3.47 \(\int \frac {(a+b x^2) \cosh (c+d x)}{x^4} \, dx\)

Optimal. Leaf size=105 \[ \frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a \cosh (c+d x)}{3 x^3}-\frac {a d \sinh (c+d x)}{6 x^2}+b d \sinh (c) \text {Chi}(d x)+b d \cosh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5287, 3297, 3303, 3298, 3301} \[ \frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a d \sinh (c+d x)}{6 x^2}-\frac {a \cosh (c+d x)}{3 x^3}+b d \sinh (c) \text {Chi}(d x)+b d \cosh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*Cosh[c + d*x])/x^4,x]

[Out]

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

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 5287

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 95, normalized size = 0.90 \[ -\frac {-d x^3 \sinh (c) \left (a d^2+6 b\right ) \text {Chi}(d x)-d x^3 \cosh (c) \left (a d^2+6 b\right ) \text {Shi}(d x)+a d^2 x^2 \cosh (c+d x)+a d x \sinh (c+d x)+2 a \cosh (c+d x)+6 b x^2 \cosh (c+d x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*Cosh[c + d*x])/x^4,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.51, size = 127, normalized size = 1.21 \[ -\frac {2 \, a d x \sinh \left (d x + c\right ) + 2 \, {\left ({\left (a d^{2} + 6 \, b\right )} x^{2} + 2 \, a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (d x\right ) - {\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \relax (c) - {\left ({\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (d x\right ) + {\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \relax (c)}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.12, size = 170, normalized size = 1.62 \[ -\frac {a d^{3} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x^{3} {\rm Ei}\left (d x\right ) e^{c} + 6 \, b d x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 6 \, b d x^{3} {\rm Ei}\left (d x\right ) e^{c} + a d^{2} x^{2} e^{\left (d x + c\right )} + a d^{2} x^{2} e^{\left (-d x - c\right )} + a d x e^{\left (d x + c\right )} + 6 \, b x^{2} e^{\left (d x + c\right )} - a d x e^{\left (-d x - c\right )} + 6 \, b x^{2} e^{\left (-d x - c\right )} + 2 \, a e^{\left (d x + c\right )} + 2 \, a e^{\left (-d x - c\right )}}{12 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+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 + 6*b*d*x^3*Ei(-d*x)*e^(-c) - 6*b*d*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) + a*d*x*e^(d*x + c) + 6*b*x^2*e^(d*x + c) - a*d*x*e^(-d*x - c)
+ 6*b*x^2*e^(-d*x - c) + 2*a*e^(d*x + c) + 2*a*e^(-d*x - c))/x^3

________________________________________________________________________________________

maple [A]  time = 0.12, size = 172, normalized size = 1.64 \[ -\frac {d^{2} a \,{\mathrm e}^{-d x -c}}{12 x}+\frac {d a \,{\mathrm e}^{-d x -c}}{12 x^{2}}-\frac {a \,{\mathrm e}^{-d x -c}}{6 x^{3}}+\frac {d^{3} a \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{12}-\frac {b \,{\mathrm e}^{-d x -c}}{2 x}+\frac {d b \,{\mathrm e}^{-c} \Ei \left (1, d x \right )}{2}-\frac {a \,{\mathrm e}^{d x +c}}{6 x^{3}}-\frac {d a \,{\mathrm e}^{d x +c}}{12 x^{2}}-\frac {d^{2} a \,{\mathrm e}^{d x +c}}{12 x}-\frac {d^{3} a \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{12}-\frac {b \,{\mathrm e}^{d x +c}}{2 x}-\frac {d b \,{\mathrm e}^{c} \Ei \left (1, -d x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+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)-1/2*b*ex
p(-d*x-c)/x+1/2*d*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)-1/1
2*d^3*a*exp(c)*Ei(1,-d*x)-1/2*b/x*exp(d*x+c)-1/2*d*b*exp(c)*Ei(1,-d*x)

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 73, normalized size = 0.70 \[ \frac {1}{6} \, {\left (a d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - a d^{2} e^{c} \Gamma \left (-2, -d x\right ) - 3 \, b {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 3 \, b {\rm Ei}\left (d x\right ) e^{c}\right )} d - \frac {{\left (3 \, b x^{2} + a\right )} \cosh \left (d x + c\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(a + b*x^2))/x^4,x)

[Out]

int((cosh(c + d*x)*(a + b*x^2))/x^4, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{2}\right ) \cosh {\left (c + d x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*cosh(d*x+c)/x**4,x)

[Out]

Integral((a + b*x**2)*cosh(c + d*x)/x**4, x)

________________________________________________________________________________________