3.218 \(\int (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx\)

Optimal. Leaf size=95 \[ -\frac{13^{n/2} \sin \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \cos ^{n+1}\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\cos ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right )}{d (n+1) \sqrt{\sin ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}} \]

[Out]

-((13^(n/2)*Cos[c + d*x - ArcTan[3/2]]^(1 + n)*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[c + d*x - ArcT
an[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]])/(d*(1 + n)*Sqrt[Sin[c + d*x - ArcTan[3/2]]^2]))

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Rubi [A]  time = 0.0489555, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3077, 2643} \[ -\frac{13^{n/2} \sin \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \cos ^{n+1}\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right )}{d (n+1) \sqrt{\sin ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Int[(2*Cos[c + d*x] + 3*Sin[c + d*x])^n,x]

[Out]

-((13^(n/2)*Cos[c + d*x - ArcTan[3/2]]^(1 + n)*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[c + d*x - ArcT
an[3/2]]^2]*Sin[c + d*x - ArcTan[3/2]])/(d*(1 + n)*Sqrt[Sin[c + d*x - ArcTan[3/2]]^2]))

Rule 3077

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^2 + b^2)^(n/2),
Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) && GtQ[
a^2 + b^2, 0]

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 (2 \cos (c+d x)+3 \sin (c+d x))^n \, dx &=13^{n/2} \int \cos ^n\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \, dx\\ &=-\frac{13^{n/2} \cos ^{1+n}\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{1+n}{2};\frac{3+n}{2};\cos ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )\right ) \sin \left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}{d (1+n) \sqrt{\sin ^2\left (c+d x-\tan ^{-1}\left (\frac{3}{2}\right )\right )}}\\ \end{align*}

Mathematica [A]  time = 0.176881, size = 88, normalized size = 0.93 \[ -\frac{\sin \left (2 \left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )\right ) \sin ^2\left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )^{-\frac{n}{2}-\frac{1}{2}} (3 \sin (c+d x)+2 \cos (c+d x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1-n}{2},\frac{3}{2},\cos ^2\left (c+d x+\tan ^{-1}\left (\frac{2}{3}\right )\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(2*Cos[c + d*x] + 3*Sin[c + d*x])^n,x]

[Out]

-(Hypergeometric2F1[1/2, (1 - n)/2, 3/2, Cos[c + d*x + ArcTan[2/3]]^2]*(2*Cos[c + d*x] + 3*Sin[c + d*x])^n*(Si
n[c + d*x + ArcTan[2/3]]^2)^(-1/2 - n/2)*Sin[2*(c + d*x + ArcTan[2/3])])/(2*d)

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Maple [F]  time = 0.416, size = 0, normalized size = 0. \begin{align*} \int \left ( 2\,\cos \left ( dx+c \right ) +3\,\sin \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*cos(d*x+c)+3*sin(d*x+c))^n,x)

[Out]

int((2*cos(d*x+c)+3*sin(d*x+c))^n,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((2*cos(d*x + c) + 3*sin(d*x + c))^n, 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((2*cos(d*x+c)+3*sin(d*x+c))**n,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (2 \, \cos \left (d x + c\right ) + 3 \, \sin \left (d x + c\right )\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*cos(d*x+c)+3*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((2*cos(d*x + c) + 3*sin(d*x + c))^n, x)