Integrand size = 33, antiderivative size = 93 \[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b (2 A+5 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^2 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}} \]
-3/10*b*(2*A+5*C)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b *sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)+3/5*A*b^2*tan(d*x+c)/d/(b*sec(d*x+ c))^(5/3)
Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 \cot (c+d x) \left (A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\sec ^2(c+d x)\right )-5 C \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )\right ) \sqrt [3]{b \sec (c+d x)} \sqrt {-\tan ^2(c+d x)}}{5 d} \]
(-3*Cot[c + d*x]*(A*Cos[c + d*x]^2*Hypergeometric2F1[-5/6, 1/2, 1/6, Sec[c + d*x]^2] - 5*C*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2])*(b*Sec[ c + d*x])^(1/3)*Sqrt[-Tan[c + d*x]^2])/(5*d)
Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2030, 4533, 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.
| \(\displaystyle \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^2 \int \frac {C \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{5/3}}dx\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle b^2 \left (\frac {(2 A+5 C) \int \sqrt [3]{b \sec (c+d x)}dx}{5 b^2}+\frac {3 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\frac {(2 A+5 C) \int \sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{5 b^2}+\frac {3 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle b^2 \left (\frac {(2 A+5 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}}dx}{5 b^2}+\frac {3 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\frac {(2 A+5 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}}dx}{5 b^2}+\frac {3 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle b^2 \left (\frac {3 A \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 (2 A+5 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{10 b d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}\right )\) |
b^2*((-3*(2*A + 5*C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[ c + d*x])/(10*b*d*(b*Sec[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*Tan[ c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)))
3.1.5.3.1 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[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
\[\int \cos \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {1}{3}} \left (A +C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
\[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt [3]{b \sec {\left (c + d x \right )}} \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}\, dx \]
\[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
\[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
Timed out. \[ \int \cos ^2(c+d x) \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \]