Integrand size = 33, antiderivative size = 95 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (5 A+2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{2/3} \tan (c+d x)}{5 b^2 d} \]
-3/5*(5*A+2*C)*hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(b* sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)+3/5*C*(b*sec(d*x+c))^(2/3)*tan(d*x+ c)/b^2/d
Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.01 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 (b \sec (c+d x))^{2/3} \left (4 A \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right )+C \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\sec ^2(c+d x)\right ) \sec (c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{8 b^2 d} \]
(3*(b*Sec[c + d*x])^(2/3)*(4*A*Cot[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/ 3, Sec[c + d*x]^2] + C*Csc[c + d*x]*Hypergeometric2F1[1/2, 4/3, 7/3, Sec[c + d*x]^2]*Sec[c + d*x])*Sqrt[-Tan[c + d*x]^2])/(8*b^2*d)
Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2030, 3042, 4534, 3042, 4259, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {\int (b \sec (c+d x))^{2/3} \left (C \sec ^2(c+d x)+A\right )dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx}{b^2}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \int (b \sec (c+d x))^{2/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b^2}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{5} (5 A+2 C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2/3}}dx+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}}{b^2}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\frac {3 C \tan (c+d x) (b \sec (c+d x))^{2/3}}{5 d}-\frac {3 b (5 A+2 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}}{b^2}\) |
((-3*b*(5*A + 2*C)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Sec[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(5*d))/b^2
3.1.16.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
\[\int \frac {\sec \left (d x +c \right )^{2} \left (A +C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]