### 3.960 $$\int e^{c+d x} \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=195 $-\frac{b e^{c+d x} \sinh (a+b x)}{8 \left (b^2-d^2\right )}+\frac{3 b e^{c+d x} \sinh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}+\frac{5 b e^{c+d x} \sinh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}+\frac{d e^{c+d x} \cosh (a+b x)}{8 \left (b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}$

[Out]

(d*E^(c + d*x)*Cosh[a + b*x])/(8*(b^2 - d^2)) - (d*E^(c + d*x)*Cosh[3*a + 3*b*x])/(16*(9*b^2 - d^2)) - (d*E^(c
+ d*x)*Cosh[5*a + 5*b*x])/(16*(25*b^2 - d^2)) - (b*E^(c + d*x)*Sinh[a + b*x])/(8*(b^2 - d^2)) + (3*b*E^(c + d
*x)*Sinh[3*a + 3*b*x])/(16*(9*b^2 - d^2)) + (5*b*E^(c + d*x)*Sinh[5*a + 5*b*x])/(16*(25*b^2 - d^2))

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Rubi [A]  time = 0.132763, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {5509, 5475} $-\frac{b e^{c+d x} \sinh (a+b x)}{8 \left (b^2-d^2\right )}+\frac{3 b e^{c+d x} \sinh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}+\frac{5 b e^{c+d x} \sinh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}+\frac{d e^{c+d x} \cosh (a+b x)}{8 \left (b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(c + d*x)*Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

(d*E^(c + d*x)*Cosh[a + b*x])/(8*(b^2 - d^2)) - (d*E^(c + d*x)*Cosh[3*a + 3*b*x])/(16*(9*b^2 - d^2)) - (d*E^(c
+ d*x)*Cosh[5*a + 5*b*x])/(16*(25*b^2 - d^2)) - (b*E^(c + d*x)*Sinh[a + b*x])/(8*(b^2 - d^2)) + (3*b*E^(c + d
*x)*Sinh[3*a + 3*b*x])/(16*(9*b^2 - d^2)) + (5*b*E^(c + d*x)*Sinh[5*a + 5*b*x])/(16*(25*b^2 - d^2))

Rule 5509

Int[Cosh[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol]
:> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sinh[d + e*x]^m*Cosh[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e,
f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 5475

Int[Cosh[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b*x))
*Cosh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sinh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2)
, x] /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 - b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int e^{c+d x} \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{1}{8} e^{c+d x} \cosh (a+b x)+\frac{1}{16} e^{c+d x} \cosh (3 a+3 b x)+\frac{1}{16} e^{c+d x} \cosh (5 a+5 b x)\right ) \, dx\\ &=\frac{1}{16} \int e^{c+d x} \cosh (3 a+3 b x) \, dx+\frac{1}{16} \int e^{c+d x} \cosh (5 a+5 b x) \, dx-\frac{1}{8} \int e^{c+d x} \cosh (a+b x) \, dx\\ &=\frac{d e^{c+d x} \cosh (a+b x)}{8 \left (b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}-\frac{d e^{c+d x} \cosh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}-\frac{b e^{c+d x} \sinh (a+b x)}{8 \left (b^2-d^2\right )}+\frac{3 b e^{c+d x} \sinh (3 a+3 b x)}{16 \left (9 b^2-d^2\right )}+\frac{5 b e^{c+d x} \sinh (5 a+5 b x)}{16 \left (25 b^2-d^2\right )}\\ \end{align*}

Mathematica [A]  time = 1.20979, size = 118, normalized size = 0.61 $\frac{1}{16} e^{c+d x} \left (\frac{3 b \sinh (3 (a+b x))-d \cosh (3 (a+b x))}{9 b^2-d^2}+\frac{5 b \sinh (5 (a+b x))-d \cosh (5 (a+b x))}{25 b^2-d^2}+\frac{2 d \cosh (a+b x)-2 b \sinh (a+b x)}{(b-d) (b+d)}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(c + d*x)*Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

(E^(c + d*x)*((2*d*Cosh[a + b*x] - 2*b*Sinh[a + b*x])/((b - d)*(b + d)) + (-(d*Cosh[3*(a + b*x)]) + 3*b*Sinh[3
*(a + b*x)])/(9*b^2 - d^2) + (-(d*Cosh[5*(a + b*x)]) + 5*b*Sinh[5*(a + b*x)])/(25*b^2 - d^2)))/16

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Maple [A]  time = 0.009, size = 278, normalized size = 1.4 \begin{align*} -{\frac{\sinh \left ( a-c+ \left ( b-d \right ) x \right ) }{16\,b-16\,d}}-{\frac{\sinh \left ( a+c+ \left ( b+d \right ) x \right ) }{16\,b+16\,d}}+{\frac{\sinh \left ( 3\,a-c+ \left ( 3\,b-d \right ) x \right ) }{96\,b-32\,d}}+{\frac{\sinh \left ( 3\,a+c+ \left ( 3\,b+d \right ) x \right ) }{96\,b+32\,d}}+{\frac{\sinh \left ( \left ( 5\,b-d \right ) x+5\,a-c \right ) }{160\,b-32\,d}}+{\frac{\sinh \left ( \left ( 5\,b+d \right ) x+5\,a+c \right ) }{160\,b+32\,d}}+{\frac{\cosh \left ( a-c+ \left ( b-d \right ) x \right ) }{16\,b-16\,d}}-{\frac{\cosh \left ( a+c+ \left ( b+d \right ) x \right ) }{16\,b+16\,d}}-{\frac{\cosh \left ( 3\,a-c+ \left ( 3\,b-d \right ) x \right ) }{96\,b-32\,d}}+{\frac{\cosh \left ( 3\,a+c+ \left ( 3\,b+d \right ) x \right ) }{96\,b+32\,d}}-{\frac{\cosh \left ( \left ( 5\,b-d \right ) x+5\,a-c \right ) }{160\,b-32\,d}}+{\frac{\cosh \left ( \left ( 5\,b+d \right ) x+5\,a+c \right ) }{160\,b+32\,d}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*sinh(a-c+(b-d)*x)/(b-d)-1/16*sinh(a+c+(b+d)*x)/(b+d)+1/32*sinh(3*a-c+(3*b-d)*x)/(3*b-d)+1/32*sinh(3*a+c+
(3*b+d)*x)/(3*b+d)+1/32/(5*b-d)*sinh((5*b-d)*x+5*a-c)+1/32/(5*b+d)*sinh((5*b+d)*x+5*a+c)+1/16*cosh(a-c+(b-d)*x
)/(b-d)-1/16*cosh(a+c+(b+d)*x)/(b+d)-1/32*cosh(3*a-c+(3*b-d)*x)/(3*b-d)+1/32*cosh(3*a+c+(3*b+d)*x)/(3*b+d)-1/3
2*cosh((5*b-d)*x+5*a-c)/(5*b-d)+1/32*cosh((5*b+d)*x+5*a+c)/(5*b+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(exp(d*x+c)*cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.11295, size = 2202, normalized size = 11.29 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(5*(9*b^4*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)*cosh(d*x + c)*sinh(b*x + a)^4 - 5*(9*b^5 - 10*b^3*d^2 + b*
d^4)*cosh(d*x + c)*sinh(b*x + a)^5 - (75*b^5 - 78*b^3*d^2 + 3*b*d^4 + 50*(9*b^5 - 10*b^3*d^2 + b*d^4)*cosh(b*x
+ a)^2)*cosh(d*x + c)*sinh(b*x + a)^3 + (10*(9*b^4*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)^3 + 3*(25*b^4*d - 26*b
^2*d^3 + d^5)*cosh(b*x + a))*cosh(d*x + c)*sinh(b*x + a)^2 + (450*b^5 - 68*b^3*d^2 + 2*b*d^4 - 25*(9*b^5 - 10*
b^3*d^2 + b*d^4)*cosh(b*x + a)^4 - 9*(25*b^5 - 26*b^3*d^2 + b*d^4)*cosh(b*x + a)^2)*cosh(d*x + c)*sinh(b*x + a
) + ((9*b^4*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)^5 + (25*b^4*d - 26*b^2*d^3 + d^5)*cosh(b*x + a)^3 - 2*(225*b^4
*d - 34*b^2*d^3 + d^5)*cosh(b*x + a))*cosh(d*x + c) + ((9*b^4*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)^5 + 5*(9*b^4
*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)*sinh(b*x + a)^4 - 5*(9*b^5 - 10*b^3*d^2 + b*d^4)*sinh(b*x + a)^5 + (25*b^
4*d - 26*b^2*d^3 + d^5)*cosh(b*x + a)^3 - (75*b^5 - 78*b^3*d^2 + 3*b*d^4 + 50*(9*b^5 - 10*b^3*d^2 + b*d^4)*cos
h(b*x + a)^2)*sinh(b*x + a)^3 + (10*(9*b^4*d - 10*b^2*d^3 + d^5)*cosh(b*x + a)^3 + 3*(25*b^4*d - 26*b^2*d^3 +
d^5)*cosh(b*x + a))*sinh(b*x + a)^2 - 2*(225*b^4*d - 34*b^2*d^3 + d^5)*cosh(b*x + a) + (450*b^5 - 68*b^3*d^2 +
2*b*d^4 - 25*(9*b^5 - 10*b^3*d^2 + b*d^4)*cosh(b*x + a)^4 - 9*(25*b^5 - 26*b^3*d^2 + b*d^4)*cosh(b*x + a)^2)*
sinh(b*x + a))*sinh(d*x + c))/((225*b^6 - 259*b^4*d^2 + 35*b^2*d^4 - d^6)*cosh(b*x + a)^6 - 3*(225*b^6 - 259*b
^4*d^2 + 35*b^2*d^4 - d^6)*cosh(b*x + a)^4*sinh(b*x + a)^2 + 3*(225*b^6 - 259*b^4*d^2 + 35*b^2*d^4 - d^6)*cosh
(b*x + a)^2*sinh(b*x + a)^4 - (225*b^6 - 259*b^4*d^2 + 35*b^2*d^4 - d^6)*sinh(b*x + a)^6)

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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(exp(d*x+c)*cosh(b*x+a)**3*sinh(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.17169, size = 178, normalized size = 0.91 \begin{align*} \frac{e^{\left (5 \, b x + d x + 5 \, a + c\right )}}{32 \,{\left (5 \, b + d\right )}} + \frac{e^{\left (3 \, b x + d x + 3 \, a + c\right )}}{32 \,{\left (3 \, b + d\right )}} - \frac{e^{\left (b x + d x + a + c\right )}}{16 \,{\left (b + d\right )}} + \frac{e^{\left (-b x + d x - a + c\right )}}{16 \,{\left (b - d\right )}} - \frac{e^{\left (-3 \, b x + d x - 3 \, a + c\right )}}{32 \,{\left (3 \, b - d\right )}} - \frac{e^{\left (-5 \, b x + d x - 5 \, a + c\right )}}{32 \,{\left (5 \, b - d\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*e^(5*b*x + d*x + 5*a + c)/(5*b + d) + 1/32*e^(3*b*x + d*x + 3*a + c)/(3*b + d) - 1/16*e^(b*x + d*x + a +
c)/(b + d) + 1/16*e^(-b*x + d*x - a + c)/(b - d) - 1/32*e^(-3*b*x + d*x - 3*a + c)/(3*b - d) - 1/32*e^(-5*b*x
+ d*x - 5*a + c)/(5*b - d)