3.2.81 \(\int \frac {\cos ^m(c+d x) (A+C \cos ^2(c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\) [181]

3.2.81.1 Optimal result

Integrand size = 33, antiderivative size = 149 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}-\frac {3 (C (1-3 m)-A (2+3 m)) \cos ^m(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

output

3*C*cos(d*x+c)^m*sin(d*x+c)/b/d/(2+3*m)/(b*cos(d*x+c))^(1/3)-3*(C*(1-3*m)- 
A*(2+3*m))*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/(-9*m^2-3*m+2)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2 
)
 

3.2.81.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=-\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left (A (5+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (-1+3 m),\frac {1}{6} (5+3 m),\cos ^2(c+d x)\right )+C (-1+3 m) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-1+3 m) (5+3 m) (b \cos (c+d x))^{4/3}} \]

input

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

output

(-3*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(A*(5 + 3*m)*Hypergeometric2F1[1/2, 
(-1 + 3*m)/6, (5 + 3*m)/6, Cos[c + d*x]^2] + C*(-1 + 3*m)*Cos[c + d*x]^2*H 
ypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[Sin 
[c + d*x]^2])/(d*(-1 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(4/3))
 

3.2.81.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2034, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {3 C \sin (c+d x) \cos ^{m-\frac {1}{3}}(c+d x)}{d (3 m+2)}-\frac {(C (1-3 m)-A (3 m+2)) \int \cos ^{m-\frac {4}{3}}(c+d x)dx}{3 m+2}\right )}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {3 C \sin (c+d x) \cos ^{m-\frac {1}{3}}(c+d x)}{d (3 m+2)}-\frac {(C (1-3 m)-A (3 m+2)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {4}{3}}dx}{3 m+2}\right )}{b \sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3122

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

input

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

output

(Cos[c + d*x]^(1/3)*((3*C*Cos[c + d*x]^(-1/3 + m)*Sin[c + d*x])/(d*(2 + 3* 
m)) - (3*(C*(1 - 3*m) - A*(2 + 3*m))*Cos[c + d*x]^(-1/3 + m)*Hypergeometri 
c2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - 
 3*m)*(2 + 3*m)*Sqrt[Sin[c + d*x]^2])))/(b*(b*Cos[c + d*x])^(1/3))
 

3.2.81.3.1 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 3493

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]
 

3.2.81.4 Maple [F]

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

input

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

output

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

3.2.81.5 Fricas [F]

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

input

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

output

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

3.2.81.6 Sympy [F]

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

input

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

output

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

3.2.81.7 Maxima [F]

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

input

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

output

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

3.2.81.8 Giac [F]

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

input

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

output

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

3.2.81.9 Mupad [F(-1)]

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

input

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

output

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