Integrand size = 34, antiderivative size = 281 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\frac {3 C (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b d}+\frac {\sqrt {2} (5 b B-3 a C) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) (a+b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} \left (5 a b B-3 a^2 C-2 b^2 C\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \sin (c+d x)}{5 b^2 d \sqrt {1+\cos (c+d x)} \sqrt [3]{a+b \cos (c+d x)}} \] Output:
3/5*C*(a+b*cos(d*x+c))^(2/3)*sin(d*x+c)/b/d+1/5*2^(1/2)*(5*B*b-3*C*a)*Appe llF1(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)*(5*B*a*b-3*C*a^2-2*C*b^2)*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)*si n(d*x+c)/b^2/d/(1+cos(d*x+c))^(1/2)/(a+b*cos(d*x+c))^(1/3)
Time = 3.68 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.94 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=-\frac {3 (a+b \cos (c+d x))^{2/3} \csc (c+d x) \left (5 \left (-5 a b B+3 a^2 C+2 b^2 C\right ) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {-\frac {b (1+\cos (c+d x))}{a-b}}+2 (5 b B-3 a C) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) \sqrt {-\frac {b (-1+\cos (c+d x))}{a+b}} \sqrt {\frac {b (1+\cos (c+d x))}{-a+b}} (a+b \cos (c+d x))-10 b^2 C \sin ^2(c+d x)\right )}{50 b^3 d} \] Input:
Integrate[(B*Cos[c + d*x] + 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*B + 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 + Cos[c + d*x]))/(a - b))] + 2*(5*b*B - 3*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*C os[c + d*x]) - 10*b^2*C*Sin[c + d*x]^2))/(50*b^3*d)
Time = 0.67 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3502, 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.
\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {3 \int \frac {2 b C+(5 b B-3 a C) \cos (c+d x)}{3 \sqrt [3]{a+b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 b C+(5 b B-3 a C) \cos (c+d x)}{\sqrt [3]{a+b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 b C+(5 b B-3 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3235 |
\(\displaystyle \frac {\frac {(5 b B-3 a C) \int (a+b \cos (c+d x))^{2/3}dx}{b}-\frac {\left (-3 a^2 C+5 a b B-2 b^2 C\right ) \int \frac {1}{\sqrt [3]{a+b \cos (c+d x)}}dx}{b}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {(5 b B-3 a C) \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{b}-\frac {\left (-3 a^2 C+5 a b B-2 b^2 C\right ) \int \frac {1}{\sqrt [3]{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 3144 |
\(\displaystyle \frac {\frac {\left (-3 a^2 C+5 a b B-2 b^2 C\right ) \sin (c+d x) \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}-\frac {(5 b B-3 a C) \sin (c+d x) \int \frac {(a+b \cos (c+d x))^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {\frac {\left (-3 a^2 C+5 a b B-2 b^2 C\right ) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}-\frac {(5 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \int \frac {\left (\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}\right )^{2/3}}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1}}d\cos (c+d x)}{b d \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\frac {\sqrt {2} (5 b B-3 a C) \sin (c+d x) (a+b \cos (c+d x))^{2/3} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-\frac {2}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \left (\frac {a+b \cos (c+d x)}{a+b}\right )^{2/3}}-\frac {\sqrt {2} \left (-3 a^2 C+5 a b B-2 b^2 C\right ) \sin (c+d x) \sqrt [3]{\frac {a+b \cos (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\cos (c+d x)),\frac {b (1-\cos (c+d x))}{a+b}\right )}{b d \sqrt {\cos (c+d x)+1} \sqrt [3]{a+b \cos (c+d x)}}}{5 b}+\frac {3 C \sin (c+d x) (a+b \cos (c+d x))^{2/3}}{5 b d}\) |
Input:
Int[(B*Cos[c + d*x] + 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) + ((Sqrt[2]*(5*b*B - 3*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])^(2/3)*Sin[c + d*x])/(b*d*Sqrt[1 + C os[c + d*x]]*((a + b*Cos[c + d*x])/(a + b))^(2/3)) - (Sqrt[2]*(5*a*b*B - 3 *a^2*C - 2*b^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*Sqrt[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_.) + (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]
\[\int \frac {B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}}{\left (a +b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
Output:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorith m="fricas")
Output:
integral((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), x )
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\sqrt [3]{a + b \cos {\left (c + d x \right )}}}\, dx \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/3),x)
Output:
Integral((B + C*cos(c + d*x))*cos(c + d*x)/(a + b*cos(c + d*x))**(1/3), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorith m="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x, algorith m="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/(b*cos(d*x + c) + a)^(1/3), x)
Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3),x)
Output:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(a + b*cos(c + d*x))^(1/3), x)
\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt [3]{a+b \cos (c+d x)}} \, dx=\left (\int \frac {\cos \left (d x +c \right )}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {\cos \left (d x +c \right )^{2}}{\left (\cos \left (d x +c \right ) b +a \right )^{\frac {1}{3}}}d x \right ) c \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/3),x)
Output:
int(cos(c + d*x)/(cos(c + d*x)*b + a)**(1/3),x)*b + int(cos(c + d*x)**2/(c os(c + d*x)*b + a)**(1/3),x)*c