Integrand size = 29, antiderivative size = 115 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=-\frac {3 A b^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \sec (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 b B \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}} \]
-3/2*A*b^2*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*B*hypergeom([-1/6, 1/2],[5/6],cos(d*x +c)^2)*(b*sec(d*x+c))^(1/3)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76 \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\frac {3 \cot (c+d x) \left (4 A \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\sec ^2(c+d x)\right )+B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\sec ^2(c+d x)\right )\right ) (b \sec (c+d x))^{4/3} \sqrt {-\tan ^2(c+d x)}}{4 d} \]
(3*Cot[c + d*x]*(4*A*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2] + B*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 = 0.47 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 2030, 4274, 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.
\(\displaystyle \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \sqrt [3]{b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )} \left (A+B \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle b \left (A \int \sqrt [3]{b \sec (c+d x)}dx+\frac {B \int (b \sec (c+d x))^{4/3}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (A \int \sqrt [3]{b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {B \int \left (b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx}{b}\right )\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle b \left (A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\cos (c+d x)}{b}\right )^{4/3}}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (A \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\sqrt [3]{\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}}}dx+\frac {B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)} \int \frac {1}{\left (\frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{b}\right )^{4/3}}dx}{b}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\frac {3 B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}-\frac {3 A b \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}\right )\) |
b*((-3*A*b*Hypergeometric2F1[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*B*Hypergeometric2F1 [-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c + d*x])/(d*S qrt[Sin[c + d*x]^2]))
3.1.15.3.1 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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
\[\int \cos \left (d x +c \right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )\right )d x\]
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \]
integral((B*b*cos(d*x + c)*sec(d*x + c)^2 + A*b*cos(d*x + c)*sec(d*x + c)) *(b*sec(d*x + c))^(1/3), x)
Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \]
\[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \cos (c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)) \, dx=\int \cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]