\(\int \frac {(a+b x^2) \cosh (c+d x)}{x^5} \, dx\) [48]

Optimal result

Integrand size = 17, antiderivative size = 149 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=-\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d^2 \cosh (c+d x)}{24 x^2}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{24} a d^4 \cosh (c) \text {Chi}(d x)-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}-\frac {a d^3 \sinh (c+d x)}{24 x}+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)+\frac {1}{24} a d^4 \sinh (c) \text {Shi}(d x) \]

[Out]

1/2*b*d^2*Chi(d*x)*cosh(c)+1/24*a*d^4*Chi(d*x)*cosh(c)-1/4*a*cosh(d*x+c)/x^4-1/2*b*cosh(d*x+c)/x^2-1/24*a*d^2*
cosh(d*x+c)/x^2+1/2*b*d^2*Shi(d*x)*sinh(c)+1/24*a*d^4*Shi(d*x)*sinh(c)-1/12*a*d*sinh(d*x+c)/x^3-1/2*b*d*sinh(d
*x+c)/x-1/24*a*d^3*sinh(d*x+c)/x

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5395, 3378, 3384, 3379, 3382} \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=\frac {1}{24} a d^4 \cosh (c) \text {Chi}(d x)+\frac {1}{24} a d^4 \sinh (c) \text {Shi}(d x)-\frac {a d^3 \sinh (c+d x)}{24 x}-\frac {a d^2 \cosh (c+d x)}{24 x^2}-\frac {a \cosh (c+d x)}{4 x^4}-\frac {a d \sinh (c+d x)}{12 x^3}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{2 x^2}-\frac {b d \sinh (c+d x)}{2 x} \]

[In]

Int[((a + b*x^2)*Cosh[c + d*x])/x^5,x]

[Out]

-1/4*(a*Cosh[c + d*x])/x^4 - (b*Cosh[c + d*x])/(2*x^2) - (a*d^2*Cosh[c + d*x])/(24*x^2) + (b*d^2*Cosh[c]*CoshI
ntegral[d*x])/2 + (a*d^4*Cosh[c]*CoshIntegral[d*x])/24 - (a*d*Sinh[c + d*x])/(12*x^3) - (b*d*Sinh[c + d*x])/(2
*x) - (a*d^3*Sinh[c + d*x])/(24*x) + (b*d^2*Sinh[c]*SinhIntegral[d*x])/2 + (a*d^4*Sinh[c]*SinhIntegral[d*x])/2
4

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^5}+\frac {b \cosh (c+d x)}{x^3}\right ) \, dx \\ & = a \int \frac {\cosh (c+d x)}{x^5} \, dx+b \int \frac {\cosh (c+d x)}{x^3} \, dx \\ & = -\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}+\frac {1}{4} (a d) \int \frac {\sinh (c+d x)}{x^4} \, dx+\frac {1}{2} (b d) \int \frac {\sinh (c+d x)}{x^2} \, dx \\ & = -\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}+\frac {1}{12} \left (a d^2\right ) \int \frac {\cosh (c+d x)}{x^3} \, dx+\frac {1}{2} \left (b d^2\right ) \int \frac {\cosh (c+d x)}{x} \, dx \\ & = -\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d^2 \cosh (c+d x)}{24 x^2}-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}+\frac {1}{24} \left (a d^3\right ) \int \frac {\sinh (c+d x)}{x^2} \, 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)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d^2 \cosh (c+d x)}{24 x^2}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}-\frac {a d^3 \sinh (c+d x)}{24 x}+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)+\frac {1}{24} \left (a d^4\right ) \int \frac {\cosh (c+d x)}{x} \, dx \\ & = -\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d^2 \cosh (c+d x)}{24 x^2}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}-\frac {a d^3 \sinh (c+d x)}{24 x}+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)+\frac {1}{24} \left (a d^4 \cosh (c)\right ) \int \frac {\cosh (d x)}{x} \, dx+\frac {1}{24} \left (a d^4 \sinh (c)\right ) \int \frac {\sinh (d x)}{x} \, dx \\ & = -\frac {a \cosh (c+d x)}{4 x^4}-\frac {b \cosh (c+d x)}{2 x^2}-\frac {a d^2 \cosh (c+d x)}{24 x^2}+\frac {1}{2} b d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{24} a d^4 \cosh (c) \text {Chi}(d x)-\frac {a d \sinh (c+d x)}{12 x^3}-\frac {b d \sinh (c+d x)}{2 x}-\frac {a d^3 \sinh (c+d x)}{24 x}+\frac {1}{2} b d^2 \sinh (c) \text {Shi}(d x)+\frac {1}{24} a d^4 \sinh (c) \text {Shi}(d x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=-\frac {6 a \cosh (c+d x)+12 b x^2 \cosh (c+d x)+a d^2 x^2 \cosh (c+d x)-d^2 \left (12 b+a d^2\right ) x^4 \cosh (c) \text {Chi}(d x)+2 a d x \sinh (c+d x)+12 b d x^3 \sinh (c+d x)+a d^3 x^3 \sinh (c+d x)-d^2 \left (12 b+a d^2\right ) x^4 \sinh (c) \text {Shi}(d x)}{24 x^4} \]

[In]

Integrate[((a + b*x^2)*Cosh[c + d*x])/x^5,x]

[Out]

-1/24*(6*a*Cosh[c + d*x] + 12*b*x^2*Cosh[c + d*x] + a*d^2*x^2*Cosh[c + d*x] - d^2*(12*b + a*d^2)*x^4*Cosh[c]*C
oshIntegral[d*x] + 2*a*d*x*Sinh[c + d*x] + 12*b*d*x^3*Sinh[c + d*x] + a*d^3*x^3*Sinh[c + d*x] - d^2*(12*b + a*
d^2)*x^4*Sinh[c]*SinhIntegral[d*x])/x^4

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.61

method	result	size
risch	\(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a \,d^{4} x^{4}+{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a \,d^{4} x^{4}+12 \,{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) b \,d^{2} x^{4}+12 \,{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) b \,d^{2} x^{4}-{\mathrm e}^{-d x -c} a \,d^{3} x^{3}+{\mathrm e}^{d x +c} a \,d^{3} x^{3}+{\mathrm e}^{-d x -c} a \,d^{2} x^{2}-12 \,{\mathrm e}^{-d x -c} b d \,x^{3}+{\mathrm e}^{d x +c} a \,d^{2} x^{2}+12 \,{\mathrm e}^{d x +c} b d \,x^{3}-2 \,{\mathrm e}^{-d x -c} a d x +12 \,{\mathrm e}^{-d x -c} b \,x^{2}+2 \,{\mathrm e}^{d x +c} a d x +12 \,{\mathrm e}^{d x +c} b \,x^{2}+6 \,{\mathrm e}^{-d x -c} a +6 a \,{\mathrm e}^{d x +c}}{48 x^{4}}\)	\(240\)
meijerg	\(-\frac {d^{2} b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+2 \ln \left (i d \right )\right )}{\sqrt {\pi }}-\frac {4 \left (\frac {9 x^{2} d^{2}}{2}+3\right )}{3 \sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \cosh \left (d x \right )}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}-\frac {4 \left (\operatorname {Chi}\left (d x \right )-\ln \left (d x \right )-\gamma \right )}{\sqrt {\pi }}\right )}{8}+\frac {i d^{2} b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}+\frac {4 i \sinh \left (d x \right )}{x^{2} d^{2} \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{8}+\frac {a \cosh \left (c \right ) \sqrt {\pi }\, d^{4} \left (-\frac {8}{\sqrt {\pi }\, x^{4} d^{4}}-\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {\frac {4 \gamma }{3}-\frac {25}{9}+\frac {4 \ln \left (x \right )}{3}+\frac {4 \ln \left (i d \right )}{3}}{\sqrt {\pi }}+\frac {\frac {25}{9} d^{4} x^{4}+8 x^{2} d^{2}+8}{\sqrt {\pi }\, x^{4} d^{4}}-\frac {8 \left (\frac {15 x^{2} d^{2}}{2}+45\right ) \cosh \left (d x \right )}{45 \sqrt {\pi }\, x^{4} d^{4}}-\frac {8 \left (\frac {15 x^{2} d^{2}}{2}+15\right ) \sinh \left (d x \right )}{45 \sqrt {\pi }\, x^{3} d^{3}}+\frac {\frac {4 \,\operatorname {Chi}\left (d x \right )}{3}-\frac {4 \ln \left (d x \right )}{3}-\frac {4 \gamma }{3}}{\sqrt {\pi }}\right )}{32}-\frac {i a \sinh \left (c \right ) \sqrt {\pi }\, d^{4} \left (-\frac {8 i \left (\frac {x^{2} d^{2}}{2}+1\right ) \cosh \left (d x \right )}{3 d^{3} x^{3} \sqrt {\pi }}-\frac {8 i \left (\frac {x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{3 d^{4} x^{4} \sqrt {\pi }}+\frac {4 i \operatorname {Shi}\left (d x \right )}{3 \sqrt {\pi }}\right )}{32}\)	\(399\)

[In]

int((b*x^2+a)*cosh(d*x+c)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/48*(exp(c)*Ei(1,-d*x)*a*d^4*x^4+exp(-c)*Ei(1,d*x)*a*d^4*x^4+12*exp(c)*Ei(1,-d*x)*b*d^2*x^4+12*exp(-c)*Ei(1,
d*x)*b*d^2*x^4-exp(-d*x-c)*a*d^3*x^3+exp(d*x+c)*a*d^3*x^3+exp(-d*x-c)*a*d^2*x^2-12*exp(-d*x-c)*b*d*x^3+exp(d*x
+c)*a*d^2*x^2+12*exp(d*x+c)*b*d*x^3-2*exp(-d*x-c)*a*d*x+12*exp(-d*x-c)*b*x^2+2*exp(d*x+c)*a*d*x+12*exp(d*x+c)*
b*x^2+6*exp(-d*x-c)*a+6*a*exp(d*x+c))/x^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=-\frac {2 \, {\left ({\left (a d^{2} + 12 \, b\right )} x^{2} + 6 \, a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{4} + 12 \, b d^{2}\right )} x^{4} {\rm Ei}\left (d x\right ) + {\left (a d^{4} + 12 \, b d^{2}\right )} x^{4} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + 2 \, {\left ({\left (a d^{3} + 12 \, b d\right )} x^{3} + 2 \, a d x\right )} \sinh \left (d x + c\right ) - {\left ({\left (a d^{4} + 12 \, b d^{2}\right )} x^{4} {\rm Ei}\left (d x\right ) - {\left (a d^{4} + 12 \, b d^{2}\right )} x^{4} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{48 \, x^{4}} \]

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="fricas")

[Out]

-1/48*(2*((a*d^2 + 12*b)*x^2 + 6*a)*cosh(d*x + c) - ((a*d^4 + 12*b*d^2)*x^4*Ei(d*x) + (a*d^4 + 12*b*d^2)*x^4*E
i(-d*x))*cosh(c) + 2*((a*d^3 + 12*b*d)*x^3 + 2*a*d*x)*sinh(d*x + c) - ((a*d^4 + 12*b*d^2)*x^4*Ei(d*x) - (a*d^4
+ 12*b*d^2)*x^4*Ei(-d*x))*sinh(c))/x^4

Sympy [F]

\[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=\int \frac {\left (a + b x^{2}\right ) \cosh {\left (c + d x \right )}}{x^{5}}\, dx \]

[In]

integrate((b*x**2+a)*cosh(d*x+c)/x**5,x)

[Out]

Integral((a + b*x**2)*cosh(c + d*x)/x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=\frac {1}{8} \, {\left (a d^{3} e^{\left (-c\right )} \Gamma \left (-3, d x\right ) + a d^{3} e^{c} \Gamma \left (-3, -d x\right ) + 2 \, b d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + 2 \, b d e^{c} \Gamma \left (-1, -d x\right )\right )} d - \frac {{\left (2 \, b x^{2} + a\right )} \cosh \left (d x + c\right )}{4 \, x^{4}} \]

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="maxima")

[Out]

1/8*(a*d^3*e^(-c)*gamma(-3, d*x) + a*d^3*e^c*gamma(-3, -d*x) + 2*b*d*e^(-c)*gamma(-1, d*x) + 2*b*d*e^c*gamma(-
1, -d*x))*d - 1/4*(2*b*x^2 + a)*cosh(d*x + c)/x^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=\frac {a d^{4} x^{4} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a d^{4} x^{4} {\rm Ei}\left (d x\right ) e^{c} + 12 \, b d^{2} x^{4} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 12 \, b d^{2} x^{4} {\rm Ei}\left (d x\right ) e^{c} - a d^{3} x^{3} e^{\left (d x + c\right )} + a d^{3} x^{3} e^{\left (-d x - c\right )} - a d^{2} x^{2} e^{\left (d x + c\right )} - 12 \, b d x^{3} e^{\left (d x + c\right )} - a d^{2} x^{2} e^{\left (-d x - c\right )} + 12 \, b d x^{3} e^{\left (-d x - c\right )} - 2 \, a d x e^{\left (d x + c\right )} - 12 \, b x^{2} e^{\left (d x + c\right )} + 2 \, a d x e^{\left (-d x - c\right )} - 12 \, b x^{2} e^{\left (-d x - c\right )} - 6 \, a e^{\left (d x + c\right )} - 6 \, a e^{\left (-d x - c\right )}}{48 \, x^{4}} \]

[In]

integrate((b*x^2+a)*cosh(d*x+c)/x^5,x, algorithm="giac")

[Out]

1/48*(a*d^4*x^4*Ei(-d*x)*e^(-c) + a*d^4*x^4*Ei(d*x)*e^c + 12*b*d^2*x^4*Ei(-d*x)*e^(-c) + 12*b*d^2*x^4*Ei(d*x)*
e^c - a*d^3*x^3*e^(d*x + c) + a*d^3*x^3*e^(-d*x - c) - a*d^2*x^2*e^(d*x + c) - 12*b*d*x^3*e^(d*x + c) - a*d^2*
x^2*e^(-d*x - c) + 12*b*d*x^3*e^(-d*x - c) - 2*a*d*x*e^(d*x + c) - 12*b*x^2*e^(d*x + c) + 2*a*d*x*e^(-d*x - c)
- 12*b*x^2*e^(-d*x - c) - 6*a*e^(d*x + c) - 6*a*e^(-d*x - c))/x^4

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^5} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^5} \,d x \]

[In]

int((cosh(c + d*x)*(a + b*x^2))/x^5,x)

[Out]

int((cosh(c + d*x)*(a + b*x^2))/x^5, x)