sinh \left ( 2\,a-c+ \left ( 2\,b-d \right ) x \right ) }{32\,b-16\,d}}+{\frac{3\,\sinh \left ( dx+c \right ) }{8\,d}}+{\frac{\sinh \left ( 3\,dx+3\,c \right ) }{24\,d}}+{\frac{3\,\sinh \left ( 2\,a+c+ \left ( 2\,b+d \right ) x \right ) }{32\,b+16\,d}}+{\frac{\sinh \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(cosh(b*x+a)^2*cosh(d*x+c)^3,x)

[Out]

1/16*sinh(2*a-3*c+(2*b-3*d)*x)/(2*b-3*d)+3/16*sinh(2*a-c+(2*b-d)*x)/(2*b-d)+3/8*sinh(d*x+c)/d+1/24*sinh(3*d*x+
3*c)/d+3/16*sinh(2*a+c+(2*b+d)*x)/(2*b+d)+1/16*sinh(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(cosh(b*x+a)^2*cosh(d*x+c)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.92245, size = 926, normalized size = 6.43 \begin{align*} \frac{36 \,{\left (4 \, b^{3} d - b d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) \sinh \left (d x + c\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} - 9 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \sinh \left (b x + a\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 12 \,{\left ({\left (4 \, b^{3} d - b d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{3} + 3 \,{\left (4 \, b^{3} d - 9 \, b d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right ) + 3 \,{\left (48 \, b^{4} - 120 \, b^{2} d^{2} + 27 \, d^{4} - 3 \,{\left (4 \, b^{2} d^{2} - 9 \, d^{4}\right )} \cosh \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 )^{2} - 3 \,{\left (4 \, b^{2} d^{2} - 9 \, d^{4} + 3 \,{\left (4 \, b^{2} d^{2} - d^{4}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (b x + a\right )^{2}\right )} \sinh \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(cosh(b*x+a)^2*cosh(d*x+c)^3,x, algorithm="fricas")

[Out]

1/24*(36*(4*b^3*d - b*d^3)*cosh(b*x + a)*cosh(d*x + c)*sinh(b*x + a)*sinh(d*x + c)^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 - 9*(4*b^2*d^2 - d^4)*sinh(b*x + a)^2)*sinh(d*x + c)^3 + 12*((4*b^
3*d - b*d^3)*cosh(b*x + a)*cosh(d*x + c)^3 + 3*(4*b^3*d - 9*b*d^3)*cosh(b*x + a)*cosh(d*x + c))*sinh(b*x + a)
+ 3*(48*b^4 - 120*b^2*d^2 + 27*d^4 - 3*(4*b^2*d^2 - 9*d^4)*cosh(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)^2 - 3*(4*b^2*d^2 - 9*d^4 + 3*(4*b^2*d^2 - d^4)*cosh(d*x + c)^
2)*sinh(b*x + a)^2)*sinh(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(cosh(b*x+a)**2*cosh(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21821, 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(cosh(b*x+a)^2*cosh(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