\(\int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [178]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 146 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (7+3 m)}-\frac {3 (C (4+3 m)+A (7+3 m)) \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {1}{6} (10+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (4+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*cos(d*x+c)^(1+m)*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/d/(7+3*m)-3*(C*(4+3*m)+A*(7+3*m))*cos(d*x+c)^(1+m)*(b*cos
(d*x+c))^(1/3)*hypergeom([1/2, 2/3+1/2*m],[5/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(9*m^2+33*m+28)/(sin(d*x+c)^2
)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {20, 3093, 2722} \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+1}(c+d x)}{d (3 m+7)}-\frac {3 \left (\frac {A}{3 m+4}+\frac {C}{3 m+7}\right ) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+4),\frac {1}{6} (3 m+10),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(3*C*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(d*(7 + 3*m)) - (3*(A/(4 + 3*m) + C/(7 + 3*m))*
Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/2, (4 + 3*m)/6, (10 + 3*m)/6, Cos[c + d*x]^2]*
Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2])

Rule 20

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

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3093

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos
[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[(A*(m + 2) + C*(m + 1))/(m + 2), Int[(b*Sin[e +
f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{b \cos (c+d x)} \int \cos ^{\frac {1}{3}+m}(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\sqrt [3]{\cos (c+d x)}} \\ & = \frac {3 C \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (7+3 m)}+\frac {\left (\left (C \left (\frac {4}{3}+m\right )+A \left (\frac {7}{3}+m\right )\right ) \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {1}{3}+m}(c+d x) \, dx}{\left (\frac {7}{3}+m\right ) \sqrt [3]{\cos (c+d x)}} \\ & = \frac {3 C \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (7+3 m)}-\frac {3 (C (4+3 m)+A (7+3 m)) \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {1}{6} (10+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (4+3 m) (7+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97 \[ \int \cos ^m(c+d x) \sqrt [3]{b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \cos ^{1+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \csc (c+d x) \left (C (4+3 m) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\cos ^2(c+d x)\right )+A (10+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (4+3 m),\frac {5}{3}+\frac {m}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (4+3 m) (10+3 m)} \]

[In]

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

[Out]

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(1/3)*Csc[c + d*x]*(C*(4 + 3*m)*Cos[c + d*x]^2*Hypergeometric2F1[1/2
, 5/3 + m/2, 8/3 + m/2, Cos[c + d*x]^2] + A*(10 + 3*m)*Hypergeometric2F1[1/2, (4 + 3*m)/6, 5/3 + m/2, Cos[c +
d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(4 + 3*m)*(10 + 3*m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(1/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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