3.274 \(\int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^2} \, dx\)

Optimal. Leaf size=89 \[ \frac {1}{2} b \cosh (2 a) \text {Chi}(2 b x)+\frac {1}{2} b \cosh (4 a) \text {Chi}(4 b x)+\frac {1}{2} b \sinh (2 a) \text {Shi}(2 b x)+\frac {1}{2} b \sinh (4 a) \text {Shi}(4 b x)-\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x} \]

[Out]

1/2*b*Chi(2*b*x)*cosh(2*a)+1/2*b*Chi(4*b*x)*cosh(4*a)+1/2*b*Shi(2*b*x)*sinh(2*a)+1/2*b*Shi(4*b*x)*sinh(4*a)-1/
4*sinh(2*b*x+2*a)/x-1/8*sinh(4*b*x+4*a)/x

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Rubi [A]  time = 0.19, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5448, 3297, 3303, 3298, 3301} \[ \frac {1}{2} b \cosh (2 a) \text {Chi}(2 b x)+\frac {1}{2} b \cosh (4 a) \text {Chi}(4 b x)+\frac {1}{2} b \sinh (2 a) \text {Shi}(2 b x)+\frac {1}{2} b \sinh (4 a) \text {Shi}(4 b x)-\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cosh[2*a]*CoshIntegral[2*b*x])/2 + (b*Cosh[4*a]*CoshIntegral[4*b*x])/2 - Sinh[2*a + 2*b*x]/(4*x) - Sinh[4*a
+ 4*b*x]/(8*x) + (b*Sinh[2*a]*SinhIntegral[2*b*x])/2 + (b*Sinh[4*a]*SinhIntegral[4*b*x])/2

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^3(a+b x) \sinh (a+b x)}{x^2} \, dx &=\int \left (\frac {\sinh (2 a+2 b x)}{4 x^2}+\frac {\sinh (4 a+4 b x)}{8 x^2}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sinh (4 a+4 b x)}{x^2} \, dx+\frac {1}{4} \int \frac {\sinh (2 a+2 b x)}{x^2} \, dx\\ &=-\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x}+\frac {1}{2} b \int \frac {\cosh (2 a+2 b x)}{x} \, dx+\frac {1}{2} b \int \frac {\cosh (4 a+4 b x)}{x} \, dx\\ &=-\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x}+\frac {1}{2} (b \cosh (2 a)) \int \frac {\cosh (2 b x)}{x} \, dx+\frac {1}{2} (b \cosh (4 a)) \int \frac {\cosh (4 b x)}{x} \, dx+\frac {1}{2} (b \sinh (2 a)) \int \frac {\sinh (2 b x)}{x} \, dx+\frac {1}{2} (b \sinh (4 a)) \int \frac {\sinh (4 b x)}{x} \, dx\\ &=\frac {1}{2} b \cosh (2 a) \text {Chi}(2 b x)+\frac {1}{2} b \cosh (4 a) \text {Chi}(4 b x)-\frac {\sinh (2 a+2 b x)}{4 x}-\frac {\sinh (4 a+4 b x)}{8 x}+\frac {1}{2} b \sinh (2 a) \text {Shi}(2 b x)+\frac {1}{2} b \sinh (4 a) \text {Shi}(4 b x)\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 80, normalized size = 0.90 \[ \frac {4 b x \cosh (2 a) \text {Chi}(2 b x)+4 b x \cosh (4 a) \text {Chi}(4 b x)+4 b x \sinh (2 a) \text {Shi}(2 b x)+4 b x \sinh (4 a) \text {Shi}(4 b x)-2 \sinh (2 (a+b x))-\sinh (4 (a+b x))}{8 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*x*Cosh[2*a]*CoshIntegral[2*b*x] + 4*b*x*Cosh[4*a]*CoshIntegral[4*b*x] - 2*Sinh[2*(a + b*x)] - Sinh[4*(a +
b*x)] + 4*b*x*Sinh[2*a]*SinhIntegral[2*b*x] + 4*b*x*Sinh[4*a]*SinhIntegral[4*b*x])/(8*x)

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fricas [A]  time = 0.85, size = 139, normalized size = 1.56 \[ -\frac {2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - {\left (b x {\rm Ei}\left (4 \, b x\right ) + b x {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) - {\left (b x {\rm Ei}\left (2 \, b x\right ) + b x {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 2 \, {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - {\left (b x {\rm Ei}\left (4 \, b x\right ) - b x {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) - {\left (b x {\rm Ei}\left (2 \, b x\right ) - b x {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*cosh(b*x + a)*sinh(b*x + a)^3 - (b*x*Ei(4*b*x) + b*x*Ei(-4*b*x))*cosh(4*a) - (b*x*Ei(2*b*x) + b*x*Ei(-
2*b*x))*cosh(2*a) + 2*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) - (b*x*Ei(4*b*x) - b*x*Ei(-4*b*x))*sinh(
4*a) - (b*x*Ei(2*b*x) - b*x*Ei(-2*b*x))*sinh(2*a))/x

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giac [A]  time = 0.12, size = 100, normalized size = 1.12 \[ \frac {4 \, b x {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} + 4 \, b x {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 4 \, b x {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 4 \, b x {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{16 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*b*x*Ei(4*b*x)*e^(4*a) + 4*b*x*Ei(2*b*x)*e^(2*a) + 4*b*x*Ei(-2*b*x)*e^(-2*a) + 4*b*x*Ei(-4*b*x)*e^(-4*a
) - e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) + 2*e^(-2*b*x - 2*a) + e^(-4*b*x - 4*a))/x

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maple [A]  time = 0.48, size = 110, normalized size = 1.24 \[ \frac {{\mathrm e}^{-4 b x -4 a}}{16 x}-\frac {b \,{\mathrm e}^{-4 a} \Ei \left (1, 4 b x \right )}{4}+\frac {{\mathrm e}^{-2 b x -2 a}}{8 x}-\frac {b \,{\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{4}-\frac {{\mathrm e}^{2 b x +2 a}}{8 x}-\frac {b \,{\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{4}-\frac {{\mathrm e}^{4 b x +4 a}}{16 x}-\frac {b \,{\mathrm e}^{4 a} \Ei \left (1, -4 b x \right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^3*sinh(b*x+a)/x^2,x)

[Out]

1/16*exp(-4*b*x-4*a)/x-1/4*b*exp(-4*a)*Ei(1,4*b*x)+1/8*exp(-2*b*x-2*a)/x-1/4*b*exp(-2*a)*Ei(1,2*b*x)-1/8*exp(2
*b*x+2*a)/x-1/4*b*exp(2*a)*Ei(1,-2*b*x)-1/16/x*exp(4*b*x+4*a)-1/4*b*exp(4*a)*Ei(1,-4*b*x)

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maxima [A]  time = 0.44, size = 53, normalized size = 0.60 \[ \frac {1}{4} \, b e^{\left (-4 \, a\right )} \Gamma \left (-1, 4 \, b x\right ) + \frac {1}{4} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x\right ) + \frac {1}{4} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x\right ) + \frac {1}{4} \, b e^{\left (4 \, a\right )} \Gamma \left (-1, -4 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)^3*sinh(b*x+a)/x^2,x, algorithm="maxima")

[Out]

1/4*b*e^(-4*a)*gamma(-1, 4*b*x) + 1/4*b*e^(-2*a)*gamma(-1, 2*b*x) + 1/4*b*e^(2*a)*gamma(-1, -2*b*x) + 1/4*b*e^
(4*a)*gamma(-1, -4*b*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,\mathrm {sinh}\left (a+b\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^3*sinh(a + b*x))/x^2,x)

[Out]

int((cosh(a + b*x)^3*sinh(a + b*x))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**3*sinh(b*x+a)/x**2,x)

[Out]

Integral(sinh(a + b*x)*cosh(a + b*x)**3/x**2, x)

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