Integrand size = 29, antiderivative size = 116 \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {3 A (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}} \]
-3/4*A*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d *x+c)/d/(sin(d*x+c)^2)^(1/2)-3/7*B*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/ 6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)
Time = 0.00 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.75 \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=-\frac {3 b \sqrt [3]{b \cos (c+d x)} \cot (c+d x) \left (7 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )+4 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{28 d} \]
(-3*b*(b*Cos[c + d*x])^(1/3)*Cot[c + d*x]*(7*A*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2] + 4*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(28*d)
Time = 0.36 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 2030, 3227, 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.
\(\displaystyle \int \sec (c+d x) (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \sqrt [3]{b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \left (A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle b \left (A \int \sqrt [3]{b \cos (c+d x)}dx+\frac {B \int (b \cos (c+d x))^{4/3}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (A \int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx}{b}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (-\frac {3 A \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 b^2 d \sqrt {\sin ^2(c+d x)}}\right )\) |
b*((-3*A*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d *x]^2]*Sin[c + d*x])/(4*b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^ (7/3)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*b ^2*d*Sqrt[Sin[c + d*x]^2]))
3.9.96.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )\right ) \sec \left (d x +c \right )d x\]
\[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
\[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
Timed out. \[ \int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{\cos \left (c+d\,x\right )} \,d x \]