Integrand size = 33, antiderivative size = 93 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}-\frac {3 (2 A+5 C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b d \sqrt {\sin ^2(c+d x)}} \]
3/5*A*b*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)/b/d/(sin(d*x+c)^2)^ (1/2)
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 b \csc (c+d x) \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{5 d (b \cos (c+d x))^{5/3}} \]
(-3*b*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])*Sq rt[Sin[c + d*x]^2])/(5*d*(b*Cos[c + d*x])^(5/3))
Time = 0.38 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle b^2 \left (\frac {(2 A+5 C) \int \frac {1}{(b \cos (c+d x))^{2/3}}dx}{5 b^2}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (\frac {(2 A+5 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{5 b^2}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b^2 \left (\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}-\frac {3 (2 A+5 C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{5 b^3 d \sqrt {\sin ^2(c+d x)}}\right )\) |
b^2*((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]))
3.2.68.3.1 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]
\[\int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]