Integrand size = 25, antiderivative size = 93 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {3 C \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{4 b d}-\frac {3 (4 A+C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 b d \sqrt {\sin ^2(c+d x)}} \] Output:
3/4*C*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/b/d-3/4*(4*A+C)*(b*cos(d*x+c))^(1/3) *hypergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/b/d/(sin(d*x+c)^2)^(1 /2)
Time = 0.01 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94 \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \cot (c+d x) \left (7 A \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+C \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{7 d (b \cos (c+d x))^{2/3}} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(2/3),x]
Output:
(-3*Cot[c + d*x]*(7*A*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + C *Cos[c + d*x]^2*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2])*Sqrt[Si n[c + d*x]^2])/(7*d*(b*Cos[c + d*x])^(2/3))
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3493, 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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 3493 |
\(\displaystyle \frac {1}{4} (4 A+C) \int \frac {1}{(b \cos (c+d x))^{2/3}}dx+\frac {3 C \sin (c+d x) \sqrt [3]{b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} (4 A+C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+\frac {3 C \sin (c+d x) \sqrt [3]{b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {3 C \sin (c+d x) \sqrt [3]{b \cos (c+d x)}}{4 b d}-\frac {3 (4 A+C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{4 b d \sqrt {\sin ^2(c+d x)}}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/(b*Cos[c + d*x])^(2/3),x]
Output:
(3*C*(b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*b*d) - (3*(4*A + C)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x ])/(4*b*d*Sqrt[Sin[c + d*x]^2])
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[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f *(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2) Int[(b*Sin[e + f*x] )^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \frac {A +C \cos \left (d x +c \right )^{2}}{\left (b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}d x\]
Input:
int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x)
Output:
int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(1/3)/(b*cos(d*x + c)), x )
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\text {Timed out} \] Input:
integrate((A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(2/3),x)
Output:
Timed out
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(2/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c))^(2/3), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(2/3),x)
Output:
int((A + C*cos(c + d*x)^2)/(b*cos(c + d*x))^(2/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {\left (\int \cos \left (d x +c \right )^{\frac {4}{3}}d x \right ) c +\left (\int \frac {1}{\cos \left (d x +c \right )^{\frac {2}{3}}}d x \right ) a}{b^{\frac {2}{3}}} \] Input:
int((A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(2/3),x)
Output:
(int(cos(c + d*x)**2/cos(c + d*x)**(2/3),x)*c + int(1/cos(c + d*x)**(2/3), x)*a)/b**(2/3)