Integrand size = 25, antiderivative size = 90 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=-\frac {3 b (2 A-C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}} \]
-3/8*b*(2*A-C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*se c(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)+3/2*C*tan(d*x+c)/d/(b*sec(d*x+c))^(1/ 3)
Time = 0.38 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\frac {3 C \tan (c+d x)-3 (2 A-C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{2 d \sqrt [3]{b \sec (c+d x)}} \]
(3*C*Tan[c + d*x] - 3*(2*A - C)*Cot[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2])/(2*d*(b*Sec[c + d*x])^(1/3))
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.
\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {1}{2} (2 A-C) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}}dx+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (2 A-C) \int \frac {1}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {1}{2} (2 A-C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \sqrt [3]{\frac {\cos (c+d x)}{b}}dx+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} (2 A-C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}dx+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}-\frac {3 b (2 A-C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}\) |
(-3*b*(2*A - C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d *x])/(8*d*(b*Sec[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x] )/(2*d*(b*Sec[c + d*x])^(1/3))
3.1.13.3.1 Defintions of rubi rules used
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]
\[\int \frac {A +C \sec \left (d x +c \right )^{2}}{\left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\sqrt [3]{b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{\left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt [3]{b \sec (c+d x)}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]