Integrand size = 31, antiderivative size = 90 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 A \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt {\sin ^2(c+d x)}} \]
3/4*A*sin(d*x+c)/d/(b*cos(d*x+c))^(4/3)-3/8*(A+4*C)*(b*cos(d*x+c))^(2/3)*h ypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1 /2)
Time = 0.01 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \csc (c+d x) \left (-A \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1}{2},\frac {1}{3},\cos ^2(c+d x)\right )+2 C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{4/3}} \]
(-3*Csc[c + d*x]*(-(A*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]) + 2*C*Cos[c + d*x]^2*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2])*Sqrt [Sin[c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(4/3))
Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 2030, 3491, 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.
\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/3}}dx\) |
\(\Big \downarrow \) 3491 |
\(\displaystyle b \left (\frac {(A+4 C) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}}dx}{4 b^2}+\frac {3 A \sin (c+d x)}{4 b d (b \cos (c+d x))^{4/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {(A+4 C) \int \frac {1}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 b^2}+\frac {3 A \sin (c+d x)}{4 b d (b \cos (c+d x))^{4/3}}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\frac {3 A \sin (c+d x)}{4 b d (b \cos (c+d x))^{4/3}}-\frac {3 (A+4 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{8 b^3 d \sqrt {\sin ^2(c+d x)}}\right )\) |
b*((3*A*Sin[c + d*x])/(4*b*d*(b*Cos[c + d*x])^(4/3)) - (3*(A + 4*C)*(b*Cos [c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^3*d*Sqrt[Sin[c + d*x]^2]))
3.2.73.3.1 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_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x _)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)) Int[(b*Sin[e + f* x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
\[\int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)/(b^2*c os(d*x + c)^2), x)
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \]