Integrand size = 33, antiderivative size = 92 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{16 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]
3/8*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(8/3)+3/16*(5*A+8*C)*hypergeom([-1/3 , 1/2],[2/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x+c))^(2/3)/(sin(d*x+c)^2 )^(1/2)
Time = 0.32 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.99 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 \left (4 C \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)+A \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},-\frac {1}{3},\cos ^2(c+d x)\right ) \sec (c+d x) \tan (c+d x)\right )}{8 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}} \]
(3*(4*C*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x] + A *Hypergeometric2F1[-4/3, 1/2, -1/3, Cos[c + d*x]^2]*Sec[c + d*x]*Tan[c + d *x]))/(8*d*(b*Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2])
Time = 0.39 (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.
\(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^3 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{11/3}}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle b^3 \left (\frac {(5 A+8 C) \int \frac {1}{(b \cos (c+d x))^{5/3}}dx}{8 b^2}+\frac {3 A \sin (c+d x)}{8 b d (b \cos (c+d x))^{8/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^3 \left (\frac {(5 A+8 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{8 b^2}+\frac {3 A \sin (c+d x)}{8 b d (b \cos (c+d x))^{8/3}}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b^3 \left (\frac {3 (5 A+8 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{16 b^3 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}+\frac {3 A \sin (c+d x)}{8 b d (b \cos (c+d x))^{8/3}}\right )\) |
b^3*((3*A*Sin[c + d*x])/(8*b*d*(b*Cos[c + d*x])^(8/3)) + (3*(5*A + 8*C)*Hy pergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*Sin[c + d*x])/(16*b^3*d*(b *Cos[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]))
3.2.69.3.1 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]
\[\int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{3}\left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^3\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]