3.4.87 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx\) [387]

3.4.87.1 Optimal result

Integrand size = 35, antiderivative size = 144 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^{2/3}} \, dx=\frac {3 (A-B+C) \sin (c+d x)}{d (a+a \cos (c+d x))^{2/3}}+\frac {3 C \sqrt [3]{a+a \cos (c+d x)} \sin (c+d x)}{4 a d}-\frac {(4 A-8 B+7 C) \sqrt [3]{a+a \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{2 \sqrt [6]{2} a d (1+\cos (c+d x))^{5/6}} \]

output

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

3.4.87.2 Mathematica [F]

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

input

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

output

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

3.4.87.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3502, 27, 3042, 3229, 3042, 3131, 3042, 3130}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {3 \int \frac {a (4 A+C)+a (4 B-3 C) \cos (c+d x)}{3 (\cos (c+d x) a+a)^{2/3}}dx}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (4 A+C)+a (4 B-3 C) \cos (c+d x)}{(\cos (c+d x) a+a)^{2/3}}dx}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3229

\(\displaystyle \frac {\frac {12 a (A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-(4 A-8 B+7 C) \int \sqrt [3]{\cos (c+d x) a+a}dx}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {12 a (A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-(4 A-8 B+7 C) \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {12 a (A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {(4 A-8 B+7 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\cos (c+d x)+1}dx}{\sqrt [3]{\cos (c+d x)+1}}}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {12 a (A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {(4 A-8 B+7 C) \sqrt [3]{a \cos (c+d x)+a} \int \sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}dx}{\sqrt [3]{\cos (c+d x)+1}}}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {12 a (A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)^{2/3}}-\frac {2^{5/6} (4 A-8 B+7 C) \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{d (\cos (c+d x)+1)^{5/6}}}{4 a}+\frac {3 C \sin (c+d x) \sqrt [3]{a \cos (c+d x)+a}}{4 a d}\)

input

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

output

(3*C*(a + a*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*a*d) + ((12*a*(A - B + C) 
*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(2/3)) - (2^(5/6)*(4*A - 8*B + 7*C) 
*(a + a*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 3/2, (1 - Cos[c + 
d*x])/2]*Sin[c + d*x])/(d*(1 + Cos[c + d*x])^(5/6)))/(4*a)
 

3.4.87.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3130

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 
 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeome 
tric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; FreeQ[{a, 
 b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]
 

rule 3131

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Sin[c + d*x])^FracPart[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

rule 3229

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f* 
x])^m/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && 
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

3.4.87.4 Maple [F]

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

input

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

output

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

3.4.87.5 Fricas [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(2/3),x, algori 
thm="fricas")
 

output

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

3.4.87.6 Sympy [F]

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

input

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

output

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/(a*(cos(c + d*x) + 1))** 
(2/3), x)
 

3.4.87.7 Maxima [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(2/3),x, algori 
thm="maxima")
 

output

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

3.4.87.8 Giac [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(2/3),x, algori 
thm="giac")
 

output

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

3.4.87.9 Mupad [F(-1)]

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

input

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

output

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