\(\int \frac {\cos ^m(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [215]

Optimal result

Integrand size = 40, antiderivative size = 173 \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 B \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 C \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \] Output:

-3*B*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d*x+c)^2) 
*sin(d*x+c)/b/d/(2+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)-3*C*cos( 
d*x+c)^(2+m)*hypergeom([1/2, 5/6+1/2*m],[11/6+1/2*m],cos(d*x+c)^2)*sin(d*x 
+c)/b/d/(5+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cos ^{2+m}(c+d x) \csc (c+d x) \left (B (5+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right )+C (2+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (2+3 m) (5+3 m) (b \cos (c+d x))^{4/3}} \] Input:

Integrate[(Cos[c + d*x]^m*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + 
d*x])^(4/3),x]
 

Output:

(-3*Cos[c + d*x]^(2 + m)*Csc[c + d*x]*(B*(5 + 3*m)*Hypergeometric2F1[1/2, 
(2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2] + C*(2 + 3*m)*Cos[c + d*x]*Hyper 
geometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[Sin[c + 
 d*x]^2])/(d*(2 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(4/3))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {2034, 3042, 3489, 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.

\(\displaystyle \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3489

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{m-\frac {1}{3}}(c+d x) (B+C \cos (c+d x))dx}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (B \int \cos ^{m-\frac {1}{3}}(c+d x)dx+C \int \cos ^{m+\frac {2}{3}}(c+d x)dx\right )}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (-\frac {3 B \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+2),\frac {1}{6} (3 m+8),\cos ^2(c+d x)\right )}{d (3 m+2) \sqrt {\sin ^2(c+d x)}}-\frac {3 C \sin (c+d x) \cos ^{m+\frac {5}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+5),\frac {1}{6} (3 m+11),\cos ^2(c+d x)\right )}{d (3 m+5) \sqrt {\sin ^2(c+d x)}}\right )}{b \sqrt [3]{b \cos (c+d x)}}\)

Input:

Int[(Cos[c + d*x]^m*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^ 
(4/3),x]
 

Output:

(Cos[c + d*x]^(1/3)*((-3*B*Cos[c + d*x]^(2/3 + m)*Hypergeometric2F1[1/2, ( 
2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(2 + 3*m)*Sqrt[S 
in[c + d*x]^2]) - (3*C*Cos[c + d*x]^(5/3 + m)*Hypergeometric2F1[1/2, (5 + 
3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 3*m)*Sqrt[Sin[ 
c + d*x]^2])))/(b*(b*Cos[c + d*x])^(1/3))
 

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]
 

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]
 

rule 3489

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + 
(C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b   Int[(b*Sin[e + f* 
x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
 

Maple [F]

\[\int \frac {\cos \left (d x +c \right )^{m} \left (B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]

Input:

int(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
 

Output:

int(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
 

Fricas [F]

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

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), 
x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(cos(d*x+c)**m*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(4/ 
3),x)
 

Output:

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

Maxima [F]

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

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), 
x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^m/(b*cos(d*x + 
c))^(4/3), x)
 

Giac [F]

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

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), 
x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^m/(b*cos(d*x + 
c))^(4/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:

int((cos(c + d*x)^m*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x))^ 
(4/3),x)
 

Output:

int((cos(c + d*x)^m*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x))^ 
(4/3), x)
 

Reduce [F]

\[ \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {\left (\int \frac {\cos \left (d x +c \right )^{m}}{\cos \left (d x +c \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \cos \left (d x +c \right )^{m} \cos \left (d x +c \right )^{\frac {2}{3}}d x \right ) c}{b^{\frac {4}{3}}} \] Input:

int(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
 

Output:

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