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]

1/8*d*exp(d*x+c)*cosh(b*x+a)/(b^2-d^2)-1/16*d*exp(d*x+c)*cosh(3*b*x+3*a)/(9*b^2-d^2)-1/16*d*exp(d*x+c)*cosh(5*
b*x+5*a)/(25*b^2-d^2)-1/8*b*exp(d*x+c)*sinh(b*x+a)/(b^2-d^2)+3/16*b*exp(d*x+c)*sinh(3*b*x+3*a)/(9*b^2-d^2)+5/1
6*b*exp(d*x+c)*sinh(5*b*x+5*a)/(25*b^2-d^2)

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Rubi [A]  time = 0.13, antiderivative size = 195, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 1.29, 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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fricas [B]  time = 0.53, size = 917, normalized size = 4.70 \[ \text {result too large to display} \]

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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giac [A]  time = 0.15, size = 132, normalized size = 0.68 \[ \frac {e^{\left (5 \, b x + d x + 5 \, a + c\right )}}{32 \, {\left (5 \, b + d\right )}} + \frac {e^{\left (3 \, b x + d x + 3 \, a + c\right )}}{32 \, {\left (3 \, b + d\right )}} - \frac {e^{\left (b x + d x + a + c\right )}}{16 \, {\left (b + d\right )}} + \frac {e^{\left (-b x + d x - a + c\right )}}{16 \, {\left (b - d\right )}} - \frac {e^{\left (-3 \, b x + d x - 3 \, a + c\right )}}{32 \, {\left (3 \, b - d\right )}} - \frac {e^{\left (-5 \, b x + d x - 5 \, a + c\right )}}{32 \, {\left (5 \, b - d\right )}} \]

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)

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maple [A]  time = 0.54, size = 278, normalized size = 1.43 \[ -\frac {\sinh \left (a -c +\left (b -d \right ) x \right )}{16 \left (b -d \right )}-\frac {\sinh \left (a +c +\left (b +d \right ) x \right )}{16 \left (b +d \right )}+\frac {\sinh \left (3 a -c +\left (3 b -d \right ) x \right )}{96 b -32 d}+\frac {\sinh \left (3 a +c +\left (3 b +d \right ) x \right )}{96 b +32 d}+\frac {\sinh \left (\left (5 b -d \right ) x +5 a -c \right )}{160 b -32 d}+\frac {\sinh \left (\left (5 b +d \right ) x +5 a +c \right )}{160 b +32 d}+\frac {\cosh \left (a -c +\left (b -d \right ) x \right )}{16 b -16 d}-\frac {\cosh \left (a +c +\left (b +d \right ) x \right )}{16 \left (b +d \right )}-\frac {\cosh \left (3 a -c +\left (3 b -d \right ) x \right )}{32 \left (3 b -d \right )}+\frac {\cosh \left (3 a +c +\left (3 b +d \right ) x \right )}{96 b +32 d}-\frac {\cosh \left (\left (5 b -d \right ) x +5 a -c \right )}{32 \left (5 b -d \right )}+\frac {\cosh \left (\left (5 b +d \right ) x +5 a +c \right )}{160 b +32 d} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-d/b>0)', see `assume?` for mo
re details)Is -d/b equal to -1?

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mupad [B]  time = 2.42, size = 393, normalized size = 2.02 \[ \frac {{\mathrm {cosh}\left (a+b\,x\right )}^5\,{\mathrm {e}}^{c+d\,x}\,\left (26\,b^4\,d-2\,b^2\,d^3\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6}+\frac {3\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {e}}^{c+d\,x}\,{\mathrm {sinh}\left (a+b\,x\right )}^3\,\left (25\,b^5-10\,b^3\,d^2+b\,d^4\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6}+\frac {2\,{\mathrm {cosh}\left (a+b\,x\right )}^4\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (a+b\,x\right )\,\left (b\,d^4-13\,b^3\,d^2\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6}-\frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {e}}^{c+d\,x}\,{\mathrm {sinh}\left (a+b\,x\right )}^2\,\left (65\,b^4\,d-18\,b^2\,d^3+d^5\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6}-\frac {6\,b^3\,{\mathrm {e}}^{c+d\,x}\,{\mathrm {sinh}\left (a+b\,x\right )}^5\,\left (5\,b^2-d^2\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6}+\frac {6\,b^2\,d\,\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {e}}^{c+d\,x}\,{\mathrm {sinh}\left (a+b\,x\right )}^4\,\left (5\,b^2-d^2\right )}{225\,b^6-259\,b^4\,d^2+35\,b^2\,d^4-d^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(a + b*x)^5*exp(c + d*x)*(26*b^4*d - 2*b^2*d^3))/(225*b^6 - d^6 + 35*b^2*d^4 - 259*b^4*d^2) + (3*cosh(a +
b*x)^2*exp(c + d*x)*sinh(a + b*x)^3*(b*d^4 + 25*b^5 - 10*b^3*d^2))/(225*b^6 - d^6 + 35*b^2*d^4 - 259*b^4*d^2)
+ (2*cosh(a + b*x)^4*exp(c + d*x)*sinh(a + b*x)*(b*d^4 - 13*b^3*d^2))/(225*b^6 - d^6 + 35*b^2*d^4 - 259*b^4*d
^2) - (cosh(a + b*x)^3*exp(c + d*x)*sinh(a + b*x)^2*(65*b^4*d + d^5 - 18*b^2*d^3))/(225*b^6 - d^6 + 35*b^2*d^4
- 259*b^4*d^2) - (6*b^3*exp(c + d*x)*sinh(a + b*x)^5*(5*b^2 - d^2))/(225*b^6 - d^6 + 35*b^2*d^4 - 259*b^4*d^2
) + (6*b^2*d*cosh(a + b*x)*exp(c + d*x)*sinh(a + b*x)^4*(5*b^2 - d^2))/(225*b^6 - d^6 + 35*b^2*d^4 - 259*b^4*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(exp(d*x+c)*cosh(b*x+a)**3*sinh(b*x+a)**2,x)

[Out]

Timed out

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