Integrand size = 41, antiderivative size = 154 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {3 (10 A+7 C) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{4/3} \sin (c+d x)}{40 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) (b \sec (c+d x))^{7/3} \sin (c+d x)}{7 b^3 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C (b \sec (c+d x))^{7/3} \tan (c+d x)}{10 b^3 d} \] Output:
3/40*(10*A+7*C)*hypergeom([-2/3, 1/2],[1/3],cos(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*B*hypergeom([-7/6, 1/2],[-1 /6],cos(d*x+c)^2)*(b*sec(d*x+c))^(7/3)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/ 2)+3/10*C*(b*sec(d*x+c))^(7/3)*tan(d*x+c)/b^3/d
Time = 0.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.77 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=-\frac {3 \csc ^3(c+d x) \left (130 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\sec ^2(c+d x)\right )+91 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\sec ^2(c+d x)\right ) \sec (c+d x)+70 C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {13}{6},\frac {19}{6},\sec ^2(c+d x)\right ) \sec ^2(c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{3/2}}{910 d (b \sec (c+d x))^{2/3}} \] Input:
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(b*Sec[ c + d*x])^(2/3),x]
Output:
(-3*Csc[c + d*x]^3*(130*A*Hypergeometric2F1[1/2, 7/6, 13/6, Sec[c + d*x]^2 ] + 91*B*Hypergeometric2F1[1/2, 5/3, 8/3, Sec[c + d*x]^2]*Sec[c + d*x] + 7 0*C*Hypergeometric2F1[1/2, 13/6, 19/6, Sec[c + d*x]^2]*Sec[c + d*x]^2)*(-T an[c + d*x]^2)^(3/2))/(910*d*(b*Sec[c + d*x])^(2/3))
Time = 0.76 (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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle \frac {\int (b \sec (c+d x))^{7/3} \left (C \sec ^2(c+d x)+B \sec (c+d x)+A\right )dx}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^3}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {\int (b \sec (c+d x))^{7/3} \left (C \sec ^2(c+d x)+A\right )dx+\frac {B \int (b \sec (c+d x))^{10/3}dx}{b}}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}dx}{b}}{b^3}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{10/3}}dx}{b}}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{10/3}}dx}{b}}{b^3}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3} \left (C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A\right )dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}}{b^3}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int (b \sec (c+d x))^{7/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b^3}\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{7/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} (10 A+7 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{7/3}}dx+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b^3}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {\frac {3 b (10 A+7 C) \sin (c+d x) (b \sec (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{40 d \sqrt {\sin ^2(c+d x)}}+\frac {3 B \sin (c+d x) (b \sec (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x) (b \sec (c+d x))^{7/3}}{10 d}}{b^3}\) |
Input:
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(b*Sec[c + d* x])^(2/3),x]
Output:
((3*b*(10*A + 7*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*(b*Se c[c + d*x])^(4/3)*Sin[c + d*x])/(40*d*Sqrt[Sin[c + d*x]^2]) + (3*B*Hyperge ometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(7/3)*Sin[c + d*x])/(7*d*Sqrt[Sin[c + d*x]^2]) + (3*C*(b*Sec[c + d*x])^(7/3)*Tan[c + d* x])/(10*d))/b^3
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]
\[\int \frac {\sec \left (d x +c \right )^{3} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x)
Output:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="fricas")
Output:
integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*(b*sec(d *x + c))^(1/3)/b, x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \] Input:
integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(b*sec(d*x+c))**( 2/3),x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(b*sec(c + d*x))**(2/3), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d* x + c))^(2/3), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3 ),x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d* x + c))^(2/3), x)
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(b/cos(c + d*x ))^(2/3)),x)
Output:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(b/cos(c + d*x ))^(2/3)), x)
\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{2/3}} \, dx=\frac {\left (\int \sec \left (d x +c \right )^{\frac {13}{3}}d x \right ) c +\left (\int \sec \left (d x +c \right )^{\frac {10}{3}}d x \right ) b +\left (\int \sec \left (d x +c \right )^{\frac {7}{3}}d x \right ) a}{b^{\frac {2}{3}}} \] Input:
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(2/3),x)
Output:
(int(sec(c + d*x)**5/sec(c + d*x)**(2/3),x)*c + int(sec(c + d*x)**4/sec(c + d*x)**(2/3),x)*b + int(sec(c + d*x)**3/sec(c + d*x)**(2/3),x)*a)/b**(2/3 )