\(\int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx\) [15]

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

Optimal result

Integrand size = 17, antiderivative size = 121 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=-\frac {a^2 \cosh (c+d x)}{2 x^2}-\frac {2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} a^2 d^2 \cosh (c) \text {Chi}(d x)+2 a b d \text {Chi}(d x) \sinh (c)-\frac {a^2 d \sinh (c+d x)}{2 x}+2 a b d \cosh (c) \text {Shi}(d x)+b^2 \sinh (c) \text {Shi}(d x)+\frac {1}{2} a^2 d^2 \sinh (c) \text {Shi}(d x) \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=\frac {1}{2} \left (\text {Chi}(d x) \left (\left (2 b^2+a^2 d^2\right ) \cosh (c)+4 a b d \sinh (c)\right )-\frac {a ((a+4 b x) \cosh (c+d x)+a d x \sinh (c+d x))}{x^2}+\left (4 a b d \cosh (c)+\left (2 b^2+a^2 d^2\right ) \sinh (c)\right ) \text {Shi}(d x)\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {7293, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} a^2 d^2 \cosh (c) \text {Chi}(d x)+\frac {1}{2} a^2 d^2 \sinh (c) \text {Shi}(d x)-\frac {a^2 \cosh (c+d x)}{2 x^2}-\frac {a^2 d \sinh (c+d x)}{2 x}+2 a b d \sinh (c) \text {Chi}(d x)+2 a b d \cosh (c) \text {Shi}(d x)-\frac {2 a b \cosh (c+d x)}{x}+b^2 \cosh (c) \text {Chi}(d x)+b^2 \sinh (c) \text {Shi}(d x)\)

Input:

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

Output:

-1/2*(a^2*Cosh[c + d*x])/x^2 - (2*a*b*Cosh[c + d*x])/x + b^2*Cosh[c]*CoshI 
ntegral[d*x] + (a^2*d^2*Cosh[c]*CoshIntegral[d*x])/2 + 2*a*b*d*CoshIntegra 
l[d*x]*Sinh[c] - (a^2*d*Sinh[c + d*x])/(2*x) + 2*a*b*d*Cosh[c]*SinhIntegra 
l[d*x] + b^2*Sinh[c]*SinhIntegral[d*x] + (a^2*d^2*Sinh[c]*SinhIntegral[d*x 
])/2
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.54

method result size
risch \(-\frac {{\mathrm e}^{c} \operatorname {expIntegral}_{1}\left (-d x \right ) a^{2} d^{2} x^{2}+{\mathrm e}^{-c} \operatorname {expIntegral}_{1}\left (d x \right ) a^{2} d^{2} x^{2}+4 \,{\mathrm e}^{c} \operatorname {expIntegral}_{1}\left (-d x \right ) a b d \,x^{2}-4 \,{\mathrm e}^{-c} \operatorname {expIntegral}_{1}\left (d x \right ) a b d \,x^{2}+2 \,{\mathrm e}^{c} \operatorname {expIntegral}_{1}\left (-d x \right ) b^{2} x^{2}+2 \,{\mathrm e}^{-c} \operatorname {expIntegral}_{1}\left (d x \right ) b^{2} x^{2}+a^{2} d x \,{\mathrm e}^{d x +c}-a^{2} d x \,{\mathrm e}^{-d x -c}+4 \,{\mathrm e}^{d x +c} a b x +4 \,{\mathrm e}^{-d x -c} a b x +{\mathrm e}^{d x +c} a^{2}+{\mathrm e}^{-d x -c} a^{2}}{4 x^{2}}\) \(186\)
meijerg \(\frac {b^{2} \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {2 \,\operatorname {Chi}\left (d x \right )-2 \ln \left (d x \right )-2 \gamma }{\sqrt {\pi }}+\frac {2 \gamma +2 \ln \left (x \right )+2 \ln \left (i d \right )}{\sqrt {\pi }}\right )}{2}+b^{2} \sinh \left (c \right ) \operatorname {Shi}\left (d x \right )+\frac {i d a b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{2}+\frac {d b a \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {4}{\sqrt {\pi }}-\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}+\frac {4 \,\operatorname {Chi}\left (d x \right )-4 \ln \left (d x \right )-4 \gamma }{\sqrt {\pi }}+\frac {4 \gamma -4+4 \ln \left (x \right )+4 \ln \left (i d \right )}{\sqrt {\pi }}\right )}{2}-\frac {a^{2} \cosh \left (c \right ) \sqrt {\pi }\, d^{2} \left (-\frac {4 \left (\frac {9 x^{2} d^{2}}{2}+3\right )}{3 \sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \cosh \left (d x \right )}{\sqrt {\pi }\, x^{2} d^{2}}+\frac {4 \sinh \left (d x \right )}{\sqrt {\pi }\, x d}-\frac {4 \left (\operatorname {Chi}\left (d x \right )-\ln \left (d x \right )-\gamma \right )}{\sqrt {\pi }}-\frac {2 \left (2 \gamma -3+2 \ln \left (x \right )+2 \ln \left (i d \right )\right )}{\sqrt {\pi }}+\frac {4}{\sqrt {\pi }\, x^{2} d^{2}}\right )}{8}+\frac {i a^{2} \sinh \left (c \right ) \sqrt {\pi }\, d^{2} \left (\frac {4 i \cosh \left (d x \right )}{d x \sqrt {\pi }}+\frac {4 i \sinh \left (d x \right )}{x^{2} d^{2} \sqrt {\pi }}-\frac {4 i \operatorname {Shi}\left (d x \right )}{\sqrt {\pi }}\right )}{8}\) \(343\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.26 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=-\frac {2 \, a^{2} d x \sinh \left (d x + c\right ) + 2 \, {\left (4 \, a b x + a^{2}\right )} \cosh \left (d x + c\right ) - {\left ({\left (a^{2} d^{2} + 4 \, a b d + 2 \, b^{2}\right )} x^{2} {\rm Ei}\left (d x\right ) + {\left (a^{2} d^{2} - 4 \, a b d + 2 \, b^{2}\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \cosh \left (c\right ) - {\left ({\left (a^{2} d^{2} + 4 \, a b d + 2 \, b^{2}\right )} x^{2} {\rm Ei}\left (d x\right ) - {\left (a^{2} d^{2} - 4 \, a b d + 2 \, b^{2}\right )} x^{2} {\rm Ei}\left (-d x\right )\right )} \sinh \left (c\right )}{4 \, x^{2}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=\frac {1}{4} \, {\left ({\left (d e^{\left (-c\right )} \Gamma \left (-1, d x\right ) + d e^{c} \Gamma \left (-1, -d x\right )\right )} a^{2} - 4 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} - {\rm Ei}\left (d x\right ) e^{c}\right )} a b - \frac {4 \, b^{2} \cosh \left (d x + c\right ) \log \left (x\right )}{d} + \frac {2 \, {\left ({\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + {\rm Ei}\left (d x\right ) e^{c}\right )} b^{2}}{d}\right )} d + \frac {1}{2} \, {\left (2 \, b^{2} \log \left (x\right ) - \frac {4 \, a b x + a^{2}}{x^{2}}\right )} \cosh \left (d x + c\right ) \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.50 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=\frac {a^{2} d^{2} x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + a^{2} d^{2} x^{2} {\rm Ei}\left (d x\right ) e^{c} - 4 \, a b d x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 4 \, a b d x^{2} {\rm Ei}\left (d x\right ) e^{c} + 2 \, b^{2} x^{2} {\rm Ei}\left (-d x\right ) e^{\left (-c\right )} + 2 \, b^{2} x^{2} {\rm Ei}\left (d x\right ) e^{c} - a^{2} d x e^{\left (d x + c\right )} + a^{2} d x e^{\left (-d x - c\right )} - 4 \, a b x e^{\left (d x + c\right )} - 4 \, a b x e^{\left (-d x - c\right )} - a^{2} e^{\left (d x + c\right )} - a^{2} e^{\left (-d x - c\right )}}{4 \, x^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.64 \[ \int \frac {(a+b x)^2 \cosh (c+d x)}{x^3} \, dx=\frac {e^{d x} \mathit {ei} \left (-d x \right ) a^{2} d^{2} x^{2}-4 e^{d x} \mathit {ei} \left (-d x \right ) a b d \,x^{2}+2 e^{d x} \mathit {ei} \left (-d x \right ) b^{2} x^{2}+e^{d x +2 c} \mathit {ei} \left (d x \right ) a^{2} d^{2} x^{2}+4 e^{d x +2 c} \mathit {ei} \left (d x \right ) a b d \,x^{2}+2 e^{d x +2 c} \mathit {ei} \left (d x \right ) b^{2} x^{2}-e^{2 d x +2 c} a^{2} d x -e^{2 d x +2 c} a^{2}-4 e^{2 d x +2 c} a b x +a^{2} d x -a^{2}-4 a b x}{4 e^{d x +c} x^{2}} \] Input:

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

Output:

(e**(d*x)*ei( - d*x)*a**2*d**2*x**2 - 4*e**(d*x)*ei( - d*x)*a*b*d*x**2 + 2 
*e**(d*x)*ei( - d*x)*b**2*x**2 + e**(2*c + d*x)*ei(d*x)*a**2*d**2*x**2 + 4 
*e**(2*c + d*x)*ei(d*x)*a*b*d*x**2 + 2*e**(2*c + d*x)*ei(d*x)*b**2*x**2 - 
e**(2*c + 2*d*x)*a**2*d*x - e**(2*c + 2*d*x)*a**2 - 4*e**(2*c + 2*d*x)*a*b 
*x + a**2*d*x - a**2 - 4*a*b*x)/(4*e**(c + d*x)*x**2)