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

Optimal. Leaf size=149 \[ \frac{3 C \sin (c+d x) \cos ^m(c+d x)}{b d (3 m+2) \sqrt [3]{b \cos (c+d x)}}-\frac{3 (C (1-3 m)-A (3 m+2)) \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m-1);\frac{1}{6} (3 m+5);\cos ^2(c+d x)\right )}{b d (1-3 m) (3 m+2) \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.122399, antiderivative size = 139, 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} \[ \frac{3 \left (\frac{A}{1-3 m}-\frac{C}{3 m+2}\right ) \sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m-1);\frac{1}{6} (3 m+5);\cos ^2(c+d x)\right )}{b d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac{3 C \sin (c+d x) \cos ^m(c+d x)}{b d (3 m+2) \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*Cos[c + d*x]^m*Sin[c + d*x])/(b*d*(2 + 3*m)*(b*Cos[c + d*x])^(1/3)) + (3*(A/(1 - 3*m) - C/(2 + 3*m))*Cos[
c + d*x]^m*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(b*Cos[c + d*x
])^(1/3)*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 \frac{\cos ^m(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx &=\frac{\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac{4}{3}+m}(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=\frac{3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}+\frac{\left (\left (C \left (-\frac{1}{3}+m\right )+A \left (\frac{2}{3}+m\right )\right ) \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac{4}{3}+m}(c+d x) \, dx}{b \left (\frac{2}{3}+m\right ) \sqrt [3]{b \cos (c+d x)}}\\ &=\frac{3 C \cos ^m(c+d x) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)}}-\frac{3 (C (1-3 m)-A (2+3 m)) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (-1+3 m);\frac{1}{6} (5+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (1-3 m) (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.25823, size = 142, normalized size = 0.95 \[ -\frac{3 \sqrt{\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (A (3 m+5) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m-1);\frac{1}{6} (3 m+5);\cos ^2(c+d x)\right )+C (3 m-1) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{6} (3 m+5);\frac{1}{6} (3 m+11);\cos ^2(c+d x)\right )\right )}{d (3 m-1) (3 m+5) (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^(1 + m)*Csc[c + d*x]*(A*(5 + 3*m)*Hypergeometric2F1[1/2, (-1 + 3*m)/6, (5 + 3*m)/6, Cos[c + d
*x]^2] + C*(-1 + 3*m)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[S
in[c + d*x]^2])/(d*(-1 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(4/3))

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Maple [F]  time = 0.283, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \cos \left (d x + c\right )^{m}}{b^{2} \cos \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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