\(\int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx\) [36]

Optimal result

Integrand size = 14, antiderivative size = 104 \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=-\frac {\cosh (c+d x)}{2 b (a+b x)^2}+\frac {d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (\frac {a d}{b}+d x\right )}{2 b^3}-\frac {d \sinh (c+d x)}{2 b^2 (a+b x)}+\frac {d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (\frac {a d}{b}+d x\right )}{2 b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3378, 3384, 3379, 3382} \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=\frac {d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (x d+\frac {a d}{b}\right )}{2 b^3}+\frac {d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (x d+\frac {a d}{b}\right )}{2 b^3}-\frac {d \sinh (c+d x)}{2 b^2 (a+b x)}-\frac {\cosh (c+d x)}{2 b (a+b x)^2} \]

[In]

Int[Cosh[c + d*x]/(a + b*x)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=\frac {d^2 \cosh \left (c-\frac {a d}{b}\right ) \text {Chi}\left (d \left (\frac {a}{b}+x\right )\right )-\frac {b (b \cosh (c+d x)+d (a+b x) \sinh (c+d x))}{(a+b x)^2}+d^2 \sinh \left (c-\frac {a d}{b}\right ) \text {Shi}\left (d \left (\frac {a}{b}+x\right )\right )}{2 b^3} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(275\) vs. \(2(98)=196\).

Time = 0.22 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.65

method	result	size
risch	\(\frac {d^{3} {\mathrm e}^{-d x -c} x}{4 b \left (x^{2} d^{2} b^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}+\frac {d^{3} {\mathrm e}^{-d x -c} a}{4 b^{2} \left (x^{2} d^{2} b^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}-\frac {d^{2} {\mathrm e}^{-d x -c}}{4 b \left (x^{2} d^{2} b^{2}+2 a b \,d^{2} x +a^{2} d^{2}\right )}-\frac {d^{2} {\mathrm e}^{\frac {d a -c b}{b}} \operatorname {Ei}_{1}\left (d x +c +\frac {d a -c b}{b}\right )}{4 b^{3}}-\frac {d^{2} {\mathrm e}^{d x +c}}{4 b^{3} \left (\frac {d a}{b}+d x \right )^{2}}-\frac {d^{2} {\mathrm e}^{d x +c}}{4 b^{3} \left (\frac {d a}{b}+d x \right )}-\frac {d^{2} {\mathrm e}^{-\frac {d a -c b}{b}} \operatorname {Ei}_{1}\left (-d x -c -\frac {d a -c b}{b}\right )}{4 b^{3}}\)	\(276\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (98) = 196\).

Time = 0.24 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.43 \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=-\frac {2 \, b^{2} \cosh \left (d x + c\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \cosh \left (-\frac {b c - a d}{b}\right ) + 2 \, {\left (b^{2} d x + a b d\right )} \sinh \left (d x + c\right ) + {\left ({\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, a b d^{2} x + a^{2} d^{2}\right )} {\rm Ei}\left (-\frac {b d x + a d}{b}\right )\right )} \sinh \left (-\frac {b c - a d}{b}\right )}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

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

[Out]

-1/4*(2*b^2*cosh(d*x + c) - ((b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*Ei((b*d*x + a*d)/b) + (b^2*d^2*x^2 + 2*a*b*
d^2*x + a^2*d^2)*Ei(-(b*d*x + a*d)/b))*cosh(-(b*c - a*d)/b) + 2*(b^2*d*x + a*b*d)*sinh(d*x + c) + ((b^2*d^2*x^
2 + 2*a*b*d^2*x + a^2*d^2)*Ei((b*d*x + a*d)/b) - (b^2*d^2*x^2 + 2*a*b*d^2*x + a^2*d^2)*Ei(-(b*d*x + a*d)/b))*s
inh(-(b*c - a*d)/b))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=\frac {d {\left (\frac {e^{\left (-c + \frac {a d}{b}\right )} E_{2}\left (\frac {{\left (b x + a\right )} d}{b}\right )}{{\left (b x + a\right )} b} - \frac {e^{\left (c - \frac {a d}{b}\right )} E_{2}\left (-\frac {{\left (b x + a\right )} d}{b}\right )}{{\left (b x + a\right )} b}\right )}}{4 \, b} - \frac {\cosh \left (d x + c\right )}{2 \, {\left (b x + a\right )}^{2} b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (98) = 196\).

Time = 0.27 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.87 \[ \int \frac {\cosh (c+d x)}{(a+b x)^3} \, dx=\frac {b^{2} d^{2} x^{2} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + b^{2} d^{2} x^{2} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + 2 \, a b d^{2} x {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + 2 \, a b d^{2} x {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} + a^{2} d^{2} {\rm Ei}\left (\frac {b d x + a d}{b}\right ) e^{\left (c - \frac {a d}{b}\right )} + a^{2} d^{2} {\rm Ei}\left (-\frac {b d x + a d}{b}\right ) e^{\left (-c + \frac {a d}{b}\right )} - b^{2} d x e^{\left (d x + c\right )} + b^{2} d x e^{\left (-d x - c\right )} - a b d e^{\left (d x + c\right )} + a b d e^{\left (-d x - c\right )} - b^{2} e^{\left (d x + c\right )} - b^{2} e^{\left (-d x - c\right )}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(cosh(c + d*x)/(a + b*x)^3,x)

[Out]

int(cosh(c + d*x)/(a + b*x)^3, x)