\(\int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx\) [19]

Optimal result

Integrand size = 17, antiderivative size = 150 \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=\frac {a \cosh (c+d x)}{b^2 d^2}-\frac {2 x \cosh (c+d x)}{b d^2}-\frac {a^3 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (\frac {a d}{b}+d x\right )}{b^4}+\frac {2 \sinh (c+d x)}{b d^3}+\frac {a^2 \sinh (c+d x)}{b^3 d}-\frac {a x \sinh (c+d x)}{b^2 d}+\frac {x^2 \sinh (c+d x)}{b d}-\frac {a^3 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (\frac {a d}{b}+d x\right )}{b^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6874, 2717, 3377, 2718, 3384, 3379, 3382} \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=-\frac {a^3 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{b^4}-\frac {a^3 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{b^4}+\frac {a^2 \sinh (c+d x)}{b^3 d}+\frac {a \cosh (c+d x)}{b^2 d^2}-\frac {a x \sinh (c+d x)}{b^2 d}+\frac {2 \sinh (c+d x)}{b d^3}-\frac {2 x \cosh (c+d x)}{b d^2}+\frac {x^2 \sinh (c+d x)}{b d} \]

[In]

Int[(x^3*Cosh[c + d*x])/(a + b*x),x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, 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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=\frac {-a^3 d^3 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (d \left (\frac {a}{b}+x\right )\right )+b \left (b d (a-2 b x) \cosh (c+d x)+\left (a^2 d^2-a b d^2 x+b^2 \left (2+d^2 x^2\right )\right ) \sinh (c+d x)\right )-a^3 d^3 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (d \left (\frac {a}{b}+x\right )\right )}{b^4 d^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.95

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {d a -c b}{b}} \operatorname {Ei}_{1}\left (-d x -c -\frac {d a -c b}{b}\right ) a^{3}}{2 b^{4}}-\frac {{\mathrm e}^{-d x -c} x^{2}}{2 d b}+\frac {{\mathrm e}^{\frac {d a -c b}{b}} \operatorname {Ei}_{1}\left (d x +c +\frac {d a -c b}{b}\right ) a^{3}}{2 b^{4}}+\frac {{\mathrm e}^{d x +c} x^{2}}{2 d b}+\frac {{\mathrm e}^{-d x -c} a x}{2 d \,b^{2}}-\frac {{\mathrm e}^{d x +c} a x}{2 d \,b^{2}}-\frac {{\mathrm e}^{-d x -c} a^{2}}{2 d \,b^{3}}-\frac {{\mathrm e}^{-d x -c} x}{d^{2} b}+\frac {a^{2} {\mathrm e}^{d x +c}}{2 d \,b^{3}}-\frac {{\mathrm e}^{d x +c} x}{d^{2} b}+\frac {{\mathrm e}^{-d x -c} a}{2 d^{2} b^{2}}+\frac {a \,{\mathrm e}^{d x +c}}{2 d^{2} b^{2}}-\frac {{\mathrm e}^{-d x -c}}{d^{3} b}+\frac {{\mathrm e}^{d x +c}}{d^{3} b}\)	\(292\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.27 \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=-\frac {2 \, {\left (2 \, b^{3} d x - a b^{2} d\right )} \cosh \left (d x + c\right ) + {\left (a^{3} d^{3} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) + a^{3} d^{3} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \cosh \left (-\frac {b c - a d}{b}\right ) - 2 \, {\left (b^{3} d^{2} x^{2} - a b^{2} d^{2} x + a^{2} b d^{2} + 2 \, b^{3}\right )} \sinh \left (d x + c\right ) - {\left (a^{3} d^{3} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) - a^{3} d^{3} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \sinh \left (-\frac {b c - a d}{b}\right )}{2 \, b^{4} d^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (151) = 302\).

Time = 0.26 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.19 \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=\frac {1}{12} \, d {\left (\frac {6 \, a^{3} {\left (\frac {e^{\left (-c + \frac {a d}{b}\right )} E_{1}\left (\frac {{\left (b x + a\right )} d}{b}\right )}{b} + \frac {e^{\left (c - \frac {a d}{b}\right )} E_{1}\left (-\frac {{\left (b x + a\right )} d}{b}\right )}{b}\right )}}{b^{3} d} - \frac {6 \, a^{2} {\left (\frac {{\left (d x e^{c} - e^{c}\right )} e^{\left (d x\right )}}{d^{2}} + \frac {{\left (d x + 1\right )} e^{\left (-d x - c\right )}}{d^{2}}\right )}}{b^{3}} + \frac {3 \, a {\left (\frac {{\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} e^{\left (d x\right )}}{d^{3}} + \frac {{\left (d^{2} x^{2} + 2 \, d x + 2\right )} e^{\left (-d x - c\right )}}{d^{3}}\right )}}{b^{2}} - \frac {2 \, {\left (\frac {{\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} e^{\left (d x\right )}}{d^{4}} + \frac {{\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} e^{\left (-d x - c\right )}}{d^{4}}\right )}}{b} + \frac {12 \, a^{3} \cosh \left (d x + c\right ) \log \left (b x + a\right )}{b^{4} d}\right )} - \frac {1}{6} \, {\left (\frac {6 \, a^{3} \log \left (b x + a\right )}{b^{4}} - \frac {2 \, b^{2} x^{3} - 3 \, a b x^{2} + 6 \, a^{2} x}{b^{3}}\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

1/12*d*(6*a^3*(e^(-c + a*d/b)*exp_integral_e(1, (b*x + a)*d/b)/b + e^(c - a*d/b)*exp_integral_e(1, -(b*x + a)*
d/b)/b)/(b^3*d) - 6*a^2*((d*x*e^c - e^c)*e^(d*x)/d^2 + (d*x + 1)*e^(-d*x - c)/d^2)/b^3 + 3*a*((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 - 2*((d^3*x^3*e^c - 3*d^2*x^2*e^c
+ 6*d*x*e^c - 6*e^c)*e^(d*x)/d^4 + (d^3*x^3 + 3*d^2*x^2 + 6*d*x + 6)*e^(-d*x - c)/d^4)/b + 12*a^3*cosh(d*x + c
)*log(b*x + a)/(b^4*d)) - 1/6*(6*a^3*log(b*x + a)/b^4 - (2*b^2*x^3 - 3*a*b*x^2 + 6*a^2*x)/b^3)*cosh(d*x + c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.71 \[ \int \frac {x^3 \cosh (c+d x)}{a+b x} \, dx=\frac {b^{3} d^{2} x^{2} e^{\left (d x + c\right )} - b^{3} d^{2} x^{2} e^{\left (-d x - c\right )} - a^{3} d^{3} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} - a^{3} d^{3} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} - a b^{2} d^{2} x e^{\left (d x + c\right )} + a b^{2} d^{2} x e^{\left (-d x - c\right )} + a^{2} b d^{2} e^{\left (d x + c\right )} - 2 \, b^{3} d x e^{\left (d x + c\right )} - a^{2} b d^{2} e^{\left (-d x - c\right )} - 2 \, b^{3} d x e^{\left (-d x - c\right )} + a b^{2} d e^{\left (d x + c\right )} + a b^{2} d e^{\left (-d x - c\right )} + 2 \, b^{3} e^{\left (d x + c\right )} - 2 \, b^{3} e^{\left (-d x - c\right )}}{2 \, b^{4} d^{3}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((x^3*cosh(c + d*x))/(a + b*x),x)

[Out]

int((x^3*cosh(c + d*x))/(a + b*x), x)