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

Optimal. Leaf size=119 \[ -\frac {1}{8} b^2 \cosh (a) \text {Chi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Chi}(3 b x)-\frac {1}{8} b^2 \sinh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \sinh (3 a) \text {Shi}(3 b x)+\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}+\frac {b \sinh (a+b x)}{8 x}-\frac {3 b \sinh (3 a+3 b x)}{8 x} \]

[Out]

-1/8*b^2*Chi(b*x)*cosh(a)+9/8*b^2*Chi(3*b*x)*cosh(3*a)+1/8*cosh(b*x+a)/x^2-1/8*cosh(3*b*x+3*a)/x^2-1/8*b^2*Shi
(b*x)*sinh(a)+9/8*b^2*Shi(3*b*x)*sinh(3*a)+1/8*b*sinh(b*x+a)/x-3/8*b*sinh(3*b*x+3*a)/x

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Rubi [A]  time = 0.22, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ -\frac {1}{8} b^2 \cosh (a) \text {Chi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Chi}(3 b x)-\frac {1}{8} b^2 \sinh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \sinh (3 a) \text {Shi}(3 b x)+\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}+\frac {b \sinh (a+b x)}{8 x}-\frac {3 b \sinh (3 a+3 b x)}{8 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cosh[a + b*x]/(8*x^2) - Cosh[3*a + 3*b*x]/(8*x^2) - (b^2*Cosh[a]*CoshIntegral[b*x])/8 + (9*b^2*Cosh[3*a]*CoshI
ntegral[3*b*x])/8 + (b*Sinh[a + b*x])/(8*x) - (3*b*Sinh[3*a + 3*b*x])/(8*x) - (b^2*Sinh[a]*SinhIntegral[b*x])/
8 + (9*b^2*Sinh[3*a]*SinhIntegral[3*b*x])/8

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 ^2(a+b x)}{x^3} \, dx &=\int \left (-\frac {\cosh (a+b x)}{4 x^3}+\frac {\cosh (3 a+3 b x)}{4 x^3}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {\cosh (a+b x)}{x^3} \, dx\right )+\frac {1}{4} \int \frac {\cosh (3 a+3 b x)}{x^3} \, dx\\ &=\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}-\frac {1}{8} b \int \frac {\sinh (a+b x)}{x^2} \, dx+\frac {1}{8} (3 b) \int \frac {\sinh (3 a+3 b x)}{x^2} \, dx\\ &=\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}+\frac {b \sinh (a+b x)}{8 x}-\frac {3 b \sinh (3 a+3 b x)}{8 x}-\frac {1}{8} b^2 \int \frac {\cosh (a+b x)}{x} \, dx+\frac {1}{8} \left (9 b^2\right ) \int \frac {\cosh (3 a+3 b x)}{x} \, dx\\ &=\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}+\frac {b \sinh (a+b x)}{8 x}-\frac {3 b \sinh (3 a+3 b x)}{8 x}-\frac {1}{8} \left (b^2 \cosh (a)\right ) \int \frac {\cosh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^2 \cosh (3 a)\right ) \int \frac {\cosh (3 b x)}{x} \, dx-\frac {1}{8} \left (b^2 \sinh (a)\right ) \int \frac {\sinh (b x)}{x} \, dx+\frac {1}{8} \left (9 b^2 \sinh (3 a)\right ) \int \frac {\sinh (3 b x)}{x} \, dx\\ &=\frac {\cosh (a+b x)}{8 x^2}-\frac {\cosh (3 a+3 b x)}{8 x^2}-\frac {1}{8} b^2 \cosh (a) \text {Chi}(b x)+\frac {9}{8} b^2 \cosh (3 a) \text {Chi}(3 b x)+\frac {b \sinh (a+b x)}{8 x}-\frac {3 b \sinh (3 a+3 b x)}{8 x}-\frac {1}{8} b^2 \sinh (a) \text {Shi}(b x)+\frac {9}{8} b^2 \sinh (3 a) \text {Shi}(3 b x)\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 107, normalized size = 0.90 \[ \frac {-b^2 x^2 \cosh (a) \text {Chi}(b x)+9 b^2 x^2 \cosh (3 a) \text {Chi}(3 b x)-b^2 x^2 \sinh (a) \text {Shi}(b x)+9 b^2 x^2 \sinh (3 a) \text {Shi}(3 b x)+b x \sinh (a+b x)-3 b x \sinh (3 (a+b x))+\cosh (a+b x)-\cosh (3 (a+b x))}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[a + b*x] - Cosh[3*(a + b*x)] - b^2*x^2*Cosh[a]*CoshIntegral[b*x] + 9*b^2*x^2*Cosh[3*a]*CoshIntegral[3*b*
x] + b*x*Sinh[a + b*x] - 3*b*x*Sinh[3*(a + b*x)] - b^2*x^2*Sinh[a]*SinhIntegral[b*x] + 9*b^2*x^2*Sinh[3*a]*Sin
hIntegral[3*b*x])/(8*x^2)

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fricas [A]  time = 0.62, size = 195, normalized size = 1.64 \[ -\frac {6 \, b x \sinh \left (b x + a\right )^{3} + 2 \, \cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) + {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) + b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \cosh \relax (a) + 2 \, {\left (9 \, b x \cosh \left (b x + a\right )^{2} - b x\right )} \sinh \left (b x + a\right ) - 9 \, {\left (b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) + {\left (b^{2} x^{2} {\rm Ei}\left (b x\right ) - b^{2} x^{2} {\rm Ei}\left (-b x\right )\right )} \sinh \relax (a) - 2 \, \cosh \left (b x + a\right )}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*b*x*sinh(b*x + a)^3 + 2*cosh(b*x + a)^3 + 6*cosh(b*x + a)*sinh(b*x + a)^2 - 9*(b^2*x^2*Ei(3*b*x) + b^
2*x^2*Ei(-3*b*x))*cosh(3*a) + (b^2*x^2*Ei(b*x) + b^2*x^2*Ei(-b*x))*cosh(a) + 2*(9*b*x*cosh(b*x + a)^2 - b*x)*s
inh(b*x + a) - 9*(b^2*x^2*Ei(3*b*x) - b^2*x^2*Ei(-3*b*x))*sinh(3*a) + (b^2*x^2*Ei(b*x) - b^2*x^2*Ei(-b*x))*sin
h(a) - 2*cosh(b*x + a))/x^2

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giac [A]  time = 0.14, size = 156, normalized size = 1.31 \[ \frac {9 \, b^{2} x^{2} {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} - b^{2} x^{2} {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 9 \, b^{2} x^{2} {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} - b^{2} x^{2} {\rm Ei}\left (b x\right ) e^{a} - 3 \, b x e^{\left (3 \, b x + 3 \, a\right )} + b x e^{\left (b x + a\right )} - b x e^{\left (-b x - a\right )} + 3 \, b x e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{16 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(9*b^2*x^2*Ei(3*b*x)*e^(3*a) - b^2*x^2*Ei(-b*x)*e^(-a) + 9*b^2*x^2*Ei(-3*b*x)*e^(-3*a) - b^2*x^2*Ei(b*x)*
e^a - 3*b*x*e^(3*b*x + 3*a) + b*x*e^(b*x + a) - b*x*e^(-b*x - a) + 3*b*x*e^(-3*b*x - 3*a) - e^(3*b*x + 3*a) +
e^(b*x + a) + e^(-b*x - a) - e^(-3*b*x - 3*a))/x^2

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maple [A]  time = 0.58, size = 169, normalized size = 1.42 \[ \frac {3 b \,{\mathrm e}^{-3 b x -3 a}}{16 x}-\frac {{\mathrm e}^{-3 b x -3 a}}{16 x^{2}}-\frac {9 b^{2} {\mathrm e}^{-3 a} \Ei \left (1, 3 b x \right )}{16}-\frac {b \,{\mathrm e}^{-b x -a}}{16 x}+\frac {{\mathrm e}^{-b x -a}}{16 x^{2}}+\frac {b^{2} {\mathrm e}^{-a} \Ei \left (1, b x \right )}{16}+\frac {{\mathrm e}^{b x +a}}{16 x^{2}}+\frac {b \,{\mathrm e}^{b x +a}}{16 x}+\frac {b^{2} {\mathrm e}^{a} \Ei \left (1, -b x \right )}{16}-\frac {{\mathrm e}^{3 b x +3 a}}{16 x^{2}}-\frac {3 b \,{\mathrm e}^{3 b x +3 a}}{16 x}-\frac {9 b^{2} {\mathrm e}^{3 a} \Ei \left (1, -3 b x \right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*b*exp(-3*b*x-3*a)/x-1/16*exp(-3*b*x-3*a)/x^2-9/16*b^2*exp(-3*a)*Ei(1,3*b*x)-1/16*b*exp(-b*x-a)/x+1/16*exp
(-b*x-a)/x^2+1/16*b^2*exp(-a)*Ei(1,b*x)+1/16/x^2*exp(b*x+a)+1/16*b/x*exp(b*x+a)+1/16*b^2*exp(a)*Ei(1,-b*x)-1/1
6/x^2*exp(3*b*x+3*a)-3/16*b/x*exp(3*b*x+3*a)-9/16*b^2*exp(3*a)*Ei(1,-3*b*x)

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maxima [A]  time = 0.43, size = 58, normalized size = 0.49 \[ -\frac {9}{8} \, b^{2} e^{\left (-3 \, a\right )} \Gamma \left (-2, 3 \, b x\right ) + \frac {1}{8} \, b^{2} e^{\left (-a\right )} \Gamma \left (-2, b x\right ) + \frac {1}{8} \, b^{2} e^{a} \Gamma \left (-2, -b x\right ) - \frac {9}{8} \, b^{2} e^{\left (3 \, a\right )} \Gamma \left (-2, -3 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/8*b^2*e^(-3*a)*gamma(-2, 3*b*x) + 1/8*b^2*e^(-a)*gamma(-2, b*x) + 1/8*b^2*e^a*gamma(-2, -b*x) - 9/8*b^2*e^(
3*a)*gamma(-2, -3*b*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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