\(\int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx\) [324]

Optimal result

Integrand size = 27, antiderivative size = 78 \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 a (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/6}}{2\ 2^{5/6} d e (a+a \sin (c+d x))^{3/2}} \]

[Out]

-3/4*a*(e*cos(d*x+c))^(4/3)*hypergeom([2/3, 5/6],[5/3],1/2-1/2*sin(d*x+c))*(1+sin(d*x+c))^(5/6)*2^(1/6)/d/e/(a
+a*sin(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2768, 72, 71} \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 a (\sin (c+d x)+1)^{5/6} (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right )}{2\ 2^{5/6} d e (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[(e*Cos[c + d*x])^(1/3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*a*(e*Cos[c + d*x])^(4/3)*Hypergeometric2F1[2/3, 5/6, 5/3, (1 - Sin[c + d*x])/2]*(1 + Sin[c + d*x])^(5/6))/
(2*2^(5/6)*d*e*(a + a*Sin[c + d*x])^(3/2))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 (e \cos (c+d x))^{4/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{a-a x} (a+a x)^{5/6}} \, dx,x,\sin (c+d x)\right )}{d e (a-a \sin (c+d x))^{2/3} (a+a \sin (c+d x))^{2/3}} \\ & = \frac {\left (a^2 (e \cos (c+d x))^{4/3} \left (\frac {a+a \sin (c+d x)}{a}\right )^{5/6}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {1}{2}+\frac {x}{2}\right )^{5/6} \sqrt [3]{a-a x}} \, dx,x,\sin (c+d x)\right )}{2^{5/6} d e (a-a \sin (c+d x))^{2/3} (a+a \sin (c+d x))^{3/2}} \\ & = -\frac {3 a (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/6}}{2\ 2^{5/6} d e (a+a \sin (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right )}{2\ 2^{5/6} d e \sqrt [6]{1+\sin (c+d x)} \sqrt {a (1+\sin (c+d x))}} \]

[In]

Integrate[(e*Cos[c + d*x])^(1/3)/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-3*(e*Cos[c + d*x])^(4/3)*Hypergeometric2F1[2/3, 5/6, 5/3, (1 - Sin[c + d*x])/2])/(2*2^(5/6)*d*e*(1 + Sin[c +
d*x])^(1/6)*Sqrt[a*(1 + Sin[c + d*x])])

Maple [F]

\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{\frac {1}{3}}}{\sqrt {a +a \sin \left (d x +c \right )}}d x\]

[In]

int((e*cos(d*x+c))^(1/3)/(a+a*sin(d*x+c))^(1/2),x)

[Out]

int((e*cos(d*x+c))^(1/3)/(a+a*sin(d*x+c))^(1/2),x)

Fricas [F]

\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(1/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((e*cos(d*x + c))^(1/3)/sqrt(a*sin(d*x + c) + a), x)

Sympy [F]

\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sqrt [3]{e \cos {\left (c + d x \right )}}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

[In]

integrate((e*cos(d*x+c))**(1/3)/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral((e*cos(c + d*x))**(1/3)/sqrt(a*(sin(c + d*x) + 1)), x)

Maxima [F]

\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(1/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(1/3)/sqrt(a*sin(d*x + c) + a), x)

Giac [F]

\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(1/3)/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(1/3)/sqrt(a*sin(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{1/3}}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

[In]

int((e*cos(c + d*x))^(1/3)/(a + a*sin(c + d*x))^(1/2),x)

[Out]

int((e*cos(c + d*x))^(1/3)/(a + a*sin(c + d*x))^(1/2), x)