\(\int \frac {(a+b x^3) \cosh (c+d x)}{x^4} \, dx\) [86]

Optimal result

Integrand size = 17, antiderivative size = 91 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {a \cosh (c+d x)}{3 x^3}-\frac {a d^2 \cosh (c+d x)}{6 x}+b \cosh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \text {Chi}(d x) \sinh (c)-\frac {a d \sinh (c+d x)}{6 x^2}+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)+b \sinh (c) \text {Shi}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5395, 3378, 3384, 3379, 3382} \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=\frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a \cosh (c+d x)}{3 x^3}-\frac {a d \sinh (c+d x)}{6 x^2}+b \cosh (c) \text {Chi}(d x)+b \sinh (c) \text {Shi}(d x) \]

[In]

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

[Out]

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

Rule 3378

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 3379

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 3382

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 3384

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 5395

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=\frac {1}{6} \left (\text {Chi}(d x) \left (6 b \cosh (c)+a d^3 \sinh (c)\right )-\frac {a \left (\left (2+d^2 x^2\right ) \cosh (c+d x)+d x \sinh (c+d x)\right )}{x^3}+\left (a d^3 \cosh (c)+6 b \sinh (c)\right ) \text {Shi}(d x)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.60

method	result	size
risch	\(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a \,d^{3} x^{3}-{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a \,d^{3} x^{3}+6 \,{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) b \,x^{3}+{\mathrm e}^{-d x -c} a \,d^{2} x^{2}+6 \,{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) b \,x^{3}+{\mathrm e}^{d x +c} a \,d^{2} x^{2}-{\mathrm e}^{-d x -c} a d x +{\mathrm e}^{d x +c} a d x +2 \,{\mathrm e}^{-d x -c} a +2 a \,{\mathrm e}^{d x +c}}{12 x^{3}}\)	\(146\)
meijerg	\(\frac {b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {2 \gamma +2 \ln \left (x \right )+2 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Chi}\left (d x \right )-2 \ln \left (d x \right )-2 \gamma }{\sqrt {\pi }}\right )}{2}+b \,\operatorname {Shi}\left (d x \right ) \sinh \left (c \right )-\frac {i a \cosh \left (c \right ) \sqrt {\pi }\, d^{3} \left (-\frac {8 i \left (x^{2} d^{2}+2\right ) \cosh \left (d x \right )}{3 d^{3} x^{3} \sqrt {\pi }}-\frac {8 i \sinh \left (d x \right )}{3 x^{2} d^{2} \sqrt {\pi }}+\frac {8 i \operatorname {Shi}\left (d x \right )}{3 \sqrt {\pi }}\right )}{16}-\frac {a \sinh \left (c \right ) \sqrt {\pi }\, d^{3} \left (\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {4 \left (2 \gamma -\frac {11}{3}+2 \ln \left (x \right )+2 \ln \left (i d \right )\right )}{3 \sqrt {\pi }}-\frac {8 \left (\frac {55 x^{2} d^{2}}{2}+45\right )}{45 \sqrt {\pi }\, x^{2} d^{2}}+\frac {8 \cosh \left (d x \right )}{3 \sqrt {\pi }\, x^{2} d^{2}}+\frac {16 \left (\frac {5 x^{2} d^{2}}{2}+5\right ) \sinh \left (d x \right )}{15 \sqrt {\pi }\, x^{3} d^{3}}-\frac {8 \left (\operatorname {Chi}\left (d x \right )-\ln \left (d x \right )-\gamma \right )}{3 \sqrt {\pi }}\right )}{16}\)	\(245\)

[In]

int((b*x^3+a)*cosh(d*x+c)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {2 \, a d x \sinh \left (d x + c\right ) + 2 \, {\left (a d^{2} x^{2} + 2 \, a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{3} + 6 \, b\right )} x^{3} {\rm Ei}\left (d x\right ) - {\left (a d^{3} - 6 \, b\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - {\left ({\left (a d^{3} + 6 \, b\right )} x^{3} {\rm Ei}\left (d x\right ) + {\left (a d^{3} - 6 \, b\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, x^{3}} \]

[In]

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

Sympy [F]

\[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=\int \frac {\left (a + b x^{3}\right ) \cosh {\left (c + d x \right )}}{x^{4}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=\frac {1}{6} \, {\left ({\left (d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - d^{2} e^{c} \Gamma \left (-2, -d x\right )\right )} a - \frac {2 \, b \cosh \left (d x + c\right ) \log \left (x^{3}\right )}{d} + \frac {3 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} b}{d}\right )} d + \frac {1}{3} \, {\left (b \log \left (x^{3}\right ) - \frac {a}{x^{3}}\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.55 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {a d^{3} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x^{3} {\rm Ei}\left (d x\right ) e^{c} + a d^{2} x^{2} e^{\left (d x + c\right )} + a d^{2} x^{2} e^{\left (-d x - c\right )} - 6 \, b x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 6 \, b x^{3} {\rm Ei}\left (d x\right ) e^{c} + a d x e^{\left (d x + c\right )} - a d x e^{\left (-d x - c\right )} + 2 \, a e^{\left (d x + c\right )} + 2 \, a e^{\left (-d x - c\right )}}{12 \, x^{3}} \]

[In]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^4} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^4} \,d x \]

[In]

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

[Out]

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