\(\int \frac {(a+b x^2) \cosh (c+d x)}{x^4} \, dx\) [47]

Optimal result

Integrand size = 17, antiderivative size = 105 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {a \cosh (c+d x)}{3 x^3}-\frac {b \cosh (c+d x)}{x}-\frac {a d^2 \cosh (c+d x)}{6 x}+b d \text {Chi}(d x) \sinh (c)+\frac {1}{6} a d^3 \text {Chi}(d x) \sinh (c)-\frac {a d \sinh (c+d x)}{6 x^2}+b d \cosh (c) \text {Shi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x) \]

[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] (verified)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5395, 3378, 3384, 3379, 3382} \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=\frac {1}{6} a d^3 \sinh (c) \text {Chi}(d x)+\frac {1}{6} a d^3 \cosh (c) \text {Shi}(d x)-\frac {a d^2 \cosh (c+d x)}{6 x}-\frac {a \cosh (c+d x)}{3 x^3}-\frac {a d \sinh (c+d x)}{6 x^2}+b d \sinh (c) \text {Chi}(d x)+b d \cosh (c) \text {Shi}(d x)-\frac {b \cosh (c+d x)}{x} \]

[In]

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

[Out]

-1/3*(a*Cosh[c + d*x])/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 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^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] (verified)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {2 a \cosh (c+d x)+6 b x^2 \cosh (c+d x)+a d^2 x^2 \cosh (c+d x)-d \left (6 b+a d^2\right ) x^3 \text {Chi}(d x) \sinh (c)+a d x \sinh (c+d x)-d \left (6 b+a d^2\right ) x^3 \cosh (c) \text {Shi}(d x)}{6 x^3} \]

[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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.67

method	result	size
risch	\(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a \,d^{3} x^{3}-{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a \,d^{3} x^{3}+6 \,{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) b d \,x^{3}-6 \,{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) b d \,x^{3}+{\mathrm e}^{-d x -c} a \,d^{2} x^{2}+{\mathrm e}^{d x +c} a \,d^{2} x^{2}-{\mathrm e}^{-d x -c} a d x +6 \,{\mathrm e}^{-d x -c} b \,x^{2}+{\mathrm e}^{d x +c} a d x +6 \,{\mathrm e}^{d x +c} b \,x^{2}+2 \,{\mathrm e}^{-d x -c} a +2 a \,{\mathrm e}^{d x +c}}{12 x^{3}}\)	\(175\)
meijerg	\(\frac {i d b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {d b \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Chi}\left (d x \right )-4 \ln \left (d x \right )-4 \gamma }{\sqrt {\pi }}\right )}{4}-\frac {i a \cosh \left (c \right ) \sqrt {\pi }\, d^{3} \left (-\frac {8 i \left (x^{2} d^{2}+2\right ) \cosh \left (d x \right )}{3 d^{3} x^{3} \sqrt {\pi }}-\frac {8 i \sinh \left (d x \right )}{3 x^{2} d^{2} \sqrt {\pi }}+\frac {8 i \operatorname {Shi}\left (d x \right )}{3 \sqrt {\pi }}\right )}{16}-\frac {a \sinh \left (c \right ) \sqrt {\pi }\, d^{3} \left (\frac {8}{\sqrt {\pi }\, x^{2} d^{2}}-\frac {4 \left (2 \gamma -\frac {11}{3}+2 \ln \left (x \right )+2 \ln \left (i d \right )\right )}{3 \sqrt {\pi }}-\frac {8 \left (\frac {55 x^{2} d^{2}}{2}+45\right )}{45 \sqrt {\pi }\, x^{2} d^{2}}+\frac {8 \cosh \left (d x \right )}{3 \sqrt {\pi }\, x^{2} d^{2}}+\frac {16 \left (\frac {5 x^{2} d^{2}}{2}+5\right ) \sinh \left (d x \right )}{15 \sqrt {\pi }\, x^{3} d^{3}}-\frac {8 \left (\operatorname {Chi}\left (d x \right )-\ln \left (d x \right )-\gamma \right )}{3 \sqrt {\pi }}\right )}{16}\)	\(297\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {2 \, a d x \sinh \left (d x + c\right ) + 2 \, {\left ({\left (a d^{2} + 6 \, b\right )} x^{2} + 2 \, a\right )} \cosh \left (d x + c\right ) - {\left ({\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (d x\right ) - {\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - {\left ({\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (d x\right ) + {\left (a d^{3} + 6 \, b d\right )} x^{3} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{12 \, x^{3}} \]

[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

Sympy [F]

\[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=\int \frac {\left (a + b x^{2}\right ) \cosh {\left (c + d x \right )}}{x^{4}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=\frac {1}{6} \, {\left (a d^{2} e^{\left (-c\right )} \Gamma \left (-2, d x\right ) - a d^{2} e^{c} \Gamma \left (-2, -d x\right ) - 3 \, b {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 3 \, b {\rm Ei}\left (d x\right ) e^{c}\right )} d - \frac {{\left (3 \, b x^{2} + a\right )} \cosh \left (d x + c\right )}{3 \, x^{3}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.62 \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=-\frac {a d^{3} x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x^{3} {\rm Ei}\left (d x\right ) e^{c} + 6 \, b d x^{3} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 6 \, b d x^{3} {\rm Ei}\left (d x\right ) e^{c} + a d^{2} x^{2} e^{\left (d x + c\right )} + a d^{2} x^{2} e^{\left (-d x - c\right )} + a d x e^{\left (d x + c\right )} + 6 \, b x^{2} e^{\left (d x + c\right )} - a d x e^{\left (-d x - c\right )} + 6 \, b x^{2} e^{\left (-d x - c\right )} + 2 \, a e^{\left (d x + c\right )} + 2 \, a e^{\left (-d x - c\right )}}{12 \, x^{3}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right ) \cosh (c+d x)}{x^4} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^2+a\right )}{x^4} \,d x \]

[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)