3.4.51 \(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [351]

3.4.51.1 Optimal result

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)
 

3.4.51.2 Mathematica [A] (verified)

Time = 0.87 (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)
 

3.4.51.3 Rubi [A] (verified)

Time = 0.54 (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
 

3.4.51.3.1 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]
 

3.4.51.4 Maple [F]

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

input

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

3.4.51.5 Fricas [F]

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

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)
 

3.4.51.6 Sympy [F(-1)]

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
 

3.4.51.7 Maxima [F]

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

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)
 

3.4.51.8 Giac [F]

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

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)
 

3.4.51.9 Mupad [F(-1)]

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)