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

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

[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])) - (b*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])

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Rubi [A]  time = 0.0640834, 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} \[ -\frac{a \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )}{d (m+1) \sqrt{\sin ^2(c+d x)}}-\frac{b \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )}{d (m+2) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^m*(a + b*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])) - (b*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+b \cos (c+d x)) \, dx &=a \int \cos ^m(c+d x) \, dx+b \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{b \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 [A]  time = 0.152915, size = 112, normalized size = 0.85 \[ -\frac{\sqrt{\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+1}(c+d x) \left (a (m+2) \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\cos ^2(c+d x)\right )+b (m+1) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};\cos ^2(c+d x)\right )\right )}{d (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.27, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a+b\cos \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((b*cos(d*x + c) + a)*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 (b \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+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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