3.315 \(\int \frac {\cosh (a+b x) \sinh ^3(a+b x)}{x^4} \, dx\)

Optimal. Leaf size=169 \[ -\frac {1}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)-\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x)+\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}+\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2} \]

[Out]

-1/3*b^3*Chi(2*b*x)*cosh(2*a)+4/3*b^3*Chi(4*b*x)*cosh(4*a)+1/12*b*cosh(2*b*x+2*a)/x^2-1/12*b*cosh(4*b*x+4*a)/x
^2-1/3*b^3*Shi(2*b*x)*sinh(2*a)+4/3*b^3*Shi(4*b*x)*sinh(4*a)+1/12*sinh(2*b*x+2*a)/x^3+1/6*b^2*sinh(2*b*x+2*a)/
x-1/24*sinh(4*b*x+4*a)/x^3-1/3*b^2*sinh(4*b*x+4*a)/x

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Rubi [A]  time = 0.29, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 14, 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} \[ -\frac {1}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)-\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x)+\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}+\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}+\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^4,x]

[Out]

(b*Cosh[2*a + 2*b*x])/(12*x^2) - (b*Cosh[4*a + 4*b*x])/(12*x^2) - (b^3*Cosh[2*a]*CoshIntegral[2*b*x])/3 + (4*b
^3*Cosh[4*a]*CoshIntegral[4*b*x])/3 + Sinh[2*a + 2*b*x]/(12*x^3) + (b^2*Sinh[2*a + 2*b*x])/(6*x) - Sinh[4*a +
4*b*x]/(24*x^3) - (b^2*Sinh[4*a + 4*b*x])/(3*x) - (b^3*Sinh[2*a]*SinhIntegral[2*b*x])/3 + (4*b^3*Sinh[4*a]*Sin
hIntegral[4*b*x])/3

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 (a+b x) \sinh ^3(a+b x)}{x^4} \, dx &=\int \left (-\frac {\sinh (2 a+2 b x)}{4 x^4}+\frac {\sinh (4 a+4 b x)}{8 x^4}\right ) \, dx\\ &=\frac {1}{8} \int \frac {\sinh (4 a+4 b x)}{x^4} \, dx-\frac {1}{4} \int \frac {\sinh (2 a+2 b x)}{x^4} \, dx\\ &=\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {1}{6} b \int \frac {\cosh (2 a+2 b x)}{x^3} \, dx+\frac {1}{6} b \int \frac {\cosh (4 a+4 b x)}{x^3} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}+\frac {\sinh (2 a+2 b x)}{12 x^3}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {1}{6} b^2 \int \frac {\sinh (2 a+2 b x)}{x^2} \, dx+\frac {1}{3} b^2 \int \frac {\sinh (4 a+4 b x)}{x^2} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}+\frac {\sinh (2 a+2 b x)}{12 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}-\frac {1}{3} b^3 \int \frac {\cosh (2 a+2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3\right ) \int \frac {\cosh (4 a+4 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}+\frac {\sinh (2 a+2 b x)}{12 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}-\frac {1}{3} \left (b^3 \cosh (2 a)\right ) \int \frac {\cosh (2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3 \cosh (4 a)\right ) \int \frac {\cosh (4 b x)}{x} \, dx-\frac {1}{3} \left (b^3 \sinh (2 a)\right ) \int \frac {\sinh (2 b x)}{x} \, dx+\frac {1}{3} \left (4 b^3 \sinh (4 a)\right ) \int \frac {\sinh (4 b x)}{x} \, dx\\ &=\frac {b \cosh (2 a+2 b x)}{12 x^2}-\frac {b \cosh (4 a+4 b x)}{12 x^2}-\frac {1}{3} b^3 \cosh (2 a) \text {Chi}(2 b x)+\frac {4}{3} b^3 \cosh (4 a) \text {Chi}(4 b x)+\frac {\sinh (2 a+2 b x)}{12 x^3}+\frac {b^2 \sinh (2 a+2 b x)}{6 x}-\frac {\sinh (4 a+4 b x)}{24 x^3}-\frac {b^2 \sinh (4 a+4 b x)}{3 x}-\frac {1}{3} b^3 \sinh (2 a) \text {Shi}(2 b x)+\frac {4}{3} b^3 \sinh (4 a) \text {Shi}(4 b x)\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 150, normalized size = 0.89 \[ -\frac {8 b^3 x^3 \cosh (2 a) \text {Chi}(2 b x)-32 b^3 x^3 \cosh (4 a) \text {Chi}(4 b x)+8 b^3 x^3 \sinh (2 a) \text {Shi}(2 b x)-32 b^3 x^3 \sinh (4 a) \text {Shi}(4 b x)-4 b^2 x^2 \sinh (2 (a+b x))+8 b^2 x^2 \sinh (4 (a+b x))-2 \sinh (2 (a+b x))+\sinh (4 (a+b x))-2 b x \cosh (2 (a+b x))+2 b x \cosh (4 (a+b x))}{24 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[a + b*x]*Sinh[a + b*x]^3)/x^4,x]

[Out]

-1/24*(-2*b*x*Cosh[2*(a + b*x)] + 2*b*x*Cosh[4*(a + b*x)] + 8*b^3*x^3*Cosh[2*a]*CoshIntegral[2*b*x] - 32*b^3*x
^3*Cosh[4*a]*CoshIntegral[4*b*x] - 2*Sinh[2*(a + b*x)] - 4*b^2*x^2*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)] + 8*b
^2*x^2*Sinh[4*(a + b*x)] + 8*b^3*x^3*Sinh[2*a]*SinhIntegral[2*b*x] - 32*b^3*x^3*Sinh[4*a]*SinhIntegral[4*b*x])
/x^3

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fricas [A]  time = 0.68, size = 264, normalized size = 1.56 \[ -\frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + 2 \, {\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} - b x \cosh \left (b x + a\right )^{2} + {\left (6 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right )^{2} - 8 \, {\left (b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right )\right )} \cosh \left (4 \, a\right ) + 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) + b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + 2 \, {\left ({\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{3} - {\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) - 8 \, {\left (b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right )\right )} \sinh \left (4 \, a\right ) + 2 \, {\left (b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) - b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^4,x, algorithm="fricas")

[Out]

-1/12*(b*x*cosh(b*x + a)^4 + b*x*sinh(b*x + a)^4 + 2*(8*b^2*x^2 + 1)*cosh(b*x + a)*sinh(b*x + a)^3 - b*x*cosh(
b*x + a)^2 + (6*b*x*cosh(b*x + a)^2 - b*x)*sinh(b*x + a)^2 - 8*(b^3*x^3*Ei(4*b*x) + b^3*x^3*Ei(-4*b*x))*cosh(4
*a) + 2*(b^3*x^3*Ei(2*b*x) + b^3*x^3*Ei(-2*b*x))*cosh(2*a) + 2*((8*b^2*x^2 + 1)*cosh(b*x + a)^3 - (2*b^2*x^2 +
1)*cosh(b*x + a))*sinh(b*x + a) - 8*(b^3*x^3*Ei(4*b*x) - b^3*x^3*Ei(-4*b*x))*sinh(4*a) + 2*(b^3*x^3*Ei(2*b*x)
- b^3*x^3*Ei(-2*b*x))*sinh(2*a))/x^3

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giac [A]  time = 0.12, size = 236, normalized size = 1.40 \[ \frac {32 \, b^{3} x^{3} {\rm Ei}\left (4 \, b x\right ) e^{\left (4 \, a\right )} - 8 \, b^{3} x^{3} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 8 \, b^{3} x^{3} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} + 32 \, b^{3} x^{3} {\rm Ei}\left (-4 \, b x\right ) e^{\left (-4 \, a\right )} - 8 \, b^{2} x^{2} e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b^{2} x^{2} e^{\left (2 \, b x + 2 \, a\right )} - 4 \, b^{2} x^{2} e^{\left (-2 \, b x - 2 \, a\right )} + 8 \, b^{2} x^{2} e^{\left (-4 \, b x - 4 \, a\right )} - 2 \, b x e^{\left (4 \, b x + 4 \, a\right )} + 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} + 2 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, b x e^{\left (-4 \, b x - 4 \, a\right )} - e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} - 2 \, e^{\left (-2 \, b x - 2 \, a\right )} + e^{\left (-4 \, b x - 4 \, a\right )}}{48 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^4,x, algorithm="giac")

[Out]

1/48*(32*b^3*x^3*Ei(4*b*x)*e^(4*a) - 8*b^3*x^3*Ei(2*b*x)*e^(2*a) - 8*b^3*x^3*Ei(-2*b*x)*e^(-2*a) + 32*b^3*x^3*
Ei(-4*b*x)*e^(-4*a) - 8*b^2*x^2*e^(4*b*x + 4*a) + 4*b^2*x^2*e^(2*b*x + 2*a) - 4*b^2*x^2*e^(-2*b*x - 2*a) + 8*b
^2*x^2*e^(-4*b*x - 4*a) - 2*b*x*e^(4*b*x + 4*a) + 2*b*x*e^(2*b*x + 2*a) + 2*b*x*e^(-2*b*x - 2*a) - 2*b*x*e^(-4
*b*x - 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^3

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maple [A]  time = 0.63, size = 246, normalized size = 1.46 \[ \frac {b^{2} {\mathrm e}^{-4 b x -4 a}}{6 x}-\frac {b \,{\mathrm e}^{-4 b x -4 a}}{24 x^{2}}+\frac {{\mathrm e}^{-4 b x -4 a}}{48 x^{3}}-\frac {2 b^{3} {\mathrm e}^{-4 a} \Ei \left (1, 4 b x \right )}{3}-\frac {b^{2} {\mathrm e}^{-2 b x -2 a}}{12 x}+\frac {b \,{\mathrm e}^{-2 b x -2 a}}{24 x^{2}}-\frac {{\mathrm e}^{-2 b x -2 a}}{24 x^{3}}+\frac {b^{3} {\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{6}+\frac {{\mathrm e}^{2 b x +2 a}}{24 x^{3}}+\frac {b \,{\mathrm e}^{2 b x +2 a}}{24 x^{2}}+\frac {b^{2} {\mathrm e}^{2 b x +2 a}}{12 x}+\frac {b^{3} {\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{6}-\frac {{\mathrm e}^{4 b x +4 a}}{48 x^{3}}-\frac {b \,{\mathrm e}^{4 b x +4 a}}{24 x^{2}}-\frac {b^{2} {\mathrm e}^{4 b x +4 a}}{6 x}-\frac {2 b^{3} {\mathrm e}^{4 a} \Ei \left (1, -4 b x \right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)^3/x^4,x)

[Out]

1/6*b^2*exp(-4*b*x-4*a)/x-1/24*b*exp(-4*b*x-4*a)/x^2+1/48*exp(-4*b*x-4*a)/x^3-2/3*b^3*exp(-4*a)*Ei(1,4*b*x)-1/
12*b^2*exp(-2*b*x-2*a)/x+1/24*b*exp(-2*b*x-2*a)/x^2-1/24*exp(-2*b*x-2*a)/x^3+1/6*b^3*exp(-2*a)*Ei(1,2*b*x)+1/2
4*exp(2*b*x+2*a)/x^3+1/24*b*exp(2*b*x+2*a)/x^2+1/12*b^2*exp(2*b*x+2*a)/x+1/6*b^3*exp(2*a)*Ei(1,-2*b*x)-1/48/x^
3*exp(4*b*x+4*a)-1/24*b/x^2*exp(4*b*x+4*a)-1/6*b^2/x*exp(4*b*x+4*a)-2/3*b^3*exp(4*a)*Ei(1,-4*b*x)

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maxima [A]  time = 0.44, size = 61, normalized size = 0.36 \[ 4 \, b^{3} e^{\left (-4 \, a\right )} \Gamma \left (-3, 4 \, b x\right ) - b^{3} e^{\left (-2 \, a\right )} \Gamma \left (-3, 2 \, b x\right ) - b^{3} e^{\left (2 \, a\right )} \Gamma \left (-3, -2 \, b x\right ) + 4 \, b^{3} e^{\left (4 \, a\right )} \Gamma \left (-3, -4 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)^3/x^4,x, algorithm="maxima")

[Out]

4*b^3*e^(-4*a)*gamma(-3, 4*b*x) - b^3*e^(-2*a)*gamma(-3, 2*b*x) - b^3*e^(2*a)*gamma(-3, -2*b*x) + 4*b^3*e^(4*a
)*gamma(-3, -4*b*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,{\mathrm {sinh}\left (a+b\,x\right )}^3}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)*sinh(a + b*x)^3)/x^4,x)

[Out]

int((cosh(a + b*x)*sinh(a + b*x)^3)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)**3/x**4,x)

[Out]

Integral(sinh(a + b*x)**3*cosh(a + b*x)/x**4, x)

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