3.4.52 \(\int \frac {\cos (c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [352]

3.4.52.1 Optimal result

Integrand size = 39, antiderivative size = 154 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 C (b \cos (c+d x))^{5/3} \sin (c+d x)}{8 b^2 d}-\frac {3 (8 A+5 C) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{40 b^2 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^3 d \sqrt {\sin ^2(c+d x)}} \]

output

3/8*C*(b*cos(d*x+c))^(5/3)*sin(d*x+c)/b^2/d-3/40*(8*A+5*C)*(b*cos(d*x+c))^ 
(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/b^2/d/(sin(d*x+ 
c)^2)^(1/2)-3/8*B*(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)
 

3.4.52.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.81 \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=-\frac {3 (b \cos (c+d x))^{2/3} \left (2 (8 A+5 C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+10 B \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-5 C \sin (2 (c+d x))\right )}{80 b d} \]

input

Integrate[(Cos[c + d*x]*(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)*(2*(8*A + 5*C)*Cot[c + d*x]*Hypergeometric2F1[1 
/2, 5/6, 11/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 10*B*Cos[c + d*x]*Co 
t[c + d*x]*Hypergeometric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sqrt[Sin[c + d 
*x]^2] - 5*C*Sin[2*(c + d*x)]))/(80*b*d)
 

3.4.52.3 Rubi [A] (verified)

Time = 0.51 (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.

\(\displaystyle \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{2/3} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{b}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {3 \int \frac {1}{3} (b \cos (c+d x))^{2/3} (b (8 A+5 C)+8 b B \cos (c+d x))dx}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int (b \cos (c+d x))^{2/3} (b (8 A+5 C)+8 b B \cos (c+d x))dx}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \left (b (8 A+5 C)+8 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {b (8 A+5 C) \int (b \cos (c+d x))^{2/3}dx+8 B \int (b \cos (c+d x))^{5/3}dx}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b (8 A+5 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}dx+8 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}dx}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {\frac {-\frac {3 (8 A+5 C) \sin (c+d x) (b \cos (c+d x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{8/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{3},\frac {7}{3},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{8 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{5/3}}{8 b d}}{b}\)

input

Int[(Cos[c + d*x]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x] 
)^(1/3),x]
 

output

((3*C*(b*Cos[c + d*x])^(5/3)*Sin[c + d*x])/(8*b*d) + ((-3*(8*A + 5*C)*(b*C 
os[c + d*x])^(5/3)*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c 
 + d*x])/(5*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(8/3)*Hypergeo 
metric2F1[1/2, 4/3, 7/3, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d 
*x]^2]))/(8*b))/b
 

3.4.52.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2030

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3122

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]
 

rule 3227

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]
 

rule 3502

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]
 

3.4.52.4 Maple [F]

\[\int \frac {\cos \left (d x +c \right ) \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (\cos \left (d x +c \right ) b \right )^{\frac {1}{3}}}d x\]

input

int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(1/3),x)
 

output

int(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(1/3),x)
 

3.4.52.5 Fricas [F]

\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), 
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, 
 x)
 

3.4.52.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\text {Timed out} \]

input

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(1/3 
),x)
 

output

Timed out
 

3.4.52.7 Maxima [F]

\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), 
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))^(1/3), x)
 

3.4.52.8 Giac [F]

\[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}} \,d x } \]

input

integrate(cos(d*x+c)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/3), 
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))^(1/3), x)
 

3.4.52.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]

input

int((cos(c + d*x)*(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)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x) 
)^(1/3), x)