\(\int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx\) [104]

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

Optimal result

Integrand size = 18, antiderivative size = 123 \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=-\frac {a}{2 d (c+d x)^2}-\frac {a \cosh (e+f x)}{2 d (c+d x)^2}+\frac {a f^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (\frac {c f}{d}+f x\right )}{2 d^3}-\frac {a f \sinh (e+f x)}{2 d^2 (c+d x)}+\frac {a f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (\frac {c f}{d}+f x\right )}{2 d^3} \]

[Out]

-1/2*a/d/(d*x+c)^2+1/2*a*f^2*Chi(c*f/d+f*x)*cosh(-e+c*f/d)/d^3-1/2*a*cosh(f*x+e)/d/(d*x+c)^2-1/2*a*f^2*Shi(c*f
/d+f*x)*sinh(-e+c*f/d)/d^3-1/2*a*f*sinh(f*x+e)/d^2/(d*x+c)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3398, 3378, 3384, 3379, 3382} \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=\frac {a f^2 \text {Chi}\left (x f+\frac {c f}{d}\right ) \cosh \left (e-\frac {c f}{d}\right )}{2 d^3}+\frac {a f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (x f+\frac {c f}{d}\right )}{2 d^3}-\frac {a f \sinh (e+f x)}{2 d^2 (c+d x)}-\frac {a \cosh (e+f x)}{2 d (c+d x)^2}-\frac {a}{2 d (c+d x)^2} \]

[In]

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

[Out]

-1/2*a/(d*(c + d*x)^2) - (a*Cosh[e + f*x])/(2*d*(c + d*x)^2) + (a*f^2*Cosh[e - (c*f)/d]*CoshIntegral[(c*f)/d +
 f*x])/(2*d^3) - (a*f*Sinh[e + f*x])/(2*d^2*(c + d*x)) + (a*f^2*Sinh[e - (c*f)/d]*SinhIntegral[(c*f)/d + f*x])
/(2*d^3)

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 3398

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73 \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=\frac {a \left (f^2 \cosh \left (e-\frac {c f}{d}\right ) \text {Chi}\left (f \left (\frac {c}{d}+x\right )\right )-\frac {d (d+d \cosh (e+f x)+f (c+d x) \sinh (e+f x))}{(c+d x)^2}+f^2 \sinh \left (e-\frac {c f}{d}\right ) \text {Shi}\left (f \left (\frac {c}{d}+x\right )\right )\right )}{2 d^3} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(295\) vs. \(2(115)=230\).

Time = 0.28 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.41

method result size
risch \(-\frac {a}{2 d \left (d x +c \right )^{2}}+\frac {f^{3} a \,{\mathrm e}^{-f x -e} x}{4 d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}+\frac {f^{3} a \,{\mathrm e}^{-f x -e} c}{4 d^{2} \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {f^{2} a \,{\mathrm e}^{-f x -e}}{4 d \left (d^{2} x^{2} f^{2}+2 c d \,f^{2} x +c^{2} f^{2}\right )}-\frac {f^{2} a \,{\mathrm e}^{\frac {c f -d e}{d}} \operatorname {Ei}_{1}\left (f x +e +\frac {c f -d e}{d}\right )}{4 d^{3}}-\frac {f^{2} a \,{\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )^{2}}-\frac {f^{2} a \,{\mathrm e}^{f x +e}}{4 d^{3} \left (\frac {c f}{d}+f x \right )}-\frac {f^{2} a \,{\mathrm e}^{-\frac {c f -d e}{d}} \operatorname {Ei}_{1}\left (-f x -e -\frac {c f -d e}{d}\right )}{4 d^{3}}\) \(296\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (115) = 230\).

Time = 0.27 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.23 \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=-\frac {2 \, a d^{2} \cosh \left (f x + e\right ) + 2 \, a d^{2} - {\left ({\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) + {\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \cosh \left (-\frac {d e - c f}{d}\right ) + 2 \, {\left (a d^{2} f x + a c d f\right )} \sinh \left (f x + e\right ) + {\left ({\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) - {\left (a d^{2} f^{2} x^{2} + 2 \, a c d f^{2} x + a c^{2} f^{2}\right )} {\rm Ei}\left (-\frac {d f x + c f}{d}\right )\right )} \sinh \left (-\frac {d e - c f}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

-1/4*(2*a*d^2*cosh(f*x + e) + 2*a*d^2 - ((a*d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)*Ei((d*f*x + c*f)/d) + (a*
d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)*Ei(-(d*f*x + c*f)/d))*cosh(-(d*e - c*f)/d) + 2*(a*d^2*f*x + a*c*d*f)*
sinh(f*x + e) + ((a*d^2*f^2*x^2 + 2*a*c*d*f^2*x + a*c^2*f^2)*Ei((d*f*x + c*f)/d) - (a*d^2*f^2*x^2 + 2*a*c*d*f^
2*x + a*c^2*f^2)*Ei(-(d*f*x + c*f)/d))*sinh(-(d*e - c*f)/d))/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=-\frac {1}{2} \, a {\left (\frac {e^{\left (-e + \frac {c f}{d}\right )} E_{3}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d} + \frac {e^{\left (e - \frac {c f}{d}\right )} E_{3}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{{\left (d x + c\right )}^{2} d}\right )} - \frac {a}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} \]

[In]

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

[Out]

-1/2*a*(e^(-e + c*f/d)*exp_integral_e(3, (d*x + c)*f/d)/((d*x + c)^2*d) + e^(e - c*f/d)*exp_integral_e(3, -(d*
x + c)*f/d)/((d*x + c)^2*d)) - 1/2*a/(d^3*x^2 + 2*c*d^2*x + c^2*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (115) = 230\).

Time = 0.29 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.57 \[ \int \frac {a+a \cosh (e+f x)}{(c+d x)^3} \, dx=\frac {a d^{2} f^{2} x^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + a d^{2} f^{2} x^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + 2 \, a c d f^{2} x {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + 2 \, a c d f^{2} x {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} + a c^{2} f^{2} {\rm Ei}\left (\frac {d f x + c f}{d}\right ) e^{\left (e - \frac {c f}{d}\right )} + a c^{2} f^{2} {\rm Ei}\left (-\frac {d f x + c f}{d}\right ) e^{\left (-e + \frac {c f}{d}\right )} - a d^{2} f x e^{\left (f x + e\right )} + a d^{2} f x e^{\left (-f x - e\right )} - a c d f e^{\left (f x + e\right )} + a c d f e^{\left (-f x - e\right )} - a d^{2} e^{\left (f x + e\right )} - a d^{2} e^{\left (-f x - e\right )} - 2 \, a d^{2}}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

1/4*(a*d^2*f^2*x^2*Ei((d*f*x + c*f)/d)*e^(e - c*f/d) + a*d^2*f^2*x^2*Ei(-(d*f*x + c*f)/d)*e^(-e + c*f/d) + 2*a
*c*d*f^2*x*Ei((d*f*x + c*f)/d)*e^(e - c*f/d) + 2*a*c*d*f^2*x*Ei(-(d*f*x + c*f)/d)*e^(-e + c*f/d) + a*c^2*f^2*E
i((d*f*x + c*f)/d)*e^(e - c*f/d) + a*c^2*f^2*Ei(-(d*f*x + c*f)/d)*e^(-e + c*f/d) - a*d^2*f*x*e^(f*x + e) + a*d
^2*f*x*e^(-f*x - e) - a*c*d*f*e^(f*x + e) + a*c*d*f*e^(-f*x - e) - a*d^2*e^(f*x + e) - a*d^2*e^(-f*x - e) - 2*
a*d^2)/(d^5*x^2 + 2*c*d^4*x + c^2*d^3)

Mupad [F(-1)]

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

[In]

int((a + a*cosh(e + f*x))/(c + d*x)^3,x)

[Out]

int((a + a*cosh(e + f*x))/(c + d*x)^3, x)