\(\int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\) [901]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 A \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{4 b^2 d \sqrt {\sin ^2(c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \cot (c+d x) \left (4 A \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{2/3}} \] Input:

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

Output:

(-3*Cot[c + d*x]*(4*A*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + B 
*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2])*Sqrt[Sin[c 
 + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle A \int \frac {1}{(b \cos (c+d x))^{2/3}}dx+\frac {B \int \sqrt [3]{b \cos (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+\frac {B \int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle -\frac {3 A \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 b^2 d \sqrt {\sin ^2(c+d x)}}\)

Input:

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

Output:

(-3*A*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x] 
^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(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^2*d*S 
qrt[Sin[c + d*x]^2])
 

Defintions of rubi rules used

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx=\frac {\left (\int \cos \left (d x +c \right )^{\frac {1}{3}}d x \right ) b +\left (\int \frac {1}{\cos \left (d x +c \right )^{\frac {2}{3}}}d x \right ) a}{b^{\frac {2}{3}}} \] Input:

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

Output:

(int(cos(c + d*x)/cos(c + d*x)**(2/3),x)*b + int(1/cos(c + d*x)**(2/3),x)* 
a)/b**(2/3)