Integrand size = 27, antiderivative size = 274 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}-\frac {3 \sqrt {2} a C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}+\frac {\sqrt {2} \left (3 a^2 C+b^2 (5 A+2 C)\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \] Output:
3/5*C*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b/d-3/5*2^(1/2)*a*C*AppellF1(1/2,- 2/3,1/2,3/2,b*(1-cos(d*x+c))/(a+b),1/2-1/2*cos(d*x+c))*(a+b*cos(d*x+c))^(2 /3)*sin(d*x+c)/b^2/d/(1+cos(d*x+c))^(1/2)/((a+b*cos(d*x+c))/(a+b))^(2/3)+1 /5*2^(1/2)*(3*a^2*C+b^2*(5*A+2*C))*AppellF1(1/2,1/3,1/2,3/2,b*(1-cos(d*x+c ))/(a+b),1/2-1/2*cos(d*x+c))*((a+b*cos(d*x+c))/(a+b))^(1/3)*sin(d*x+c)/b^2 /d/(1+cos(d*x+c))^(1/2)/(a+b*cos(d*x+c))^(1/3)
Time = 2.10 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.93 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=-\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (5 \left (5 A b^2+3 a^2 C+2 b^2 C\right ) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}}-6 a C \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-10 b^2 C \sin ^2(c+d x)\right )}{50 b^3 d} \] Input:
Integrate[(A + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(1/3),x]
Output:
(-3*(a + b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(5*(5*A*b^2 + 3*a^2*C + 2*b^2* C)*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Co s[c + d*x]))/(-a + b)] - 6*a*C*AppellF1[5/3, 1/2, 1/2, 8/3, (a + b*Cos[c + d*x])/(a - b), (a + b*Cos[c + d*x])/(a + b)]*Sqrt[-((b*(-1 + Cos[c + d*x] ))/(a + b))]*Sqrt[(b*(1 + Cos[c + d*x]))/(-a + b)]*(a + b*Cos[c + d*x]) - 10*b^2*C*Sin[c + d*x]^2))/(50*b^3*d)
Time = 0.65 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3503, 27, 3042, 3235, 3042, 3144, 156, 155}
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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3503 |
\(\displaystyle \frac {3 \int \frac {b (5 A+2 C)-3 a C \cos (c+d x)}{3 \sqrt [3]{a+b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b (5 A+2 C)-3 a C \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {b (5 A+2 C)-3 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (5 A+2 C)\right ) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}}dx}{b}-\frac {3 a C \int (a+b \cos (c+d x))^{2/3}dx}{b}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (5 A+2 C)\right ) \int \frac {1}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {3 a C \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\frac {3 a C \sin (c+d x) \int \frac {(a+b \cos (c+d x))^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {\left (3 a^2 C+b^2 (5 A+2 C)\right ) \sin (c+d x) \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\frac {3 a C \sin (c+d x) (a+b \cos (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\left (3 a^2 C+b^2 (5 A+2 C)\right ) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\frac {\sqrt {2} \left (3 a^2 C+b^2 (5 A+2 C)\right ) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}-\frac {3 \sqrt {2} a C \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
Input:
Int[(A + C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^(1/3),x]
Output:
(3*C*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b*d) + ((-3*Sqrt[2]*a*C*A ppellF1[1/2, 1/2, -2/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/ (a + b)]*(a + b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d* x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) + (Sqrt[2]*(3*a^2*C + b^2*(5*A + 2*C))*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Cos[c + d*x])/2, (b*(1 - Cos[c + d*x]))/(a + b)]*((a + b*Cos[c + d*x])/(a + b))^(1/3)*Sin[c + d*x])/(b*d*Sq rt[1 + Cos[c + d*x]]*(a + b*Cos[c + d*x])^(1/3)))/(5*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]]) Subst[Int[(a + b*x )^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)/b Int[(a + b*Sin[e + f*x])^m, x], x] + Simp[d/b Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a , b, e, f, A, C, m}, x] && !LtQ[m, -1]
\[\int \frac {A +C \cos \left (d x +c \right )^{2}}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
Output:
int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c) + a)^(1/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\sqrt [3]{a + b \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/3),x)
Output:
Integral((A + C*cos(c + d*x)**2)/(a + b*cos(c + d*x))**(1/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c) + a)^(1/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + A)/(b*cos(d*x + c) + a)^(1/3), x)
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:
int((A + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3),x)
Output:
int((A + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\left (\int \frac {\cos \left (d x +c \right )^{2}}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) c +\left (\int \frac {1}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) a \] Input:
int((A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
Output:
int(cos(c + d*x)**2/(cos(c + d*x)*b + a)**(1/3),x)*c + int(1/(cos(c + d*x) *b + a)**(1/3),x)*a