Integrand size = 31, antiderivative size = 114 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A b \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3} \sqrt {\sin ^2(c+d x)}}+\frac {3 B \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3} \sqrt {\sin ^2(c+d x)}} \]
3/7*A*b*hypergeom([-7/6, 1/2],[-1/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*cos(d*x +c))^(7/3)/(sin(d*x+c)^2)^(1/2)+3/4*B*hypergeom([-2/3, 1/2],[1/3],cos(d*x+ c)^2)*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)
Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 b^2 \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )+7 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{28 d (b \cos (c+d x))^{10/3}} \]
(3*b^2*Cot[c + d*x]*(4*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2 ] + 7*B*Cos[c + d*x]*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2])*Sq rt[Sin[c + d*x]^2])/(28*d*(b*Cos[c + d*x])^(10/3))
Time = 0.38 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^2 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{10/3}}dx\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle b^2 \left (A \int \frac {1}{(b \cos (c+d x))^{10/3}}dx+\frac {B \int \frac {1}{(b \cos (c+d x))^{7/3}}dx}{b}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^2 \left (A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{10/3}}dx+\frac {B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}}dx}{b}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b^2 \left (\frac {3 A \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {7}{6},\frac {1}{2},-\frac {1}{6},\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{7/3}}+\frac {3 B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}}\right )\) |
b^2*((3*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*Sin[c + d*x]) /(7*b*d*(b*Cos[c + d*x])^(7/3)*Sqrt[Sin[c + d*x]^2]) + (3*B*Hypergeometric 2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*b^2*d*(b*Cos[c + d*x] )^(4/3)*Sqrt[Sin[c + d*x]^2]))
3.10.9.3.1 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]
\[\int \frac {\left (A +B \cos \left (d x +c \right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)^2/(b^2*c os(d*x + c)^2), x)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \]