3.400 \(\int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx\)

Optimal. Leaf size=131 \[ -\frac{a \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{a \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) \sqrt{\sin ^2(c+d x)}} \]

[Out]

-((a*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m
)*Sqrt[Sin[c + d*x]^2])) - (a*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2
]*Sin[c + d*x])/(d*(2 + m)*Sqrt[Sin[c + d*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0637415, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2748, 2643} \[ -\frac{a \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{a \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + m
)*Sqrt[Sin[c + d*x]^2])) - (a*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (2 + m)/2, (4 + m)/2, Cos[c + d*x]^2
]*Sin[c + d*x])/(d*(2 + m)*Sqrt[Sin[c + d*x]^2])

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 \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx &=a \int \cos ^m(c+d x) \, dx+a \int \cos ^{1+m}(c+d x) \, dx\\ &=-\frac{a \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)}{d (1+m) \sqrt{\sin ^2(c+d x)}}-\frac{a \cos ^{2+m}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2+m}{2};\frac{4+m}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.997566, size = 208, normalized size = 1.59 \[ \frac{i a 2^{-m-2} \left (e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )\right )^{m+1} (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left ((m-1) m \, _2F_1\left (1,\frac{m+1}{2};\frac{1-m}{2};-e^{2 i (c+d x)}\right )+(m+1) e^{i (c+d x)} \left (2 (m-1) \, _2F_1\left (1,\frac{m+2}{2};1-\frac{m}{2};-e^{2 i (c+d x)}\right )+m e^{i (c+d x)} \, _2F_1\left (1,\frac{m+3}{2};\frac{3-m}{2};-e^{2 i (c+d x)}\right )\right )\right )}{d (m-1) m (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*2^(-2 - m)*a*((1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x)))^(1 + m)*(1 + Cos[c + d*x])*((-1 + m)*m*Hypergeomet
ric2F1[1, (1 + m)/2, (1 - m)/2, -E^((2*I)*(c + d*x))] + E^(I*(c + d*x))*(1 + m)*(2*(-1 + m)*Hypergeometric2F1[
1, (2 + m)/2, 1 - m/2, -E^((2*I)*(c + d*x))] + E^(I*(c + d*x))*m*Hypergeometric2F1[1, (3 + m)/2, (3 - m)/2, -E
^((2*I)*(c + d*x))]))*Sec[(c + d*x)/2]^2)/(d*(-1 + m)*m*(1 + m))

________________________________________________________________________________________

Maple [F]  time = 1.062, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a+\cos \left ( dx+c \right ) a \right ) \, 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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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