Integrand size = 27, antiderivative size = 78 \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 a (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{5/6}}{2\ 2^{5/6} d e (a+a \sin (c+d x))^{3/2}} \]
-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)
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=-\frac {3 (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right )}{2\ 2^{5/6} d e \sqrt [6]{1+\sin (c+d x)} \sqrt {a (1+\sin (c+d x))}} \]
(-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]) ])
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3168, 80, 27, 79}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a \sin (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a \sin (c+d x)+a}}dx\) |
\(\Big \downarrow \) 3168 |
\(\displaystyle \frac {a^2 (e \cos (c+d x))^{4/3} \int \frac {1}{\sqrt [3]{a-a \sin (c+d x)} (\sin (c+d x) a+a)^{5/6}}d\sin (c+d x)}{d e (a-a \sin (c+d x))^{2/3} (a \sin (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a^2 (\sin (c+d x)+1)^{5/6} (e \cos (c+d x))^{4/3} \int \frac {2^{5/6}}{(\sin (c+d x)+1)^{5/6} \sqrt [3]{a-a \sin (c+d x)}}d\sin (c+d x)}{2^{5/6} d e (a-a \sin (c+d x))^{2/3} (a \sin (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 (\sin (c+d x)+1)^{5/6} (e \cos (c+d x))^{4/3} \int \frac {1}{(\sin (c+d x)+1)^{5/6} \sqrt [3]{a-a \sin (c+d x)}}d\sin (c+d x)}{d e (a-a \sin (c+d x))^{2/3} (a \sin (c+d x)+a)^{3/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {3 a (\sin (c+d x)+1)^{5/6} (e \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},\frac {1}{2} (1-\sin (c+d x))\right )}{2\ 2^{5/6} d e (a \sin (c+d x)+a)^{3/2}}\) |
(-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))
3.4.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(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] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.), x_Symbol] :> Simp[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] /; Fre eQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{\frac {1}{3}}}{\sqrt {a +a \sin \left (d x +c \right )}}d x\]
\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {\sqrt [3]{e \cos {\left (c + d x \right )}}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sin \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^{1/3}}{\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]