\(\int \sinh ^2(\frac {a}{c+d x}) \, dx\) [290]

Optimal result

Integrand size = 12, antiderivative size = 39 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}-\frac {a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \]

[Out]

-a*Shi(2*a/(d*x+c))/d+(d*x+c)*sinh(a/(d*x+c))^2/d

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5418, 5410, 3394, 12, 3379} \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}-\frac {a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \]

[In]

Int[Sinh[a/(c + d*x)]^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3394

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]^
n/(d*(m + 1))), x] - Dist[f*(n/(d*(m + 1))), Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]
^(n - 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] && LtQ[m, -1]

Rule 5410

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^2
, x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && ILtQ[n, 0] && IntegerQ[p]

Rule 5418

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(u_)^(n_)])^(p_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1], Subst[Int[(
a + b*Sinh[c + d*x^n])^p, x], x, u], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[p] && LinearQ[u, x] && NeQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \sinh ^2\left (\frac {a}{x}\right ) \, dx,x,c+d x\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {\sinh ^2(a x)}{x^2} \, dx,x,\frac {1}{c+d x}\right )}{d} \\ & = \frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}+\frac {(2 i a) \text {Subst}\left (\int \frac {i \sinh (2 a x)}{2 x} \, dx,x,\frac {1}{c+d x}\right )}{d} \\ & = \frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}-\frac {a \text {Subst}\left (\int \frac {\sinh (2 a x)}{x} \, dx,x,\frac {1}{c+d x}\right )}{d} \\ & = \frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )}{d}-\frac {a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.95 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {(c+d x) \sinh ^2\left (\frac {a}{c+d x}\right )-a \text {Shi}\left (\frac {2 a}{c+d x}\right )}{d} \]

[In]

Integrate[Sinh[a/(c + d*x)]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.28

method	result	size
derivativedivides	\(-\frac {a \left (\frac {d x +c}{2 a}-\frac {\left (d x +c \right ) \cosh \left (\frac {2 a}{d x +c}\right )}{2 a}+\operatorname {Shi}\left (\frac {2 a}{d x +c}\right )\right )}{d}\)	\(50\)
default	\(-\frac {a \left (\frac {d x +c}{2 a}-\frac {\left (d x +c \right ) \cosh \left (\frac {2 a}{d x +c}\right )}{2 a}+\operatorname {Shi}\left (\frac {2 a}{d x +c}\right )\right )}{d}\)	\(50\)
risch	\(-\frac {x}{2}+\frac {{\mathrm e}^{-\frac {2 a}{d x +c}} x}{4}+\frac {{\mathrm e}^{-\frac {2 a}{d x +c}} c}{4 d}-\frac {a \,\operatorname {Ei}_{1}\left (\frac {2 a}{d x +c}\right )}{2 d}+\frac {{\mathrm e}^{\frac {2 a}{d x +c}} x}{4}+\frac {{\mathrm e}^{\frac {2 a}{d x +c}} c}{4 d}+\frac {a \,\operatorname {Ei}_{1}\left (-\frac {2 a}{d x +c}\right )}{2 d}\)	\(103\)

[In]

int(sinh(1/(d*x+c)*a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/d*a*(1/2*(d*x+c)/a-1/2*(d*x+c)/a*cosh(2/(d*x+c)*a)+Shi(2/(d*x+c)*a))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.87 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\frac {{\left (d x + c\right )} \cosh \left (\frac {a}{d x + c}\right )^{2} + {\left (d x + c\right )} \sinh \left (\frac {a}{d x + c}\right )^{2} - d x - a {\rm Ei}\left (\frac {2 \, a}{d x + c}\right ) + a {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\int \sinh ^{2}{\left (\frac {a}{c + d x} \right )}\, dx \]

[In]

integrate(sinh(a/(d*x+c))**2,x)

[Out]

Integral(sinh(a/(c + d*x))**2, x)

Maxima [F]

\[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=\int { \sinh \left (\frac {a}{d x + c}\right )^{2} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (39) = 78\).

Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.49 \[ \int \sinh ^2\left (\frac {a}{c+d x}\right ) \, dx=-\frac {{\left (\frac {2 \, a^{3} {\rm Ei}\left (\frac {2 \, a}{d x + c}\right )}{d x + c} - \frac {2 \, a^{3} {\rm Ei}\left (-\frac {2 \, a}{d x + c}\right )}{d x + c} - a^{2} e^{\left (\frac {2 \, a}{d x + c}\right )} - a^{2} e^{\left (-\frac {2 \, a}{d x + c}\right )} + 2 \, a^{2}\right )} {\left (d x + c\right )}}{4 \, a^{2} d} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(sinh(a/(c + d*x))^2,x)

[Out]

int(sinh(a/(c + d*x))^2, x)