3.36 \(\int \frac {1}{(b \sinh (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=58 \[ -\frac {3 \cosh (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};-\sinh ^2(c+d x)\right )}{b d \sqrt {\cosh ^2(c+d x)} \sqrt [3]{b \sinh (c+d x)}} \]

[Out]

-3*cosh(d*x+c)*hypergeom([-1/6, 1/2],[5/6],-sinh(d*x+c)^2)/b/d/(b*sinh(d*x+c))^(1/3)/(cosh(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2643} \[ -\frac {3 \cosh (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};-\sinh ^2(c+d x)\right )}{b d \sqrt {\cosh ^2(c+d x)} \sqrt [3]{b \sinh (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(-4/3),x]

[Out]

(-3*Cosh[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, -Sinh[c + d*x]^2])/(b*d*Sqrt[Cosh[c + d*x]^2]*(b*Sinh[c +
d*x])^(1/3))

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(b \sinh (c+d x))^{4/3}} \, dx &=-\frac {3 \cosh (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};-\sinh ^2(c+d x)\right )}{b d \sqrt {\cosh ^2(c+d x)} \sqrt [3]{b \sinh (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.95 \[ -\frac {3 \sqrt {\cosh ^2(c+d x)} \tanh (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};-\sinh ^2(c+d x)\right )}{d (b \sinh (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(-4/3),x]

[Out]

(-3*Sqrt[Cosh[c + d*x]^2]*Hypergeometric2F1[-1/6, 1/2, 5/6, -Sinh[c + d*x]^2]*Tanh[c + d*x])/(d*(b*Sinh[c + d*
x])^(4/3))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (b \sinh \left (d x + c\right )\right )^{\frac {2}{3}}}{b^{2} \sinh \left (d x + c\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((b*sinh(d*x + c))^(2/3)/(b^2*sinh(d*x + c)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((b*sinh(d*x + c))^(-4/3), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(d*x+c))^(4/3),x)

[Out]

int(1/(b*sinh(d*x+c))^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((b*sinh(d*x + c))^(-4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(c + d*x))^(4/3),x)

[Out]

int(1/(b*sinh(c + d*x))^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))**(4/3),x)

[Out]

Integral((b*sinh(c + d*x))**(-4/3), x)

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