\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx\) [899]

Optimal result

Integrand size = 31, antiderivative size = 119 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 A (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^4 d \sqrt {\sin ^2(c+d x)}} \] Output:

-3/7*A*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin( 
d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)-3/10*B*(b*cos(d*x+c))^(10/3)*hypergeom([ 
1/2, 5/3],[8/3],cos(d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.79 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \cos ^2(c+d x) \cot (c+d x) \left (10 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )+7 B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{70 d (b \cos (c+d x))^{2/3}} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(2/3),x]
 

Output:

(-3*Cos[c + d*x]^2*Cot[c + d*x]*(10*A*Hypergeometric2F1[1/2, 7/6, 13/6, Co 
s[c + d*x]^2] + 7*B*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + 
d*x]^2])*Sqrt[Sin[c + d*x]^2])/(70*d*(b*Cos[c + d*x])^(2/3))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {2030, 3042, 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.

\(\displaystyle \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x))dx}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{b^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {A \int (b \cos (c+d x))^{4/3}dx+\frac {B \int (b \cos (c+d x))^{7/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx+\frac {B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/3}dx}{b}}{b^2}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {-\frac {3 A \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{10/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3},\frac {8}{3},\cos ^2(c+d x)\right )}{10 b^2 d \sqrt {\sin ^2(c+d x)}}}{b^2}\)

Input:

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(b*Cos[c + d*x])^(2/3),x]
 

Output:

((-3*A*(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*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^( 
10/3)*Hypergeometric2F1[1/2, 5/3, 8/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*b 
^2*d*Sqrt[Sin[c + d*x]^2]))/b^2
 

Defintions of rubi rules used

rule 2030

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

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]
 

rule 3227

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]
 

Maple [F]

\[\int \frac {\cos \left (d x +c \right )^{2} \left (A +B \cos \left (d x +c \right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(2/3),x)
 

Output:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(cos(d*x+c)*b)^(2/3),x)
 

Fricas [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(2/3),x, algorithm= 
"fricas")
 

Output:

integral((B*cos(d*x + c)^2 + A*cos(d*x + c))*(b*cos(d*x + c))^(1/3)/b, x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(b*cos(d*x+c))**(2/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(2/3),x, algorithm= 
"maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d*x + c))^(2/3), x)
 

Giac [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(2/3),x, algorithm= 
"giac")
 

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d*x + c))^(2/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \] Input:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(2/3),x)
 

Output:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(b*cos(c + d*x))^(2/3), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(b \cos (c+d x))^{2/3}} \, dx=\frac {\left (\int \cos \left (d x +c \right )^{\frac {4}{3}}d x \right ) a +\left (\int \cos \left (d x +c \right )^{\frac {7}{3}}d x \right ) b}{b^{\frac {2}{3}}} \] Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(b*cos(d*x+c))^(2/3),x)
 

Output:

(int(cos(c + d*x)**3/cos(c + d*x)**(2/3),x)*b + int(cos(c + d*x)**2/cos(c 
+ d*x)**(2/3),x)*a)/b**(2/3)