3.401 \(\int \frac{\cos ^m(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=156 \[ -\frac{\sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};\cos ^2(c+d x)\right )}{a d \sqrt{\sin ^2(c+d x)}}+\frac{m \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{a d (m+1) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \cos ^m(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

(Cos[c + d*x]^m*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) - (Cos[c + d*x]^m*Hypergeometric2F1[1/2, m/2, (2 + m)/2
, Cos[c + d*x]^2]*Sin[c + d*x])/(a*d*Sqrt[Sin[c + d*x]^2]) + (m*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a*d*(1 + m)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.125815, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2769, 2748, 2643} \[ -\frac{\sin (c+d x) \cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};\cos ^2(c+d x)\right )}{a d \sqrt{\sin ^2(c+d x)}}+\frac{m \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{a d (m+1) \sqrt{\sin ^2(c+d x)}}+\frac{\sin (c+d x) \cos ^m(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m/(a + a*Cos[c + d*x]),x]

[Out]

(Cos[c + d*x]^m*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) - (Cos[c + d*x]^m*Hypergeometric2F1[1/2, m/2, (2 + m)/2
, Cos[c + d*x]^2]*Sin[c + d*x])/(a*d*Sqrt[Sin[c + d*x]^2]) + (m*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(a*d*(1 + m)*Sqrt[Sin[c + d*x]^2])

Rule 2769

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(a*f*(a + b*Sin[e + f*x])), x] + Dist[(d*n)/(a*b), Int[(c + d*Sin[e + f*
x])^(n - 1)*(a - b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && NeQ[c^2 - d^2, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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)}{a+a \cos (c+d x)} \, dx &=\frac{\cos ^m(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{m \int \cos ^{-1+m}(c+d x) (a-a \cos (c+d x)) \, dx}{a^2}\\ &=\frac{\cos ^m(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac{m \int \cos ^{-1+m}(c+d x) \, dx}{a}-\frac{m \int \cos ^m(c+d x) \, dx}{a}\\ &=\frac{\cos ^m(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac{\cos ^m(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{2+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{a d \sqrt{\sin ^2(c+d x)}}+\frac{m \cos ^{1+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{a d (1+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [F]  time = 0.815791, size = 0, normalized size = 0. \[ \int \frac{\cos ^m(c+d x)}{a+a \cos (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Cos[c + d*x]^m/(a + a*Cos[c + d*x]),x]

[Out]

Integrate[Cos[c + d*x]^m/(a + a*Cos[c + d*x]), x]

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Maple [F]  time = 0.845, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{m}}{a+\cos \left ( dx+c \right ) a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m/(a+cos(d*x+c)*a),x)

[Out]

int(cos(d*x+c)^m/(a+cos(d*x+c)*a),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^m/(a*cos(d*x + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^m/(a*cos(d*x + c) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{m}{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**m/(a+a*cos(d*x+c)),x)

[Out]

Integral(cos(c + d*x)**m/(cos(c + d*x) + 1), x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^m/(a*cos(d*x + c) + a), x)