Integrand size = 27, antiderivative size = 272 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\frac {3 C \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{4 b d}-\frac {3 a C \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{a+b \cos (c+d x)} \sin (c+d x)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \sin (c+d x)}{2 \sqrt {2} b^2 d \sqrt {1+\cos (c+d x)} (a+b \cos (c+d x))^{2/3}} \]
3/4*C*(a+b*cos(d*x+c))^(1/3)*sin(d*x+c)/b/d-3/4*a*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))^(1/3)*sin( d*x+c)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(1/3)*2^(1/2)/(1+cos(d*x+c))^(1/2)+1 /4*(3*a^2*C+b^2*(4*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))/(a+b))^(2/3)*sin(d*x+c)/b^2/d/(a+b*co s(d*x+c))^(2/3)*2^(1/2)/(1+cos(d*x+c))^(1/2)
Time = 1.46 (sec) , antiderivative size = 256, normalized size of antiderivative = 0.94 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=-\frac {3 \sqrt [3]{a+b \cos (c+d x)} \csc (c+d x) \left (4 \left (4 A b^2+\left (3 a^2+b^2\right ) C\right ) \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}}+C \left (-3 a \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-4 b^2 \sin ^2(c+d x)\right )\right )}{16 b^3 d} \]
(-3*(a + b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(4*(4*A*b^2 + (3*a^2 + b^2)*C) *AppellF1[1/3, 1/2, 1/2, 4/3, (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)] + C*(-3*a*AppellF1[4/3, 1/2, 1/2, 7/3, (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]) - 4*b^2*Sin[c + d*x]^2)))/(16*b^3*d)
Time = 0.63 (sec) , antiderivative size = 275, 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.
\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
\(\Big \downarrow \) 3503 |
\(\displaystyle \frac {3 \int \frac {b (4 A+C)-3 a C \cos (c+d x)}{3 (a+b \cos (c+d x))^{2/3}}dx}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {b (4 A+C)-3 a C \cos (c+d x)}{(a+b \cos (c+d x))^{2/3}}dx}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {b (4 A+C)-3 a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \int \frac {1}{(a+b \cos (c+d x))^{2/3}}dx}{b}-\frac {3 a C \int \sqrt [3]{a+b \cos (c+d x)}dx}{b}}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \int \frac {1}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{b}-\frac {3 a C \int \sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\frac {3 a C \sin (c+d x) \int \frac {\sqrt [3]{a+b \cos (c+d x)}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\frac {3 a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \int \frac {\sqrt [3]{\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{2/3}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\frac {\sqrt {2} \left (3 a^2 C+b^2 (4 A+C)\right ) \sin (c+d x) \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} (a+b \cos (c+d x))^{2/3}}-\frac {3 \sqrt {2} a C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}}}}{4 b}+\frac {3 C \sin (c+d x) \sqrt [3]{a+b \cos (c+d x)}}{4 b d}\) |
(3*C*(a + b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(4*b*d) + ((-3*Sqrt[2]*a*C*A ppellF1[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])^(1/3)*Sin[c + d*x])/(b*d*Sqrt[1 + Cos[c + d* x]]*((a + b*Cos[c + d*x])/(a + b))^(1/3)) + (Sqrt[2]*(3*a^2*C + b^2*(4*A + C))*AppellF1[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])/(a + b))^(2/3)*Sin[c + d*x])/(b*d*Sqrt [1 + Cos[c + d*x]]*(a + b*Cos[c + d*x])^(2/3)))/(4*b)
3.3.6.3.1 Defintions of rubi rules used
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]
\[\int \frac {A +C \left (\cos ^{2}\left (d x +c \right )\right )}{\left (a +\cos \left (d x +c \right ) b \right )^{\frac {2}{3}}}d x\]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^{2/3}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{2/3}} \,d x \]