3.35 \(\int \frac {1}{(b \sinh (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=58 \[ \frac {3 \cosh (c+d x) \sqrt [3]{b \sinh (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\sinh ^2(c+d x)\right )}{b d \sqrt {\cosh ^2(c+d x)}} \]

[Out]

3*cosh(d*x+c)*hypergeom([1/6, 1/2],[7/6],-sinh(d*x+c)^2)*(b*sinh(d*x+c))^(1/3)/b/d/(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} \[ \frac {3 \cosh (c+d x) \sqrt [3]{b \sinh (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\sinh ^2(c+d x)\right )}{b d \sqrt {\cosh ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sinh[c + d*x])^(-2/3),x]

[Out]

(3*Cosh[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, -Sinh[c + d*x]^2]*(b*Sinh[c + d*x])^(1/3))/(b*d*Sqrt[Cosh[c
+ d*x]^2])

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))^{2/3}} \, dx &=\frac {3 \cosh (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\sinh ^2(c+d x)\right ) \sqrt [3]{b \sinh (c+d x)}}{b d \sqrt {\cosh ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 55, normalized size = 0.95 \[ \frac {3 \sqrt {\cosh ^2(c+d x)} \tanh (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};-\sinh ^2(c+d x)\right )}{d (b \sinh (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Sinh[c + d*x])^(-2/3),x]

[Out]

(3*Sqrt[Cosh[c + d*x]^2]*Hypergeometric2F1[1/6, 1/2, 7/6, -Sinh[c + d*x]^2]*Tanh[c + d*x])/(d*(b*Sinh[c + d*x]
)^(2/3))

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (b \sinh \left (d x + c\right )\right )^{\frac {1}{3}}}{b \sinh \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(2/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(d*x+c))^(2/3),x)

[Out]

int(1/(b*sinh(d*x+c))^(2/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh \left (d x + c\right )\right )^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{2/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sinh(c + d*x))^(2/3),x)

[Out]

int(1/(b*sinh(c + d*x))^(2/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sinh {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sinh(d*x+c))**(2/3),x)

[Out]

Integral((b*sinh(c + d*x))**(-2/3), x)

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