Integrand size = 41, antiderivative size = 154 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 (7 A+4 C) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{4/3} \tan (c+d x)}{7 b^2 d} \]
3/7*(7*A+4*C)*hypergeom([-1/6, 1/2],[5/6],cos(d*x+c)^2)*(b*sec(d*x+c))^(1/ 3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/4*B*hypergeom([-2/3, 1/2],[1/3],c os(d*x+c)^2)*(b*sec(d*x+c))^(4/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)+3/ 7*C*(b*sec(d*x+c))^(4/3)*tan(d*x+c)/b^2/d
Time = 0.99 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 \csc (c+d x) \sqrt [3]{b \sec (c+d x)} \left (35 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )+2 \sec (c+d x) \left (10 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )+7 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right ) \sec (c+d x)\right )\right ) \sqrt {-\tan ^2(c+d x)}}{140 b d} \]
(3*Csc[c + d*x]*(b*Sec[c + d*x])^(1/3)*(35*A*Hypergeometric2F1[1/2, 2/3, 5 /3, Sec[c + d*x]^2] + 2*Sec[c + d*x]*(10*B*Hypergeometric2F1[1/2, 7/6, 13/ 6, Sec[c + d*x]^2] + 7*C*Hypergeometric2F1[1/2, 5/3, 8/3, Sec[c + d*x]^2]* Sec[c + d*x]))*Sqrt[-Tan[c + d*x]^2])/(140*b*d)
Time = 0.78 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {2030, 3042, 4535, 3042, 4259, 3042, 3122, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{4/3} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^2}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {\int (b \sec (c+d x))^{4/3} \left (C \sec ^2(c+d x)+A\right )dx+\frac {B \int (b \sec (c+d x))^{7/3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{7/3}}dx}{b}}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}}{b^2}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {\frac {1}{7} (7 A+4 C) \int (b \sec (c+d x))^{4/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{7} (7 A+4 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\frac {1}{7} (7 A+4 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{7} (7 A+4 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {3 b (7 A+4 C) \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{4/3}}{7 d}}{b^2}\) |
((3*b*(7*A + 4*C)*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec [c + d*x])^(1/3)*Sin[c + d*x])/(7*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeom etric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(4/3)*Sin[c + d* x])/(4*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(4/3)*Tan[c + d*x]) /(7*d))/b^2
3.1.53.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
\[\int \frac {\sec \left (d x +c \right )^{2} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="fricas")
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(b*sec(c + d*x))**(2/3), x)
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="maxima")
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="giac")
Timed out. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]