Integrand size = 18, antiderivative size = 98 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {4 \cosh (a+b x)}{15 b^2 \sinh ^{\frac {3}{2}}(a+b x)}+\frac {4 i \operatorname {EllipticF}\left (\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right ),2\right ) \sqrt {i \sinh (a+b x)}}{15 b^2 \sqrt {\sinh (a+b x)}} \] Output:
-2/5*x/b/sinh(b*x+a)^(5/2)-4/15*cosh(b*x+a)/b^2/sinh(b*x+a)^(3/2)+4/15*I*I nverseJacobiAM(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)
Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.68 \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=-\frac {2 \left (3 b x-2 i \operatorname {EllipticF}\left (\frac {1}{4} (-2 i a+\pi -2 i b x),2\right ) (i \sinh (a+b x))^{5/2}+\sinh (2 (a+b x))\right )}{15 b^2 \sinh ^{\frac {5}{2}}(a+b x)} \] Input:
Integrate[(x*Cosh[a + b*x])/Sinh[a + b*x]^(7/2),x]
Output:
(-2*(3*b*x - (2*I)*EllipticF[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*(I*Sinh[a + b*x])^(5/2) + Sinh[2*(a + b*x)]))/(15*b^2*Sinh[a + b*x]^(5/2))
Time = 0.51 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5895, 3042, 3116, 3042, 3121, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx\) |
\(\Big \downarrow \) 5895 |
\(\displaystyle \frac {2 \int \frac {1}{\sinh ^{\frac {5}{2}}(a+b x)}dx}{5 b}-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \int \frac {1}{(-i \sin (i a+i b x))^{5/2}}dx}{5 b}\) |
\(\Big \downarrow \) 3116 |
\(\displaystyle \frac {2 \left (-\frac {1}{3} \int \frac {1}{\sqrt {\sinh (a+b x)}}dx-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}\right )}{5 b}-\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \left (-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {1}{3} \int \frac {1}{\sqrt {-i \sin (i a+i b x)}}dx\right )}{5 b}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \left (-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {i \sinh (a+b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\right )}{5 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \left (-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}-\frac {\sqrt {i \sinh (a+b x)} \int \frac {1}{\sqrt {\sin (i a+i b x)}}dx}{3 \sqrt {\sinh (a+b x)}}\right )}{5 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {2 x}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {2 \left (-\frac {2 \cosh (a+b x)}{3 b \sinh ^{\frac {3}{2}}(a+b x)}+\frac {2 i \sqrt {i \sinh (a+b x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right ),2\right )}{3 b \sqrt {\sinh (a+b x)}}\right )}{5 b}\) |
Input:
Int[(x*Cosh[a + b*x])/Sinh[a + b*x]^(7/2),x]
Output:
(2*((-2*Cosh[a + b*x])/(3*b*Sinh[a + b*x]^(3/2)) + (((2*I)/3)*EllipticF[(I *a - Pi/2 + I*b*x)/2, 2]*Sqrt[I*Sinh[a + b*x]])/(b*Sqrt[Sinh[a + b*x]])))/ (5*b) - (2*x)/(5*b*Sinh[a + b*x]^(5/2))
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
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]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
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] - Simp[(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]
\[\int \frac {x \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{\frac {7}{2}}}d x\]
Input:
int(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x)
Output:
int(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x)
Exception generated. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Timed out. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\text {Timed out} \] Input:
integrate(x*cosh(b*x+a)/sinh(b*x+a)**(7/2),x)
Output:
Timed out
\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x, algorithm="maxima")
Output:
integrate(x*cosh(b*x + a)/sinh(b*x + a)^(7/2), x)
\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int { \frac {x \cosh \left (b x + a\right )}{\sinh \left (b x + a\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x, algorithm="giac")
Output:
integrate(x*cosh(b*x + a)/sinh(b*x + a)^(7/2), x)
Timed out. \[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\int \frac {x\,\mathrm {cosh}\left (a+b\,x\right )}{{\mathrm {sinh}\left (a+b\,x\right )}^{7/2}} \,d x \] Input:
int((x*cosh(a + b*x))/sinh(a + b*x)^(7/2),x)
Output:
int((x*cosh(a + b*x))/sinh(a + b*x)^(7/2), x)
\[ \int \frac {x \cosh (a+b x)}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx=\frac {-\frac {2 \sqrt {\sinh \left (b x +a \right )}\, x}{5}+\frac {2 \left (\int \frac {\sqrt {\sinh \left (b x +a \right )}}{\sinh \left (b x +a \right )^{3}}d x \right ) \sinh \left (b x +a \right )^{3}}{5}}{\sinh \left (b x +a \right )^{3} b} \] Input:
int(x*cosh(b*x+a)/sinh(b*x+a)^(7/2),x)
Output:
(2*( - sqrt(sinh(a + b*x))*x + int(sqrt(sinh(a + b*x))/sinh(a + b*x)**3,x) *sinh(a + b*x)**3))/(5*sinh(a + b*x)**3*b)