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

Optimal. Leaf size=80 \[ \frac {1}{4} b \cosh (a) \text {Chi}(b x)+\frac {3}{4} b \cosh (3 a) \text {Chi}(3 b x)+\frac {1}{4} b \sinh (a) \text {Shi}(b x)+\frac {3}{4} b \sinh (3 a) \text {Shi}(3 b x)-\frac {\sinh (a+b x)}{4 x}-\frac {\sinh (3 a+3 b x)}{4 x} \]

[Out]

1/4*b*Chi(b*x)*cosh(a)+3/4*b*Chi(3*b*x)*cosh(3*a)+1/4*b*Shi(b*x)*sinh(a)+3/4*b*Shi(3*b*x)*sinh(3*a)-1/4*sinh(b
*x+a)/x-1/4*sinh(3*b*x+3*a)/x

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Rubi [A]  time = 0.19, antiderivative size = 80, 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} \[ \frac {1}{4} b \cosh (a) \text {Chi}(b x)+\frac {3}{4} b \cosh (3 a) \text {Chi}(3 b x)+\frac {1}{4} b \sinh (a) \text {Shi}(b x)+\frac {3}{4} b \sinh (3 a) \text {Shi}(3 b x)-\frac {\sinh (a+b x)}{4 x}-\frac {\sinh (3 a+3 b x)}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Cosh[a]*CoshIntegral[b*x])/4 + (3*b*Cosh[3*a]*CoshIntegral[3*b*x])/4 - Sinh[a + b*x]/(4*x) - Sinh[3*a + 3*b
*x]/(4*x) + (b*Sinh[a]*SinhIntegral[b*x])/4 + (3*b*Sinh[3*a]*SinhIntegral[3*b*x])/4

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

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Mathematica [A]  time = 0.16, size = 70, normalized size = 0.88 \[ \frac {b x \cosh (a) \text {Chi}(b x)+3 b x \cosh (3 a) \text {Chi}(3 b x)+b x \sinh (a) \text {Shi}(b x)+3 b x \sinh (3 a) \text {Shi}(3 b x)-\sinh (a+b x)-\sinh (3 (a+b x))}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Cosh[a]*CoshIntegral[b*x] + 3*b*x*Cosh[3*a]*CoshIntegral[3*b*x] - Sinh[a + b*x] - Sinh[3*(a + b*x)] + b*x
*Sinh[a]*SinhIntegral[b*x] + 3*b*x*Sinh[3*a]*SinhIntegral[3*b*x])/(4*x)

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fricas [A]  time = 0.55, size = 124, normalized size = 1.55 \[ -\frac {2 \, \sinh \left (b x + a\right )^{3} - 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) + b x {\rm Ei}\left (-3 \, b x\right )\right )} \cosh \left (3 \, a\right ) - {\left (b x {\rm Ei}\left (b x\right ) + b x {\rm Ei}\left (-b x\right )\right )} \cosh \relax (a) + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right ) - 3 \, {\left (b x {\rm Ei}\left (3 \, b x\right ) - b x {\rm Ei}\left (-3 \, b x\right )\right )} \sinh \left (3 \, a\right ) - {\left (b x {\rm Ei}\left (b x\right ) - b x {\rm Ei}\left (-b x\right )\right )} \sinh \relax (a)}{8 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 90, normalized size = 1.12 \[ \frac {3 \, b x {\rm Ei}\left (3 \, b x\right ) e^{\left (3 \, a\right )} + b x {\rm Ei}\left (-b x\right ) e^{\left (-a\right )} + 3 \, b x {\rm Ei}\left (-3 \, b x\right ) e^{\left (-3 \, a\right )} + b x {\rm Ei}\left (b x\right ) e^{a} - e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )} + e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{8 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b*x*Ei(3*b*x)*e^(3*a) + b*x*Ei(-b*x)*e^(-a) + 3*b*x*Ei(-3*b*x)*e^(-3*a) + b*x*Ei(b*x)*e^a - e^(3*b*x +
3*a) - e^(b*x + a) + e^(-b*x - a) + e^(-3*b*x - 3*a))/x

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maple [A]  time = 0.36, size = 104, normalized size = 1.30 \[ \frac {{\mathrm e}^{-3 b x -3 a}}{8 x}-\frac {3 b \,{\mathrm e}^{-3 a} \Ei \left (1, 3 b x \right )}{8}+\frac {{\mathrm e}^{-b x -a}}{8 x}-\frac {b \,{\mathrm e}^{-a} \Ei \left (1, b x \right )}{8}-\frac {{\mathrm e}^{b x +a}}{8 x}-\frac {b \,{\mathrm e}^{a} \Ei \left (1, -b x \right )}{8}-\frac {{\mathrm e}^{3 b x +3 a}}{8 x}-\frac {3 b \,{\mathrm e}^{3 a} \Ei \left (1, -3 b x \right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*exp(-3*b*x-3*a)/x-3/8*b*exp(-3*a)*Ei(1,3*b*x)+1/8*exp(-b*x-a)/x-1/8*b*exp(-a)*Ei(1,b*x)-1/8/x*exp(b*x+a)-1
/8*b*exp(a)*Ei(1,-b*x)-1/8/x*exp(3*b*x+3*a)-3/8*b*exp(3*a)*Ei(1,-3*b*x)

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maxima [A]  time = 0.54, size = 50, normalized size = 0.62 \[ \frac {3}{8} \, b e^{\left (-3 \, a\right )} \Gamma \left (-1, 3 \, b x\right ) + \frac {1}{8} \, b e^{\left (-a\right )} \Gamma \left (-1, b x\right ) + \frac {1}{8} \, b e^{a} \Gamma \left (-1, -b x\right ) + \frac {3}{8} \, b e^{\left (3 \, a\right )} \Gamma \left (-1, -3 \, b x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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