\(\int \sqrt [3]{b \sinh (c+d x)} \, dx\) [33]

Optimal result

Integrand size = 12, antiderivative size = 60 \[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\frac {3 \cosh (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\sinh ^2(c+d x)\right ) (b \sinh (c+d x))^{4/3}}{4 b d \sqrt {\cosh ^2(c+d x)}} \]

[Out]

3/4*cosh(d*x+c)*hypergeom([1/2, 2/3],[5/3],-sinh(d*x+c)^2)*(b*sinh(d*x+c))^(4/3)/b/d/(cosh(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2722} \[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\frac {3 \cosh (c+d x) (b \sinh (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\sinh ^2(c+d x)\right )}{4 b d \sqrt {\cosh ^2(c+d x)}} \]

[In]

Int[(b*Sinh[c + d*x])^(1/3),x]

[Out]

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

Rule 2722

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
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \cosh (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\sinh ^2(c+d x)\right ) (b \sinh (c+d x))^{4/3}}{4 b d \sqrt {\cosh ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.95 \[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\frac {3 \sqrt {\cosh ^2(c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\sinh ^2(c+d x)\right ) \sqrt [3]{b \sinh (c+d x)} \tanh (c+d x)}{4 d} \]

[In]

Integrate[(b*Sinh[c + d*x])^(1/3),x]

[Out]

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

Maple [F]

\[\int \left (b \sinh \left (d x +c \right )\right )^{\frac {1}{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

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

[Out]

integral((b*sinh(d*x + c))^(1/3), x)

Sympy [F]

\[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\int \sqrt [3]{b \sinh {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral((b*sinh(c + d*x))**(1/3), x)

Maxima [F]

\[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(1/3), x)

Giac [F]

\[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\int { \left (b \sinh \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \]

[In]

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

[Out]

integrate((b*sinh(d*x + c))^(1/3), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{b \sinh (c+d x)} \, dx=\int {\left (b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^{1/3} \,d x \]

[In]

int((b*sinh(c + d*x))^(1/3),x)

[Out]

int((b*sinh(c + d*x))^(1/3), x)