3.4.66 \(\int \frac {\cos ^m(c+d x) (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt [3]{b \cos (c+d x)}} \, dx\) [366]

3.4.66.1 Optimal result

Integrand size = 41, antiderivative size = 229 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 C \cos ^{1+m}(c+d x) \sin (c+d x)}{d (5+3 m) \sqrt [3]{b \cos (c+d x)}}-\frac {3 (C (2+3 m)+A (5+3 m)) \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+3 m) (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}} \]

output

3*C*cos(d*x+c)^(1+m)*sin(d*x+c)/d/(5+3*m)/(b*cos(d*x+c))^(1/3)-3*(C*(2+3*m 
)+A*(5+3*m))*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d 
*x+c)^2)*sin(d*x+c)/d/(9*m^2+21*m+10)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^ 
(1/2)-3*B*cos(d*x+c)^(2+m)*hypergeom([1/2, 5/6+1/2*m],[11/6+1/2*m],cos(d*x 
+c)^2)*sin(d*x+c)/d/(5+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)
 

3.4.66.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt [3]{b \cos (c+d x)}} \, dx=\frac {3 \cos ^{1+m}(c+d x) \csc (c+d x) \left (-\left ((C (2+3 m)+A (5+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (2+3 m),\frac {1}{6} (8+3 m),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )+(2+3 m) \left (C \sin ^2(c+d x)-B \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (5+3 m),\frac {1}{6} (11+3 m),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )\right )}{d (2+3 m) (5+3 m) \sqrt [3]{b \cos (c+d x)}} \]

input

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

output

(3*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(-((C*(2 + 3*m) + A*(5 + 3*m))*Hyperg 
eometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d* 
x]^2]) + (2 + 3*m)*(C*Sin[c + d*x]^2 - B*Cos[c + d*x]*Hypergeometric2F1[1/ 
2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2])))/(d*( 
2 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(1/3))
 

3.4.66.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2034, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {3 \int \frac {1}{3} \cos ^{m-\frac {1}{3}}(c+d x) \left (3 C \left (m+\frac {2}{3}\right )+3 A \left (m+\frac {5}{3}\right )+B (3 m+5) \cos (c+d x)\right )dx}{3 m+5}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x)}{d (3 m+5)}\right )}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {\int \cos ^{m-\frac {1}{3}}(c+d x) (C (3 m+2)+A (3 m+5)+B (3 m+5) \cos (c+d x))dx}{3 m+5}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x)}{d (3 m+5)}\right )}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}} \left (C (3 m+2)+A (3 m+5)+B (3 m+5) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{3 m+5}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x)}{d (3 m+5)}\right )}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {(A (3 m+5)+C (3 m+2)) \int \cos ^{m-\frac {1}{3}}(c+d x)dx+B (3 m+5) \int \cos ^{m+\frac {2}{3}}(c+d x)dx}{3 m+5}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x)}{d (3 m+5)}\right )}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [3]{\cos (c+d x)} \left (\frac {(A (3 m+5)+C (3 m+2)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m-\frac {1}{3}}dx+B (3 m+5) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {2}{3}}dx}{3 m+5}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {2}{3}}(c+d x)}{d (3 m+5)}\right )}{\sqrt [3]{b \cos (c+d x)}}\)

\(\Big \downarrow \) 3122

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

input

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

output

(Cos[c + d*x]^(1/3)*((3*C*Cos[c + d*x]^(2/3 + m)*Sin[c + d*x])/(d*(5 + 3*m 
)) + ((-3*(C*(2 + 3*m) + A*(5 + 3*m))*Cos[c + d*x]^(2/3 + m)*Hypergeometri 
c2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(2 + 
3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(5/3 + m)*Hypergeometric2F1 
[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqrt[Sin 
[c + d*x]^2]))/(5 + 3*m)))/(b*Cos[c + d*x])^(1/3)
 

3.4.66.3.1 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 2034

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart 
[n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n]))   Int[(a*v)^(m + n 
)*Fx, x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && 
  !IntegerQ[m + n]
 

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]
 

3.4.66.4 Maple [F]

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

input

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

3.4.66.5 Fricas [F]

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

input

integrate(cos(d*x+c)^m*(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)*co 
s(d*x + c)^m/(b*cos(d*x + c)), x)
 

3.4.66.6 Sympy [F]

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

input

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

output

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

3.4.66.7 Maxima [F]

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

input

integrate(cos(d*x+c)^m*(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)^m/(b*cos(d* 
x + c))^(1/3), x)
 

3.4.66.8 Giac [F]

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

input

integrate(cos(d*x+c)^m*(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)^m/(b*cos(d* 
x + c))^(1/3), x)
 

3.4.66.9 Mupad [F(-1)]

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

input

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