3.2.77 \(\int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} (A+C \cos ^2(c+d x)) \, dx\) [177]

Optimal. Leaf size=146 \[ \frac {3 C \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \sin (c+d x)}{d (8+3 m)}-\frac {3 (C (5+3 m)+A (8+3 m)) \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5+3 m);\frac {1}{6} (11+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+3 m) (8+3 m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

3*C*cos(d*x+c)^(1+m)*(b*cos(d*x+c))^(2/3)*sin(d*x+c)/d/(8+3*m)-3*(C*(5+3*m)+A*(8+3*m))*cos(d*x+c)^(1+m)*(b*cos
(d*x+c))^(2/3)*hypergeom([1/2, 5/6+1/2*m],[11/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(9*m^2+39*m+40)/(sin(d*x+c)^
2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 136, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {20, 3093, 2722} \begin {gather*} \frac {3 C \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+1}(c+d x)}{d (3 m+8)}-\frac {3 \left (\frac {A}{3 m+5}+\frac {C}{3 m+8}\right ) \sin (c+d x) (b \cos (c+d x))^{2/3} \cos ^{m+1}(c+d x) \, _2F_1\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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*Sin[c + d*x])/(d*(8 + 3*m)) - (3*(A/(5 + 3*m) + C/(8 + 3*m))*
Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*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])

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*} \int \cos ^m(c+d x) (b \cos (c+d x))^{2/3} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\frac {(b \cos (c+d x))^{2/3} \int \cos ^{\frac {2}{3}+m}(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{\cos ^{\frac {2}{3}}(c+d x)}\\ &=\frac {3 C \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \sin (c+d x)}{d (8+3 m)}+\frac {\left (\left (C \left (\frac {5}{3}+m\right )+A \left (\frac {8}{3}+m\right )\right ) (b \cos (c+d x))^{2/3}\right ) \int \cos ^{\frac {2}{3}+m}(c+d x) \, dx}{\left (\frac {8}{3}+m\right ) \cos ^{\frac {2}{3}}(c+d x)}\\ &=\frac {3 C \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \sin (c+d x)}{d (8+3 m)}-\frac {3 (C (5+3 m)+A (8+3 m)) \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \, _2F_1\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) (8+3 m) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 142, normalized size = 0.97 \begin {gather*} -\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{2/3} \csc (c+d x) \left (A (11+3 m) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5+3 m);\frac {1}{6} (11+3 m);\cos ^2(c+d x)\right )+C (5+3 m) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (11+3 m);\frac {1}{6} (17+3 m);\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (5+3 m) (11+3 m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(2/3)*Csc[c + d*x]*(A*(11 + 3*m)*Hypergeometric2F1[1/2, (5 + 3*m)/6,
(11 + 3*m)/6, Cos[c + d*x]^2] + C*(5 + 3*m)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (11 + 3*m)/6, (17 + 3*m)/6,
Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(5 + 3*m)*(11 + 3*m))

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^{2/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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