\(\int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx\) [7]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.13 (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} \[ \int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx=\frac {b^2 \sinh \left (a-\frac {b c}{d}\right ) \text {Chi}\left (\frac {b c}{d}+b x\right )}{2 d^3}+\frac {b^2 \cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (\frac {b c}{d}+b x\right )}{2 d^3}-\frac {b \cosh (a+b x)}{2 d^2 (c+d x)}-\frac {\sinh (a+b x)}{2 d (c+d x)^2} \]

[In]

Int[Sinh[a + b*x]/(c + d*x)^3,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85 \[ \int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx=\frac {b^2 \text {Chi}\left (b \left (\frac {c}{d}+x\right )\right ) \sinh \left (a-\frac {b c}{d}\right )-\frac {d (b (c+d x) \cosh (a+b x)+d \sinh (a+b x))}{(c+d x)^2}+b^2 \cosh \left (a-\frac {b c}{d}\right ) \text {Shi}\left (b \left (\frac {c}{d}+x\right )\right )}{2 d^3} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(276\) vs. \(2(96)=192\).

Time = 0.90 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.66

method	result	size
risch	\(-\frac {b^{3} {\mathrm e}^{-b x -a} x}{4 d \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}-\frac {b^{3} {\mathrm e}^{-b x -a} c}{4 d^{2} \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}+\frac {b^{2} {\mathrm e}^{-b x -a}}{4 d \left (b^{2} d^{2} x^{2}+2 b^{2} c d x +b^{2} c^{2}\right )}+\frac {b^{2} {\mathrm e}^{-\frac {a d -b c}{d}} \operatorname {Ei}_{1}\left (b x +a -\frac {a d -b c}{d}\right )}{4 d^{3}}-\frac {b^{2} {\mathrm e}^{b x +a}}{4 d^{3} \left (\frac {b c}{d}+b x \right )^{2}}-\frac {b^{2} {\mathrm e}^{b x +a}}{4 d^{3} \left (\frac {b c}{d}+b x \right )}-\frac {b^{2} {\mathrm e}^{\frac {a d -b c}{d}} \operatorname {Ei}_{1}\left (-b x -a -\frac {-a d +b c}{d}\right )}{4 d^{3}}\)	\(277\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (96) = 192\).

Time = 0.24 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.44 \[ \int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx=-\frac {2 \, d^{2} \sinh \left (b x + a\right ) + 2 \, {\left (b d^{2} x + b c d\right )} \cosh \left (b x + a\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (-\frac {b d x + b c}{d}\right )\right )} \cosh \left (-\frac {b c - a d}{d}\right ) - {\left ({\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} {\rm Ei}\left (-\frac {b d x + b c}{d}\right )\right )} \sinh \left (-\frac {b c - a d}{d}\right )}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx=-\frac {b {\left (\frac {e^{\left (-a + \frac {b c}{d}\right )} E_{2}\left (\frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )} d} + \frac {e^{\left (a - \frac {b c}{d}\right )} E_{2}\left (-\frac {{\left (d x + c\right )} b}{d}\right )}{{\left (d x + c\right )} d}\right )}}{4 \, d} - \frac {\sinh \left (b x + a\right )}{2 \, {\left (d x + c\right )}^{2} d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (96) = 192\).

Time = 0.27 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.89 \[ \int \frac {\sinh (a+b x)}{(c+d x)^3} \, dx=\frac {b^{2} d^{2} x^{2} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) e^{\left (a - \frac {b c}{d}\right )} - b^{2} d^{2} x^{2} {\rm Ei}\left (-\frac {b d x + b c}{d}\right ) e^{\left (-a + \frac {b c}{d}\right )} + 2 \, b^{2} c d x {\rm Ei}\left (\frac {b d x + b c}{d}\right ) e^{\left (a - \frac {b c}{d}\right )} - 2 \, b^{2} c d x {\rm Ei}\left (-\frac {b d x + b c}{d}\right ) e^{\left (-a + \frac {b c}{d}\right )} + b^{2} c^{2} {\rm Ei}\left (\frac {b d x + b c}{d}\right ) e^{\left (a - \frac {b c}{d}\right )} - b^{2} c^{2} {\rm Ei}\left (-\frac {b d x + b c}{d}\right ) e^{\left (-a + \frac {b c}{d}\right )} - b d^{2} x e^{\left (b x + a\right )} - b d^{2} x e^{\left (-b x - a\right )} - b c d e^{\left (b x + a\right )} - b c d e^{\left (-b x - a\right )} - d^{2} e^{\left (b x + a\right )} + d^{2} e^{\left (-b x - a\right )}}{4 \, {\left (d^{5} x^{2} + 2 \, c d^{4} x + c^{2} d^{3}\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sinh(a + b*x)/(c + d*x)^3,x)

[Out]

int(sinh(a + b*x)/(c + d*x)^3, x)