Integrand size = 39, antiderivative size = 154 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {3 C (b \cos (c+d x))^{2/3} \sin (c+d x)}{5 b^2 d}-\frac {3 (5 A+2 C) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{10 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^3 d \sqrt {\sin ^2(c+d x)}} \] Output:
3/5*C*(b*cos(d*x+c))^(2/3)*sin(d*x+c)/b^2/d-3/10*(5*A+2*C)*(b*cos(d*x+c))^ (2/3)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+c )^2)^(1/2)-3/5*B*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+ c)^2)*sin(d*x+c)/b^3/d/(sin(d*x+c)^2)^(1/2)
Time = 0.02 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {-3 (5 A+2 C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-6 B \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+3 C \sin (2 (c+d x))}{10 b d \sqrt [3]{b \cos (c+d x)}} \] Input:
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(4/3),x]
Output:
(-3*(5*A + 2*C)*Cot[c + d*x]*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x] ^2]*Sqrt[Sin[c + d*x]^2] - 6*B*Cos[c + d*x]*Cot[c + d*x]*Hypergeometric2F1 [1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 3*C*Sin[2*(c + d*x )])/(10*b*d*(b*Cos[c + d*x])^(1/3))
Time = 0.49 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, 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.
\(\displaystyle \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle \frac {\int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\sqrt [3]{b \cos (c+d x)}}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {3 \int \frac {b (5 A+2 C)+5 b B \cos (c+d x)}{3 \sqrt [3]{b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {b (5 A+2 C)+5 b B \cos (c+d x)}{\sqrt [3]{b \cos (c+d x)}}dx}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {b (5 A+2 C)+5 b B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle \frac {\frac {b (5 A+2 C) \int \frac {1}{\sqrt [3]{b \cos (c+d x)}}dx+5 B \int (b \cos (c+d x))^{2/3}dx}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (5 A+2 C) \int \frac {1}{\sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+5 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\frac {-\frac {3 (5 A+2 C) \sin (c+d x) (b \cos (c+d x))^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{5 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3}}{5 b d}}{b}\) |
Input:
Int[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x] )^(4/3),x]
Output:
((3*C*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(5*b*d) + ((-3*(5*A + 2*C)*(b*C os[c + d*x])^(2/3)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(2*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(5/3)*Hypergeom etric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d *x]^2]))/(5*b))/b
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]
\[\int \frac {\cos \left (d x +c \right ) \left (A +B \cos \left (d x +c \right )+C \cos \left (d x +c \right )^{2}\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}d x\]
Input:
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
Output:
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), x, algorithm="fricas")
Output:
integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)/(b ^2*cos(d*x + c)), x)
Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(4/3 ),x)
Output:
Timed out
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), x, algorithm="maxima")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/(b*cos(d*x + c))^(4/3), x)
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}} \,d x } \] Input:
integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3), x, algorithm="giac")
Output:
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*cos(d*x + c)/(b*cos(d*x + c))^(4/3), x)
Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \] Input:
int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x) )^(4/3),x)
Output:
int((cos(c + d*x)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x) )^(4/3), x)
\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx=\frac {\left (\int \cos \left (d x +c \right )^{\frac {2}{3}}d x \right ) b +\left (\int \cos \left (d x +c \right )^{\frac {5}{3}}d x \right ) c +\left (\int \frac {1}{\cos \left (d x +c \right )^{\frac {1}{3}}}d x \right ) a}{b^{\frac {4}{3}}} \] Input:
int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)
Output:
(int(cos(c + d*x)/cos(c + d*x)**(1/3),x)*b + int(cos(c + d*x)**2/cos(c + d *x)**(1/3),x)*c + int(1/cos(c + d*x)**(1/3),x)*a)/(b**(1/3)*b)