Integrand size = 41, antiderivative size = 154 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (13 A+10 C) \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/3} \sin (c+d x)}{91 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{10/3} \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{10/3} \tan (c+d x)}{13 b^2 d} \]
3/91*(13*A+10*C)*hypergeom([-7/6, 1/2],[-1/6],cos(d*x+c)^2)*(b*sec(d*x+c)) ^(7/3)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)+3/10*B*hypergeom([-5/3, 1/2],[- 2/3],cos(d*x+c)^2)*(b*sec(d*x+c))^(10/3)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^( 1/2)+3/13*C*(b*sec(d*x+c))^(10/3)*tan(d*x+c)/b^2/d
Time = 1.53 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.78 \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \csc (c+d x) (b \sec (c+d x))^{7/3} \left (104 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right )+80 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\sec ^2(c+d x)\right ) \sec (c+d x)+65 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {8}{3},\frac {11}{3},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{1040 b d} \]
(3*Csc[c + d*x]*(b*Sec[c + d*x])^(7/3)*(104*A*Hypergeometric2F1[1/2, 5/3, 8/3, Sec[c + d*x]^2] + 80*B*Hypergeometric2F1[1/2, 13/6, 19/6, Sec[c + d*x ]^2]*Sec[c + d*x] + 65*C*Hypergeometric2F1[1/2, 8/3, 11/3, Sec[c + d*x]^2] *Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/(1040*b*d)
Time = 0.80 (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.
| \(\displaystyle \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {\int (b \sec (c+d x))^{10/3} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^2}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {\int (b \sec (c+d x))^{10/3} \left (C \sec ^2(c+d x)+A\right )dx+\frac {B \int (b \sec (c+d x))^{13/3}dx}{b}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{13/3}dx}{b}}{b^2}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{13/3}}dx}{b}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{13/3}}dx}{b}}{b^2}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}}{b^2}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\frac {1}{13} (13 A+10 C) \int (b \sec (c+d x))^{10/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{13} (13 A+10 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 d}}{b^2}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\frac {1}{13} (13 A+10 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{13} (13 A+10 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{10/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 d}}{b^2}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\frac {3 b (13 A+10 C) \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{91 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},\frac {1}{2},-\frac {2}{3},\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{10/3}}{13 d}}{b^2}\) |
((3*b*(13*A + 10*C)*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b* Sec[c + d*x])^(7/3)*Sin[c + d*x])/(91*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hyper geometric2F1[-5/3, 1/2, -2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(10/3)*Sin[ c + d*x])/(10*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(10/3)*Tan[c + d*x])/(13*d))/b^2
3.1.47.3.1 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]
\[\int \sec \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="fricas")
integral((C*b*sec(d*x + c)^5 + B*b*sec(d*x + c)^4 + A*b*sec(d*x + c)^3)*(b *sec(d*x + c))^(1/3), x)
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{2} \,d x } \]
integrate(sec(d*x+c)^2*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \sec ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \]