\(\int \frac {(a+b x^3) \cosh (c+d x)}{x^2} \, dx\) [84]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 17, antiderivative size = 55 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=-\frac {b \cosh (c+d x)}{d^2}-\frac {a \cosh (c+d x)}{x}+a d \text {Chi}(d x) \sinh (c)+\frac {b x \sinh (c+d x)}{d}+a d \cosh (c) \text {Shi}(d x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5395, 3378, 3384, 3379, 3382, 3377, 2718} \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=a d \sinh (c) \text {Chi}(d x)+a d \cosh (c) \text {Shi}(d x)-\frac {a \cosh (c+d x)}{x}-\frac {b \cosh (c+d x)}{d^2}+\frac {b x \sinh (c+d x)}{d} \]

[In]

Int[((a + b*x^3)*Cosh[c + d*x])/x^2,x]

[Out]

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=-\frac {b \cosh (c+d x)}{d^2}-\frac {a \cosh (c+d x)}{x}+a d \text {Chi}(d x) \sinh (c)+\frac {b x \sinh (c+d x)}{d}+a d \cosh (c) \text {Shi}(d x) \]

[In]

Integrate[((a + b*x^3)*Cosh[c + d*x])/x^2,x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(55)=110\).

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 2.07

method result size
risch \(-\frac {{\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right ) a \,d^{3} x -{\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right ) a \,d^{3} x +{\mathrm e}^{-d x -c} b d \,x^{2}-{\mathrm e}^{d x +c} b d \,x^{2}+d^{2} {\mathrm e}^{-d x -c} a +a \,d^{2} {\mathrm e}^{d x +c}+{\mathrm e}^{-d x -c} b x +{\mathrm e}^{d x +c} b x}{2 d^{2} x}\) \(114\)
meijerg \(-\frac {2 b \cosh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {b \sinh \left (c \right ) \left (\cosh \left (d x \right ) x d -\sinh \left (d x \right )\right )}{d^{2}}+\frac {i a \cosh \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{4}+\frac {a \sinh \left (c \right ) \sqrt {\pi }\, d \left (\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Chi}\left (d x \right )-4 \ln \left (d x \right )-4 \gamma }{\sqrt {\pi }}\right )}{4}\) \(167\)

[In]

int((b*x^3+a)*cosh(d*x+c)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.64 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=\frac {2 \, b d x^{2} \sinh \left (d x + c\right ) - 2 \, {\left (a d^{2} + b x\right )} \cosh \left (d x + c\right ) + {\left (a d^{3} x {\rm Ei}\left (d x\right ) - a d^{3} x {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) + {\left (a d^{3} x {\rm Ei}\left (d x\right ) + a d^{3} x {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{2 \, d^{2} x} \]

[In]

integrate((b*x^3+a)*cosh(d*x+c)/x^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=\int \frac {\left (a + b x^{3}\right ) \cosh {\left (c + d x \right )}}{x^{2}}\, dx \]

[In]

integrate((b*x**3+a)*cosh(d*x+c)/x**2,x)

[Out]

Integral((a + b*x**3)*cosh(c + d*x)/x**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=-\frac {1}{4} \, {\left (2 \, a {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - 2 \, a {\rm Ei}\left (d x\right ) e^{c} + \frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} b e^{\left (d x\right )}}{d^{3}} + \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} b e^{\left (-d x - c\right )}}{d^{3}}\right )} d + \frac {1}{2} \, {\left (b x^{2} - \frac {2 \, a}{x}\right )} \cosh \left (d x + c\right ) \]

[In]

integrate((b*x^3+a)*cosh(d*x+c)/x^2,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (55) = 110\).

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=-\frac {a d^{3} x {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - a d^{3} x {\rm Ei}\left (d x\right ) e^{c} - b d x^{2} e^{\left (d x + c\right )} + b d x^{2} e^{\left (-d x - c\right )} + a d^{2} e^{\left (d x + c\right )} + a d^{2} e^{\left (-d x - c\right )} + b x e^{\left (d x + c\right )} + b x e^{\left (-d x - c\right )}}{2 \, d^{2} x} \]

[In]

integrate((b*x^3+a)*cosh(d*x+c)/x^2,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^3\right ) \cosh (c+d x)}{x^2} \, dx=\int \frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (b\,x^3+a\right )}{x^2} \,d x \]

[In]

int((cosh(c + d*x)*(a + b*x^3))/x^2,x)

[Out]

int((cosh(c + d*x)*(a + b*x^3))/x^2, x)