\(\int \frac {(a+b x^2)^2 \cosh (c+d x)}{x} \, dx\) [52]

Optimal result

Integrand size = 19, antiderivative size = 110 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=-\frac {6 b^2 \cosh (c+d x)}{d^4}-\frac {2 a b \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+a^2 \cosh (c) \text {Chi}(d x)+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {2 a b x \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}+a^2 \sinh (c) \text {Shi}(d x) \]

[Out]

a^2*Chi(d*x)*cosh(c)-6*b^2*cosh(d*x+c)/d^4-2*a*b*cosh(d*x+c)/d^2-3*b^2*x^2*cosh(d*x+c)/d^2+a^2*Shi(d*x)*sinh(c
)+6*b^2*x*sinh(d*x+c)/d^3+2*a*b*x*sinh(d*x+c)/d+b^2*x^3*sinh(d*x+c)/d

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {5395, 3384, 3379, 3382, 3377, 2718} \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=a^2 \cosh (c) \text {Chi}(d x)+a^2 \sinh (c) \text {Shi}(d x)-\frac {2 a b \cosh (c+d x)}{d^2}+\frac {2 a b x \sinh (c+d x)}{d}-\frac {6 b^2 \cosh (c+d x)}{d^4}+\frac {6 b^2 x \sinh (c+d x)}{d^3}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {b^2 x^3 \sinh (c+d x)}{d} \]

[In]

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

[Out]

(-6*b^2*Cosh[c + d*x])/d^4 - (2*a*b*Cosh[c + d*x])/d^2 - (3*b^2*x^2*Cosh[c + d*x])/d^2 + a^2*Cosh[c]*CoshInteg
ral[d*x] + (6*b^2*x*Sinh[c + d*x])/d^3 + (2*a*b*x*Sinh[c + d*x])/d + (b^2*x^3*Sinh[c + d*x])/d + a^2*Sinh[c]*S
inhIntegral[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 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^2 \cosh (c+d x)}{x}+2 a b x \cosh (c+d x)+b^2 x^3 \cosh (c+d x)\right ) \, dx \\ & = a^2 \int \frac {\cosh (c+d x)}{x} \, dx+(2 a b) \int x \cosh (c+d x) \, dx+b^2 \int x^3 \cosh (c+d x) \, dx \\ & = \frac {2 a b x \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}-\frac {(2 a b) \int \sinh (c+d x) \, dx}{d}-\frac {\left (3 b^2\right ) \int x^2 \sinh (c+d x) \, dx}{d}+\left (a^2 \cosh (c)\right ) \int \frac {\cosh (d x)}{x} \, dx+\left (a^2 \sinh (c)\right ) \int \frac {\sinh (d x)}{x} \, dx \\ & = -\frac {2 a b \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+a^2 \cosh (c) \text {Chi}(d x)+\frac {2 a b x \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}+a^2 \sinh (c) \text {Shi}(d x)+\frac {\left (6 b^2\right ) \int x \cosh (c+d x) \, dx}{d^2} \\ & = -\frac {2 a b \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+a^2 \cosh (c) \text {Chi}(d x)+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {2 a b x \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}+a^2 \sinh (c) \text {Shi}(d x)-\frac {\left (6 b^2\right ) \int \sinh (c+d x) \, dx}{d^3} \\ & = -\frac {6 b^2 \cosh (c+d x)}{d^4}-\frac {2 a b \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+a^2 \cosh (c) \text {Chi}(d x)+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {2 a b x \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}+a^2 \sinh (c) \text {Shi}(d x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=-\frac {b \left (2 a d^2+3 b \left (2+d^2 x^2\right )\right ) \cosh (c+d x)}{d^4}+a^2 \cosh (c) \text {Chi}(d x)+\frac {b x \left (2 a d^2+b \left (6+d^2 x^2\right )\right ) \sinh (c+d x)}{d^3}+a^2 \sinh (c) \text {Shi}(d x) \]

[In]

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

[Out]

-((b*(2*a*d^2 + 3*b*(2 + d^2*x^2))*Cosh[c + d*x])/d^4) + a^2*Cosh[c]*CoshIntegral[d*x] + (b*x*(2*a*d^2 + b*(6
+ d^2*x^2))*Sinh[c + d*x])/d^3 + a^2*Sinh[c]*SinhIntegral[d*x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(225\) vs. \(2(110)=220\).

Time = 0.22 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.05

method	result	size
risch	\(-\frac {{\mathrm e}^{-d x -c} b^{2} x^{3}}{2 d}+\frac {{\mathrm e}^{d x +c} b^{2} x^{3}}{2 d}-\frac {a^{2} {\mathrm e}^{c} \operatorname {Ei}_{1}\left (-d x \right )}{2}-\frac {a^{2} {\mathrm e}^{-c} \operatorname {Ei}_{1}\left (d x \right )}{2}-\frac {{\mathrm e}^{-d x -c} a b x}{d}-\frac {3 \,{\mathrm e}^{-d x -c} b^{2} x^{2}}{2 d^{2}}+\frac {{\mathrm e}^{d x +c} a b x}{d}-\frac {3 \,{\mathrm e}^{d x +c} b^{2} x^{2}}{2 d^{2}}-\frac {{\mathrm e}^{-d x -c} a b}{d^{2}}-\frac {3 \,{\mathrm e}^{-d x -c} b^{2} x}{d^{3}}-\frac {{\mathrm e}^{d x +c} a b}{d^{2}}+\frac {3 \,{\mathrm e}^{d x +c} b^{2} x}{d^{3}}-\frac {3 \,{\mathrm e}^{-d x -c} b^{2}}{d^{4}}-\frac {3 \,{\mathrm e}^{d x +c} b^{2}}{d^{4}}\)	\(226\)
meijerg	\(\frac {8 b^{2} \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (\frac {3 x^{2} d^{2}}{2}+3\right ) \cosh \left (d x \right )}{4 \sqrt {\pi }}+\frac {d x \left (\frac {x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}-\frac {8 i b^{2} \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \left (\frac {5 x^{2} d^{2}}{2}+15\right ) \cosh \left (d x \right )}{20 \sqrt {\pi }}-\frac {i \left (\frac {15 x^{2} d^{2}}{2}+15\right ) \sinh \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}-\frac {4 b a \cosh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {2 b a \sinh \left (c \right ) \left (\cosh \left (d x \right ) x d -\sinh \left (d x \right )\right )}{d^{2}}+\frac {a^{2} \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {2 \gamma +2 \ln \left (x \right )+2 \ln \left (i d \right )}{\sqrt {\pi }}+\frac {2 \,\operatorname {Chi}\left (d x \right )-2 \ln \left (d x \right )-2 \gamma }{\sqrt {\pi }}\right )}{2}+a^{2} \operatorname {Shi}\left (d x \right ) \sinh \left (c \right )\)	\(238\)

[In]

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

[Out]

-1/2/d*exp(-d*x-c)*b^2*x^3+1/2/d*exp(d*x+c)*b^2*x^3-1/2*a^2*exp(c)*Ei(1,-d*x)-1/2*a^2*exp(-c)*Ei(1,d*x)-1/d*ex
p(-d*x-c)*a*b*x-3/2/d^2*exp(-d*x-c)*b^2*x^2+1/d*exp(d*x+c)*a*b*x-3/2/d^2*exp(d*x+c)*b^2*x^2-1/d^2*exp(-d*x-c)*
a*b-3/d^3*exp(-d*x-c)*b^2*x-1/d^2*exp(d*x+c)*a*b+3/d^3*exp(d*x+c)*b^2*x-3/d^4*exp(-d*x-c)*b^2-3/d^4*exp(d*x+c)
*b^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=-\frac {2 \, {\left (3 \, b^{2} d^{2} x^{2} + 2 \, a b d^{2} + 6 \, b^{2}\right )} \cosh \left (d x + c\right ) - {\left (a^{2} d^{4} {\rm Ei}\left (d x\right ) + a^{2} d^{4} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - 2 \, {\left (b^{2} d^{3} x^{3} + 2 \, {\left (a b d^{3} + 3 \, b^{2} d\right )} x\right )} \sinh \left (d x + c\right ) - {\left (a^{2} d^{4} {\rm Ei}\left (d x\right ) - a^{2} d^{4} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{2 \, d^{4}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.84 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=a^{2} \sinh {\left (c \right )} \operatorname {Shi}{\left (d x \right )} + a^{2} \cosh {\left (c \right )} \operatorname {Chi}\left (d x\right ) + 2 a b \left (\begin {cases} \frac {x \sinh {\left (c + d x \right )}}{d} - \frac {\cosh {\left (c + d x \right )}}{d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{2} \cosh {\left (c \right )}}{2} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \frac {x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\frac {x^{4} \cosh {\left (c \right )}}{4} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*sinh(c)*Shi(d*x) + a**2*cosh(c)*Chi(d*x) + 2*a*b*Piecewise((x*sinh(c + d*x)/d - cosh(c + d*x)/d**2, Ne(d,
0)), (x**2*cosh(c)/2, True)) + b**2*Piecewise((x**3*sinh(c + d*x)/d - 3*x**2*cosh(c + d*x)/d**2 + 6*x*sinh(c
+ d*x)/d**3 - 6*cosh(c + d*x)/d**4, Ne(d, 0)), (x**4*cosh(c)/4, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (110) = 220\).

Time = 0.26 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.14 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=-\frac {1}{8} \, {\left (4 \, a b {\left (\frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} e^{\left (d x\right )}}{d^{3}} + \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} e^{\left (-d x - c\right )}}{d^{3}}\right )} + b^{2} {\left (\frac {{\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} e^{\left (d x\right )}}{d^{5}} + \frac {{\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} e^{\left (-d x - c\right )}}{d^{5}}\right )} + \frac {4 \, a^{2} \cosh \left (d x + c\right ) \log \left (x^{2}\right )}{d} - \frac {4 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} a^{2}}{d}\right )} d + \frac {1}{4} \, {\left (b^{2} x^{4} + 4 \, a b x^{2} + 2 \, a^{2} \log \left (x^{2}\right )\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

-1/8*(4*a*b*((d^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*e^(d*x)/d^3 + (d^2*x^2 + 2*d*x + 2)*e^(-d*x - c)/d^3) + b^2*((d
^4*x^4*e^c - 4*d^3*x^3*e^c + 12*d^2*x^2*e^c - 24*d*x*e^c + 24*e^c)*e^(d*x)/d^5 + (d^4*x^4 + 4*d^3*x^3 + 12*d^2
*x^2 + 24*d*x + 24)*e^(-d*x - c)/d^5) + 4*a^2*cosh(d*x + c)*log(x^2)/d - 4*(Ei(-d*x)*e^(-c) + Ei(d*x)*e^c)*a^2
/d)*d + 1/4*(b^2*x^4 + 4*a*b*x^2 + 2*a^2*log(x^2))*cosh(d*x + c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (110) = 220\).

Time = 0.27 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.02 \[ \int \frac {\left (a+b x^2\right )^2 \cosh (c+d x)}{x} \, dx=\frac {b^{2} d^{3} x^{3} e^{\left (d x + c\right )} - b^{2} d^{3} x^{3} e^{\left (-d x - c\right )} + a^{2} d^{4} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{4} {\rm Ei}\left (d x\right ) e^{c} + 2 \, a b d^{3} x e^{\left (d x + c\right )} - 3 \, b^{2} d^{2} x^{2} e^{\left (d x + c\right )} - 2 \, a b d^{3} x e^{\left (-d x - c\right )} - 3 \, b^{2} d^{2} x^{2} e^{\left (-d x - c\right )} - 2 \, a b d^{2} e^{\left (d x + c\right )} + 6 \, b^{2} d x e^{\left (d x + c\right )} - 2 \, a b d^{2} e^{\left (-d x - c\right )} - 6 \, b^{2} d x e^{\left (-d x - c\right )} - 6 \, b^{2} e^{\left (d x + c\right )} - 6 \, b^{2} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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