\(\int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx\) [400]

Optimal result

Integrand size = 19, antiderivative size = 131 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=-\frac {a \cos ^{1+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+m) \sqrt {\sin ^2(c+d x)}}-\frac {a \cos ^{2+m}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d (2+m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-a*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(1+m)/(sin(d*x+c)^2)^(1/
2)-a*cos(d*x+c)^(2+m)*hypergeom([1/2, 1+1/2*m],[2+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(2+m)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2827, 2722} \[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=-\frac {a \sin (c+d x) \cos ^{m+1}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{2},\frac {m+3}{2},\cos ^2(c+d x)\right )}{d (m+1) \sqrt {\sin ^2(c+d x)}}-\frac {a \sin (c+d x) \cos ^{m+2}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+2}{2},\frac {m+4}{2},\cos ^2(c+d x)\right )}{d (m+2) \sqrt {\sin ^2(c+d x)}} \]

[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 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 2827

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]

Rubi steps \begin{align*} \text {integral}& = 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) \operatorname {Hypergeometric2F1}\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) \operatorname {Hypergeometric2F1}\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 [A] (verified)

Time = 0.17 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.85 \[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=-\frac {a \cos ^{1+m}(c+d x) \csc (c+d x) \left ((2+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{2},\frac {3+m}{2},\cos ^2(c+d x)\right )+(1+m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+m}{2},\frac {4+m}{2},\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (1+m) (2+m)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a +\cos \left (d x +c \right ) a \right )d x\]

[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)

Fricas [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{m} \,d x } \]

[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]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=a \left (\int \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}\, dx + \int \cos ^{m}{\left (c + d x \right )}\, dx\right ) \]

[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))

Maxima [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{m} \,d x } \]

[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)

Giac [F]

\[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=\int { {\left (a \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{m} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \cos ^m(c+d x) (a+a \cos (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^m\,\left (a+a\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

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

[Out]

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