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

Optimal. Leaf size=148 \[ \frac{3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x)}{d (3 m+10)}-\frac{3 b (A (3 m+10)+C (3 m+7)) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d (3 m+7) (3 m+10) \sqrt{\sin ^2(c+d x)}} \]

[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*(C*(7 + 3*m) + A*(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*(7 + 3*m)*(10 + 3*m)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.113143, antiderivative size = 138, 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, 3014, 2643} \[ \frac{3 b C \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x)}{d (3 m+10)}-\frac{3 b \left (\frac{A}{3 m+7}+\frac{C}{3 m+10}\right ) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[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 3014

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]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+C \cos ^2(c+d x)\right ) \, dx &=\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)} \, _2F_1\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]  time = 0.279265, size = 142, normalized size = 0.96 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^{4/3} \cos ^{m+1}(c+d x) \left (A (3 m+13) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+7);\frac{1}{6} (3 m+13);\cos ^2(c+d x)\right )+C (3 m+7) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+13);\frac{1}{6} (3 m+19);\cos ^2(c+d x)\right )\right )}{d (3 m+7) (3 m+13)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.309, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + A b \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \cos \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)