3.8 \(\int \frac {(a+b x) \cosh (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=132 \[ \frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a \cosh (c+d x)}{3 x^3}-\frac {a d \sinh (c+d x)}{6 x^2}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{2 x^2}-\frac {b d \sinh (c+d x)}{2 x} \]

[Out]

1/2*b*d^2*Chi(d*x)*cosh(c)-1/3*a*cosh(d*x+c)/x^3-1/2*b*cosh(d*x+c)/x^2-1/6*a*d^2*cosh(d*x+c)/x+1/6*a*d^3*cosh(
c)*Shi(d*x)+1/6*a*d^3*Chi(d*x)*sinh(c)+1/2*b*d^2*Shi(d*x)*sinh(c)-1/6*a*d*sinh(d*x+c)/x^2-1/2*b*d*sinh(d*x+c)/
x

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Rubi [A]  time = 0.34, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6742, 3297, 3303, 3298, 3301} \[ \frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a d \sinh (c+d x)}{6 x^2}-\frac {a \cosh (c+d x)}{3 x^3}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{2 x^2}-\frac {b d \sinh (c+d x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*Cosh[c + d*x])/x^4,x]

[Out]

-(a*Cosh[c + d*x])/(3*x^3) - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(6*x) + (b*d^2*Cosh[c]*CoshInte
gral[d*x])/2 + (a*d^3*CoshIntegral[d*x]*Sinh[c])/6 - (a*d*Sinh[c + d*x])/(6*x^2) - (b*d*Sinh[c + d*x])/(2*x) +
(a*d^3*Cosh[c]*SinhIntegral[d*x])/6 + (b*d^2*Sinh[c]*SinhIntegral[d*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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 110, normalized size = 0.83 \[ -\frac {-d^2 x^3 \text {Chi}(d x) (a d \sinh (c)+3 b \cosh (c))-d^2 x^3 \text {Shi}(d x) (a d \cosh (c)+3 b \sinh (c))+a d^2 x^2 \cosh (c+d x)+a d x \sinh (c+d x)+2 a \cosh (c+d x)+3 b d x^2 \sinh (c+d x)+3 b x \cosh (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*Cosh[c + d*x])/x^4,x]

[Out]

-1/6*(2*a*Cosh[c + d*x] + 3*b*x*Cosh[c + d*x] + a*d^2*x^2*Cosh[c + d*x] - d^2*x^3*CoshIntegral[d*x]*(3*b*Cosh[
c] + a*d*Sinh[c]) + a*d*x*Sinh[c + d*x] + 3*b*d*x^2*Sinh[c + d*x] - d^2*x^3*(a*d*Cosh[c] + 3*b*Sinh[c])*SinhIn
tegral[d*x])/x^3

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fricas [A]  time = 0.49, size = 143, normalized size = 1.08 \[ -\frac {2 \, {\left (a d^{2} x^{2} + 3 \, b x + 2 \, a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{3} + 3 \, b d^{2}\right )} x^{3} {\rm Ei}\left (d x\right ) - {\left (a d^{3} - 3 \, b d^{2}\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) + 2 \, {\left (3 \, b d x^{2} + a d x\right )} \sinh \left (d x + c\right ) - {\left ({\left (a d^{3} + 3 \, b d^{2}\right )} x^{3} {\rm Ei}\left (d x\right ) + {\left (a d^{3} - 3 \, b d^{2}\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*(a*d^2*x^2 + 3*b*x + 2*a)*cosh(d*x + c) - ((a*d^3 + 3*b*d^2)*x^3*Ei(d*x) - (a*d^3 - 3*b*d^2)*x^3*Ei(-
d*x))*cosh(c) + 2*(3*b*d*x^2 + a*d*x)*sinh(d*x + c) - ((a*d^3 + 3*b*d^2)*x^3*Ei(d*x) + (a*d^3 - 3*b*d^2)*x^3*E
i(-d*x))*sinh(c))/x^3

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giac [A]  time = 0.15, size = 199, normalized size = 1.51 \[ -\frac {a d^{3} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x^{3} {\rm Ei}\left (d x\right ) e^{c} - 3 \, b d^{2} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 3 \, b d^{2} x^{3} {\rm Ei}\left (d x\right ) e^{c} + a d^{2} x^{2} e^{\left (d x + c\right )} + a d^{2} x^{2} e^{\left (-d x - c\right )} + 3 \, b d x^{2} e^{\left (d x + c\right )} - 3 \, b d x^{2} e^{\left (-d x - c\right )} + a d x e^{\left (d x + c\right )} - a d x e^{\left (-d x - c\right )} + 3 \, b x e^{\left (d x + c\right )} + 3 \, b x e^{\left (-d x - c\right )} + 2 \, a e^{\left (d x + c\right )} + 2 \, a e^{\left (-d x - c\right )}}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(a*d^3*x^3*Ei(-d*x)*e^(-c) - a*d^3*x^3*Ei(d*x)*e^c - 3*b*d^2*x^3*Ei(-d*x)*e^(-c) - 3*b*d^2*x^3*Ei(d*x)*e
^c + a*d^2*x^2*e^(d*x + c) + a*d^2*x^2*e^(-d*x - c) + 3*b*d*x^2*e^(d*x + c) - 3*b*d*x^2*e^(-d*x - c) + a*d*x*e
^(d*x + c) - a*d*x*e^(-d*x - c) + 3*b*x*e^(d*x + c) + 3*b*x*e^(-d*x - c) + 2*a*e^(d*x + c) + 2*a*e^(-d*x - c))
/x^3

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maple [A]  time = 0.12, size = 205, normalized size = 1.55 \[ -\frac {d^{2} a \,{\mathrm e}^{-d x -c}}{12 x}+\frac {d a \,{\mathrm e}^{-d x -c}}{12 x^{2}}-\frac {a \,{\mathrm e}^{-d x -c}}{6 x^{3}}+\frac {d^{3} a \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{12}+\frac {d b \,{\mathrm e}^{-d x -c}}{4 x}-\frac {b \,{\mathrm e}^{-d x -c}}{4 x^{2}}-\frac {d^{2} b \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{4}-\frac {a \,{\mathrm e}^{d x +c}}{6 x^{3}}-\frac {d a \,{\mathrm e}^{d x +c}}{12 x^{2}}-\frac {d^{2} a \,{\mathrm e}^{d x +c}}{12 x}-\frac {d^{3} a \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{12}-\frac {b \,{\mathrm e}^{d x +c}}{4 x^{2}}-\frac {d b \,{\mathrm e}^{d x +c}}{4 x}-\frac {d^{2} b \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*cosh(d*x+c)/x^4,x)

[Out]

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

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maxima [A]  time = 0.41, size = 78, normalized size = 0.59 \[ \frac {1}{12} \, {\left (2 \, a d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - 2 \, a d^{2} e^{c} \Gamma \left (-2, -d x\right ) + 3 \, b d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 3 \, b d e^{c} \Gamma \left (-1, -d x\right )\right )} d - \frac {{\left (3 \, b x + 2 \, a\right )} \cosh \left (d x + c\right )}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*a*d^2*e^(-c)*gamma(-2, d*x) - 2*a*d^2*e^c*gamma(-2, -d*x) + 3*b*d*e^(-c)*gamma(-1, d*x) + 3*b*d*e^c*ga
mma(-1, -d*x))*d - 1/6*(3*b*x + 2*a)*cosh(d*x + c)/x^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(a + b*x))/x^4,x)

[Out]

int((cosh(c + d*x)*(a + b*x))/x^4, x)

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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((b*x+a)*cosh(d*x+c)/x**4,x)

[Out]

Timed out

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