3.4.86 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx\) [386]

3.4.86.1 Optimal result

Integrand size = 35, antiderivative size = 144 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {3 (5 B-3 C) \sin (c+d x)}{10 d \sqrt [3]{a+a \cos (c+d x)}}+\frac {3 C (a+a \cos (c+d x))^{2/3} \sin (c+d x)}{5 a d}+\frac {(10 A-5 B+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right ) \sin (c+d x)}{5\ 2^{5/6} d \sqrt [6]{1+\cos (c+d x)} \sqrt [3]{a+a \cos (c+d x)}} \]

output

3/10*(5*B-3*C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/3)+3/5*C*(a+a*cos(d*x+c))^ 
(2/3)*sin(d*x+c)/a/d+1/10*(10*A-5*B+7*C)*hypergeom([1/2, 5/6],[3/2],1/2-1/ 
2*cos(d*x+c))*sin(d*x+c)*2^(1/6)/d/(1+cos(d*x+c))^(1/6)/(a+a*cos(d*x+c))^( 
1/3)
 

3.4.86.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+a \cos (c+d x)}} \, dx=\frac {-3 i (10 A-5 B+7 C) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-e^{i (c+d x)}\right ) (1+\cos (c+d x)+i \sin (c+d x))^{2/3}+3 (5 B-C+2 C \cos (c+d x)) \sin (c+d x)}{10 d \sqrt [3]{a (1+\cos (c+d x))}} \]

input

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^(1/ 
3),x]
 

output

((-3*I)*(10*A - 5*B + 7*C)*Hypergeometric2F1[1/3, 2/3, 4/3, -E^(I*(c + d*x 
))]*(1 + Cos[c + d*x] + I*Sin[c + d*x])^(2/3) + 3*(5*B - C + 2*C*Cos[c + d 
*x])*Sin[c + d*x])/(10*d*(a*(1 + Cos[c + d*x]))^(1/3))
 

3.4.86.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05, 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, 3230, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {3 \int \frac {a (5 A+2 C)+a (5 B-3 C) \cos (c+d x)}{3 \sqrt [3]{\cos (c+d x) a+a}}dx}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (5 A+2 C)+a (5 B-3 C) \cos (c+d x)}{\sqrt [3]{\cos (c+d x) a+a}}dx}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (5 A+2 C)+a (5 B-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3230

\(\displaystyle \frac {\frac {1}{2} a (10 A-5 B+7 C) \int \frac {1}{\sqrt [3]{\cos (c+d x) a+a}}dx+\frac {3 a (5 B-3 C) \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{2} a (10 A-5 B+7 C) \int \frac {1}{\sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {3 a (5 B-3 C) \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3131

\(\displaystyle \frac {\frac {a (10 A-5 B+7 C) \sqrt [3]{\cos (c+d x)+1} \int \frac {1}{\sqrt [3]{\cos (c+d x)+1}}dx}{2 \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 a (5 B-3 C) \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (10 A-5 B+7 C) \sqrt [3]{\cos (c+d x)+1} \int \frac {1}{\sqrt [3]{\sin \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{2 \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 a (5 B-3 C) \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

\(\Big \downarrow \) 3130

\(\displaystyle \frac {\frac {a (10 A-5 B+7 C) \sin (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x))\right )}{2^{5/6} d \sqrt [6]{\cos (c+d x)+1} \sqrt [3]{a \cos (c+d x)+a}}+\frac {3 a (5 B-3 C) \sin (c+d x)}{2 d \sqrt [3]{a \cos (c+d x)+a}}}{5 a}+\frac {3 C \sin (c+d x) (a \cos (c+d x)+a)^{2/3}}{5 a d}\)

input

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(a + a*Cos[c + d*x])^(1/3),x]
 

output

(3*C*(a + a*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*a*d) + ((3*a*(5*B - 3*C)* 
Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(1/3)) + (a*(10*A - 5*B + 7*C)*Hyp 
ergeometric2F1[1/2, 5/6, 3/2, (1 - Cos[c + d*x])/2]*Sin[c + d*x])/(2^(5/6) 
*d*(1 + Cos[c + d*x])^(1/6)*(a + a*Cos[c + d*x])^(1/3)))/(5*a)
 

3.4.86.3.1 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 3230

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

3.4.86.4 Maple [F]

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

input

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

output

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

3.4.86.5 Fricas [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/3),x, algori 
thm="fricas")
 

output

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

3.4.86.6 Sympy [F]

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

input

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

output

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

3.4.86.7 Maxima [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/3),x, algori 
thm="maxima")
 

output

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

3.4.86.8 Giac [F]

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

input

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(1/3),x, algori 
thm="giac")
 

output

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

3.4.86.9 Mupad [F(-1)]

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

input

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

output

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