Integrand size = 41, antiderivative size = 154 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 C (b \cos (c+d x))^{8/3} \sin (c+d x)}{11 b^3 d}-\frac {3 (11 A+8 C) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{88 b^3 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{11/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{11 b^4 d \sqrt {\sin ^2(c+d x)}} \] Output:
3/11*C*(b*cos(d*x+c))^(8/3)*sin(d*x+c)/b^3/d-3/88*(11*A+8*C)*(b*cos(d*x+c) )^(8/3)*hypergeom([1/2, 4/3],[7/3],cos(d*x+c)^2)*sin(d*x+c)/b^3/d/(sin(d*x +c)^2)^(1/2)-3/11*B*(b*cos(d*x+c))^(11/3)*hypergeom([1/2, 11/6],[17/6],cos (d*x+c)^2)*sin(d*x+c)/b^4/d/(sin(d*x+c)^2)^(1/2)
Time = 1.40 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 (b \cos (c+d x))^{2/3} \sin (c+d x) \left ((11 A+8 C) \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+8 \cos (c+d x) \left (-C \cos (c+d x)+B \cot ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{88 b d} \] Input:
Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[ c + d*x])^(1/3),x]
Output:
(-3*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x]*((11*A + 8*C)*Cot[c + d*x]^2*Hyper geometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 8*Cos[c + d*x]*(-(C*Cos[c + d*x]) + B*Cot[c + d*x]^2*Hypergeometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(88*b*d)
Time = 0.52 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2030, 3042, 3502, 27, 3042, 3227, 3042, 3122}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int (b \cos (c+d x))^{5/3} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b^2}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{3} (b \cos (c+d x))^{5/3} (b (11 A+8 C)+11 b B \cos (c+d x))dx}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int (b \cos (c+d x))^{5/3} (b (11 A+8 C)+11 b B \cos (c+d x))dx}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3} \left (b (11 A+8 C)+11 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {b (11 A+8 C) \int (b \cos (c+d x))^{5/3}dx+11 B \int (b \cos (c+d x))^{8/3}dx}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (11 A+8 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx+11 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{8/3}dx}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {-\frac {3 (11 A+8 C) \sin (c+d x) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )}{8 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{11/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {11}{6},\frac {17}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{11 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{8/3}}{11 b d}}{b^2}\) |
Input:
Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d* x])^(1/3),x]
Output:
((3*C*(b*Cos[c + d*x])^(8/3)*Sin[c + d*x])/(11*b*d) + ((-3*(11*A + 8*C)*(b *Cos[c + d*x])^(8/3)*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[ c + d*x])/(8*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(11/3)*Hyperg eometric2F1[1/2, 11/6, 17/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]))/(11*b))/b^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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]
\[\int \frac {\cos \left (d x +c \right )^{2} \left (A +B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Input:
int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3),x)
Output:
int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3),x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3 ),x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^3 + B*cos(d*x + c)^2 + A*cos(d*x + c))*(b*cos(d*x + c))^(2/3)/b, x)
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**( 1/3),x)
Output:
Timed out
\[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3 ),x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d* x + c))^(1/3), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3 ),x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d* x + c))^(1/3), x)
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \] Input:
int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d* x))^(1/3),x)
Output:
int((cos(c + d*x)^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d* x))^(1/3), x)
\[ \int \frac {\cos ^2(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {\left (\int \cos \left (d x +c \right )^{\frac {5}{3}}d x \right ) a +\left (\int \cos \left (d x +c \right )^{\frac {11}{3}}d x \right ) c +\left (\int \cos \left (d x +c \right )^{\frac {8}{3}}d x \right ) b}{b^{\frac {1}{3}}} \] Input:
int(cos(d*x+c)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3),x)
Output:
(int(cos(c + d*x)**4/cos(c + d*x)**(1/3),x)*c + int(cos(c + d*x)**3/cos(c + d*x)**(1/3),x)*b + int(cos(c + d*x)**2/cos(c + d*x)**(1/3),x)*a)/b**(1/3 )