\(\int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx\) [548]

Optimal result

Integrand size = 18, antiderivative size = 71 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 x}{b \sqrt {\sinh (a+b x)}}-\frac {4 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{b^2 \sqrt {\sinh (a+b x)}} \]

[Out]

-2*x/b/sinh(b*x+a)^(1/2)+4*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*x)^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*x)*EllipticF(c
os(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*(I*sinh(b*x+a))^(1/2)/b^2/sinh(b*x+a)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5480, 2721, 2720} \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=-\frac {2 x}{b \sqrt {\sinh (a+b x)}}-\frac {4 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{b^2 \sqrt {\sinh (a+b x)}} \]

[In]

Int[(x*Cosh[a + b*x])/Sinh[a + b*x]^(3/2),x]

[Out]

(-2*x)/(b*Sqrt[Sinh[a + b*x]]) - ((4*I)*EllipticF[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b^2*Sqrt[
Sinh[a + b*x]])

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 5480

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[x^(m - n
+ 1)*(Sinh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x}{b \sqrt {\sinh (a+b x)}}+\frac {2 \int \frac {1}{\sqrt {\sinh (a+b x)}} \, dx}{b} \\ & = -\frac {2 x}{b \sqrt {\sinh (a+b x)}}+\frac {\left (2 \sqrt {i \sinh (a+b x)}\right ) \int \frac {1}{\sqrt {i \sinh (a+b x)}} \, dx}{b \sqrt {\sinh (a+b x)}} \\ & = -\frac {2 x}{b \sqrt {\sinh (a+b x)}}-\frac {4 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{b^2 \sqrt {\sinh (a+b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.79 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\frac {-2 b x+4 i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) \sqrt {i \sinh (a+b x)}}{b^2 \sqrt {\sinh (a+b x)}} \]

[In]

Integrate[(x*Cosh[a + b*x])/Sinh[a + b*x]^(3/2),x]

[Out]

(-2*b*x + (4*I)*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*Sqrt[I*Sinh[a + b*x]])/(b^2*Sqrt[Sinh[a + b*x]])

Maple [F]

\[\int \frac {x \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{\frac {3}{2}}}d x\]

[In]

int(x*cosh(b*x+a)/sinh(b*x+a)^(3/2),x)

[Out]

int(x*cosh(b*x+a)/sinh(b*x+a)^(3/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*cosh(b*x+a)/sinh(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {x \cosh {\left (a + b x \right )}}{\sinh ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \]

[In]

integrate(x*cosh(b*x+a)/sinh(b*x+a)**(3/2),x)

[Out]

Integral(x*cosh(a + b*x)/sinh(a + b*x)**(3/2), x)

Maxima [F]

\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x*cosh(b*x+a)/sinh(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x*cosh(b*x + a)/sinh(b*x + a)^(3/2), x)

Giac [F]

\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(x*cosh(b*x+a)/sinh(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x*cosh(b*x + a)/sinh(b*x + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {x\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int((x*cosh(a + b*x))/sinh(a + b*x)^(3/2),x)

[Out]

int((x*cosh(a + b*x))/sinh(a + b*x)^(3/2), x)