Integrand size = 41, antiderivative size = 150 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 b^3 (A-2 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d (b \sec (c+d x))^{5/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}} \]
-3/5*b^3*(A-2*C)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b *sec(d*x+c))^(5/3)/(sin(d*x+c)^2)^(1/2)-3/2*b^2*B*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* b^2*C*tan(d*x+c)/d/(b*sec(d*x+c))^(2/3)
Time = 0.17 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {3 \cot (c+d x) \left (2 A \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\sec ^2(c+d x)\right )-4 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )-C \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{4/3} \sqrt {-\tan ^2(c+d x)}}{4 d} \]
(-3*Cot[c + d*x]*(2*A*Cos[c + d*x]^2*Hypergeometric2F1[-1/3, 1/2, 2/3, Sec [c + d*x]^2] - 4*B*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[c + d *x]^2] - C*Hypergeometric2F1[1/2, 2/3, 5/3, Sec[c + d*x]^2])*(b*Sec[c + d* x])^(4/3)*Sqrt[-Tan[c + d*x]^2])/(4*d)
Time = 0.78 (sec) , antiderivative size = 146, 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 = {3042, 2030, 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.
| \(\displaystyle \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^2 \int \frac {C \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{2/3}}dx\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle b^2 \left (\int \frac {C \csc ^2\left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{2/3}}dx+\frac {B \int \sqrt [3]{b \sec (c+d x)}dx}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+\frac {B \int \sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle b^2 \left (\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}}dx}{b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left (\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}}dx}{b}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle b^2 \left (\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+A}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle b^2 \left ((A-2 C) \int \frac {1}{(b \sec (c+d x))^{2/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left ((A-2 C) \int \frac {1}{\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 4259 |
| \(\displaystyle b^2 \left ((A-2 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \left (\frac {\cos (c+d x)}{b}\right )^{2/3}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^2 \left ((A-2 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{2/3}dx-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle b^2 \left (-\frac {3 b (A-2 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}}-\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}\right )\) |
b^2*((-3*b*(A - 2*C)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin [c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*Sqrt[Sin[c + d*x]^2]) - (3*B*Hyperg eometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*(b*Sec[c + d *x])^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^ (2/3)))
3.1.51.3.1 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]
\[\int \cos \left (d x +c \right )^{2} \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \sec \left (d x +c \right )^{2}\right )d x\]
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(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*cos(d*x + c)^2*sec(d*x + c)^3 + B*b*cos(d*x + c)^2*sec(d*x + c)^2 + A*b*cos(d*x + c)^2*sec(d*x + c))*(b*sec(d*x + c))^(1/3), x)
Timed out. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(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")
\[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{2} \,d x } \]
integrate(cos(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. \[ \int \cos ^2(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]