Integrand size = 31, antiderivative size = 90 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}+\frac {3 (A-2 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]
3/2*A*sin(d*x+c)/d/(b*cos(d*x+c))^(2/3)+3/8*(A-2*C)*(b*cos(d*x+c))^(4/3)*h ypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1 /2)
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \left (-2 A \csc (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )+C \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{2/3}} \]
(-3*(-2*A*Csc[c + d*x]*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2] + C*Cos[c + d*x]*Cot[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x] ^2])*Sqrt[Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))
Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, 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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{5/3}}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle b \left (\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}-\frac {(A-2 C) \int \sqrt [3]{b \cos (c+d x)}dx}{2 b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}-\frac {(A-2 C) \int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b^2}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\frac {3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{8 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
b*((3*A*Sin[c + d*x])/(2*b*d*(b*Cos[c + d*x])^(2/3)) + (3*(A - 2*C)*(b*Cos [c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^3*d*Sqrt[Sin[c + d*x]^2]))
3.2.67.3.1 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]
\[\int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]