\(\int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} (A+C \cos ^2(c+d x)) \, dx\) [176]

Optimal result

Integrand size = 33, antiderivative size = 148 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}-\frac {3 b (C (7+3 m)+A (10+3 m)) \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) (10+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*b*C*cos(d*x+c)^(2+m)*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/d/(10+3*m)-3*b*(C*(7+3*m)+A*(10+3*m))*cos(d*x+c)^(2+m)*
(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 7/6+1/2*m],[13/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(9*m^2+51*m+70)/(sin(d
*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 138, 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} \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x)}{d (3 m+10)}-\frac {3 b \left (\frac {A}{3 m+7}+\frac {C}{3 m+10}\right ) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+7),\frac {1}{6} (3 m+13),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}} \]

[In]

Int[Cos[c + d*x]^m*(b*Cos[c + d*x])^(4/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(3*b*C*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Sin[c + d*x])/(d*(10 + 3*m)) - (3*b*(A/(7 + 3*m) + C/(10 +
3*m))*Cos[c + d*x]^(2 + m)*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 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 {\left (b \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\sqrt [3]{\cos (c+d x)}} \\ & = \frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}+\frac {\left (b \left (C \left (\frac {7}{3}+m\right )+A \left (\frac {10}{3}+m\right )\right ) \sqrt [3]{b \cos (c+d x)}\right ) \int \cos ^{\frac {4}{3}+m}(c+d x) \, dx}{\left (\frac {10}{3}+m\right ) \sqrt [3]{\cos (c+d x)}} \\ & = \frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}-\frac {3 b (C (7+3 m)+A (10+3 m)) \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) (10+3 m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.96 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{4/3} \csc (c+d x) \left (A (13+3 m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right )+C (7+3 m) \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (13+3 m),\frac {1}{6} (19+3 m),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+3 m) (13+3 m)} \]

[In]

Integrate[Cos[c + d*x]^m*(b*Cos[c + d*x])^(4/3)*(A + C*Cos[c + d*x]^2),x]

[Out]

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(4/3)*Csc[c + d*x]*(A*(13 + 3*m)*Hypergeometric2F1[1/2, (7 + 3*m)/6,
(13 + 3*m)/6, Cos[c + d*x]^2] + C*(7 + 3*m)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (13 + 3*m)/6, (19 + 3*m)/6,
Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(7 + 3*m)*(13 + 3*m))

Maple [F]

\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]

[In]

int(cos(d*x+c)^m*(cos(d*x+c)*b)^(4/3)*(A+C*cos(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^m*(cos(d*x+c)*b)^(4/3)*(A+C*cos(d*x+c)^2),x)

Fricas [F]

\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**m*(b*cos(d*x+c))**(4/3)*(A+C*cos(d*x+c)**2),x)

[Out]

Timed out

Maxima [F]

\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(4/3)*cos(d*x + c)^m, x)

Giac [F]

\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(4/3)*cos(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3} \,d x \]

[In]

int(cos(c + d*x)^m*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(4/3),x)

[Out]

int(cos(c + d*x)^m*(A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(4/3), x)