Integrand size = 33, antiderivative size = 92 \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {3 A b^2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}-\frac {3 (2 A+5 C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}} \]
3/5*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(5/3)-3/5*(2*A+5*C)*(b*cos(d*x+c))^( 1/3)*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^ (1/2)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {3 \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right ) \sec ^2(c+d x) \sqrt {\sin ^2(c+d x)}}{5 d} \]
(-3*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(-(A*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]) + 5*C*Cos[c + d*x]^2*Hypergeometric2F1[1/6, 1/2, 7/6 , Cos[c + d*x]^2])*Sec[c + d*x]^2*Sqrt[Sin[c + d*x]^2])/(5*d)
Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3042, 2030, 3491, 3042, 3122}
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.
\(\displaystyle \int \sec ^3(c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^3 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle b^3 \left (\frac {(2 A+5 C) \int \frac {1}{(b \cos (c+d x))^{2/3}}dx}{5 b^2}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^3 \left (\frac {(2 A+5 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{5 b^2}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b^3 \left (\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}-\frac {3 (2 A+5 C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{5 b^3 d \sqrt {\sin ^2(c+d x)}}\right )\) |
b^3*((3*A*Sin[c + d*x])/(5*b*d*(b*Cos[c + d*x])^(5/3)) - (3*(2*A + 5*C)*(b *Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[ c + d*x])/(5*b^3*d*Sqrt[Sin[c + d*x]^2]))
3.2.45.3.1 Defintions of rubi rules used
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{3}\left (d x +c \right )\right )d x\]
\[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
\[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
\[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}}{{\cos \left (c+d\,x\right )}^3} \,d x \]