Integrand size = 39, antiderivative size = 149 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=-\frac {3 (2 A-C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d (b \sec (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt [3]{b \sec (c+d x)} \sqrt {\sin ^2(c+d x)}}+\frac {3 C \tan (c+d x)}{2 b d \sqrt [3]{b \sec (c+d x)}} \]
-3/8*(2*A-C)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec( d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)-3*B*hypergeom([1/6, 1/2],[7/6],cos(d*x+ c)^2)*sin(d*x+c)/b/d/(b*sec(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)+3/2*C*tan(d *x+c)/b/d/(b*sec(d*x+c))^(1/3)
Time = 0.09 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\frac {3 C \tan (c+d x)-3 (2 A-C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}+3 B \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{2 b d \sqrt [3]{b \sec (c+d x)}} \]
(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] + 3*B*Csc[c + d*x]*Hypergeometr ic2F1[1/3, 1/2, 4/3, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2])/(2*b*d*(b*Sec[ c + d*x])^(1/3))
Time = 0.71 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, 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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int \frac {C \sec ^2(c+d x)+B \sec (c+d x)+A}{\sqrt [3]{b \sec (c+d x)}}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {\int \frac {C \sec ^2(c+d x)+A}{\sqrt [3]{b \sec (c+d x)}}dx+\frac {B \int (b \sec (c+d x))^{2/3}dx}{b}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}}{b}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{2/3}}dx}{b}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {B \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2/3}}dx}{b}}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}}{b}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {\frac {1}{2} (2 A-C) \int \frac {1}{\sqrt [3]{b \sec (c+d x)}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} (2 A-C) \int \frac {1}{\sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\frac {1}{2} (2 A-C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \sqrt [3]{\frac {\cos (c+d x)}{b}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} (2 A-C) \left (\frac {\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3} \int \sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {-\frac {3 b (2 A-C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \sec (c+d x)}}+\frac {3 C \tan (c+d x)}{2 d \sqrt [3]{b \sec (c+d x)}}}{b}\) |
((-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*B*Hypergeomet ric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(b*Sec[c + d*x])^(1 /3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(2*d*(b*Sec[c + d*x])^(1/3) ))/b
3.1.62.3.1 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]
\[\int \frac {\sec \left (d x +c \right ) \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )}{\left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3), x, algorithm="fricas")
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)/(b ^2*sec(d*x + c)), x)
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3), x, algorithm="maxima")
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )}{\left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(b*sec(d*x+c))^(4/3), x, algorithm="giac")
Timed out. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\cos \left (c+d\,x\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}} \,d x \]