\(\int \frac {\cos ^m(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [237]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int \frac {\cos ^m(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:

3*cos(d*x+c)^m*hypergeom([1/2, -1/6+1/2*m],[5/6+1/2*m],cos(d*x+c)^2)*sin(d 
*x+c)/b/d/(1-3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^m(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {\cos ^{1+m}(c+d x) \csc (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} \left (-\frac {1}{3}+m\right ),\frac {1}{2} \left (\frac {5}{3}+m\right ),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}}{d \left (-\frac {1}{3}+m\right ) (b \cos (c+d x))^{4/3}} \] Input:

Integrate[Cos[c + d*x]^m/(b*Cos[c + d*x])^(4/3),x]
 

Output:

-((Cos[c + d*x]^(1 + m)*Csc[c + d*x]*Hypergeometric2F1[1/2, (-1/3 + m)/2, 
(5/3 + m)/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])/(d*(-1/3 + m)*(b*Cos[c 
+ d*x])^(4/3)))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2034, 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.

\(\displaystyle \int \frac {\cos ^m(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 2034

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{m-\frac {4}{3}}(c+d x)dx}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}}dx}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3122

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

Input:

Int[Cos[c + d*x]^m/(b*Cos[c + d*x])^(4/3),x]
 

Output:

(3*Cos[c + d*x]^m*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Cos[c 
+ d*x]^2]*Sin[c + d*x])/(b*d*(1 - 3*m)*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + 
 d*x]^2])
 

Defintions of rubi rules used

rule 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

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]
 

Maple [F]

\[\int \frac {\cos \left (d x +c \right )^{m}}{\left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}}}d x\]

Input:

int(cos(d*x+c)^m/(cos(d*x+c)*b)^(4/3),x)
 

Output:

int(cos(d*x+c)^m/(cos(d*x+c)*b)^(4/3),x)
 

Fricas [F]

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

integrate(cos(d*x+c)^m/(b*cos(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

integral((b*cos(d*x + c))^(2/3)*cos(d*x + c)^m/(b^2*cos(d*x + c)^2), x)
 

Sympy [F]

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

integrate(cos(d*x+c)**m/(b*cos(d*x+c))**(4/3),x)
 

Output:

Integral(cos(c + d*x)**m/(b*cos(c + d*x))**(4/3), x)
 

Maxima [F]

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

integrate(cos(d*x+c)^m/(b*cos(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

integrate(cos(d*x + c)^m/(b*cos(d*x + c))^(4/3), x)
 

Giac [F]

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

integrate(cos(d*x+c)^m/(b*cos(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

integrate(cos(d*x + c)^m/(b*cos(d*x + c))^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^m(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:

int(cos(c + d*x)^m/(b*cos(c + d*x))^(4/3),x)
 

Output:

int(cos(c + d*x)^m/(b*cos(c + d*x))^(4/3), x)
 

Reduce [F]

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

int(cos(d*x+c)^m/(b*cos(d*x+c))^(4/3),x)
 

Output:

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