3.14 \(\int \frac {1}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=103 \[ -\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {6 \cosh (a+b x)}{5 b \sqrt {\sinh (a+b x)}}+\frac {6 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{5 b \sqrt {i \sinh (a+b x)}} \]

[Out]

-2/5*cosh(b*x+a)/b/sinh(b*x+a)^(5/2)+6/5*cosh(b*x+a)/b/sinh(b*x+a)^(1/2)-6/5*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)*EllipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b*x),2^(1/2))*sinh(b*x+a)^(1/2)/b/(I
*sinh(b*x+a))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2636, 2640, 2639} \[ -\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {6 \cosh (a+b x)}{5 b \sqrt {\sinh (a+b x)}}+\frac {6 i \sqrt {\sinh (a+b x)} E\left (\left .\frac {1}{2} \left (i a+i b x-\frac {\pi }{2}\right )\right |2\right )}{5 b \sqrt {i \sinh (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x]^(-7/2),x]

[Out]

(-2*Cosh[a + b*x])/(5*b*Sinh[a + b*x]^(5/2)) + (6*Cosh[a + b*x])/(5*b*Sqrt[Sinh[a + b*x]]) + (((6*I)/5)*Ellipt
icE[(I*a - Pi/2 + I*b*x)/2, 2]*Sqrt[Sinh[a + b*x]])/(b*Sqrt[I*Sinh[a + b*x]])

Rule 2636

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] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sinh ^{\frac {7}{2}}(a+b x)} \, dx &=-\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}-\frac {3}{5} \int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx\\ &=-\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {6 \cosh (a+b x)}{5 b \sqrt {\sinh (a+b x)}}-\frac {3}{5} \int \sqrt {\sinh (a+b x)} \, dx\\ &=-\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {6 \cosh (a+b x)}{5 b \sqrt {\sinh (a+b x)}}-\frac {\left (3 \sqrt {\sinh (a+b x)}\right ) \int \sqrt {i \sinh (a+b x)} \, dx}{5 \sqrt {i \sinh (a+b x)}}\\ &=-\frac {2 \cosh (a+b x)}{5 b \sinh ^{\frac {5}{2}}(a+b x)}+\frac {6 \cosh (a+b x)}{5 b \sqrt {\sinh (a+b x)}}+\frac {6 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b x\right )\right |2\right ) \sqrt {\sinh (a+b x)}}{5 b \sqrt {i \sinh (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 73, normalized size = 0.71 \[ \frac {3 \sinh (2 (a+b x))-2 \coth (a+b x)+6 i (i \sinh (a+b x))^{3/2} E\left (\left .\frac {1}{4} (-2 i a-2 i b x+\pi )\right |2\right )}{5 b \sinh ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x]^(-7/2),x]

[Out]

(-2*Coth[a + b*x] + (6*I)*EllipticE[((-2*I)*a + Pi - (2*I)*b*x)/4, 2]*(I*Sinh[a + b*x])^(3/2) + 3*Sinh[2*(a +
b*x)])/(5*b*Sinh[a + b*x]^(3/2))

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sinh \left (b x + a\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(b*x+a)^(7/2),x, algorithm="fricas")

[Out]

integral(sinh(b*x + a)^(-7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sinh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate(sinh(b*x + a)^(-7/2), x)

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maple [A]  time = 0.07, size = 192, normalized size = 1.86 \[ -\frac {6 \sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (b x +a \right )+i\right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \left (\sinh ^{2}\left (b x +a \right )\right ) \EllipticE \left (\sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (b x +a \right )+i\right )}\, \sqrt {i \sinh \left (b x +a \right )}\, \left (\sinh ^{2}\left (b x +a \right )\right ) \EllipticF \left (\sqrt {-i \left (\sinh \left (b x +a \right )+i\right )}, \frac {\sqrt {2}}{2}\right )-6 \left (\sinh ^{4}\left (b x +a \right )\right )-4 \left (\sinh ^{2}\left (b x +a \right )\right )+2}{5 \sinh \left (b x +a \right )^{\frac {5}{2}} \cosh \left (b x +a \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(b*x+a)^(7/2),x)

[Out]

-1/5/sinh(b*x+a)^(5/2)*(6*(-I*(sinh(b*x+a)+I))^(1/2)*2^(1/2)*(-I*(-sinh(b*x+a)+I))^(1/2)*(I*sinh(b*x+a))^(1/2)
*sinh(b*x+a)^2*EllipticE((-I*(sinh(b*x+a)+I))^(1/2),1/2*2^(1/2))-3*(-I*(sinh(b*x+a)+I))^(1/2)*2^(1/2)*(-I*(-si
nh(b*x+a)+I))^(1/2)*(I*sinh(b*x+a))^(1/2)*sinh(b*x+a)^2*EllipticF((-I*(sinh(b*x+a)+I))^(1/2),1/2*2^(1/2))-6*si
nh(b*x+a)^4-4*sinh(b*x+a)^2+2)/cosh(b*x+a)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sinh \left (b x + a\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(sinh(b*x + a)^(-7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {sinh}\left (a+b\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sinh(a + b*x)^(7/2),x)

[Out]

int(1/sinh(a + b*x)^(7/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sinh ^{\frac {7}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sinh(b*x+a)**(7/2),x)

[Out]

Integral(sinh(a + b*x)**(-7/2), x)

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