3.14 \(\int \sin ^n(c+d x) (a+a \sin (c+d x))^{-2-n} (-1-n-(-2-n) \sin (c+d x)) \, dx\)

Optimal. Leaf size=37 \[ -\frac{\cos (c+d x) \sin ^{n+1}(c+d x) (a \sin (c+d x)+a)^{-n-2}}{d} \]

[Out]

-((Cos[c + d*x]*Sin[c + d*x]^(1 + n)*(a + a*Sin[c + d*x])^(-2 - n))/d)

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Rubi [A]  time = 0.119467, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.023, Rules used = {2974} \[ -\frac{\cos (c+d x) \sin ^{n+1}(c+d x) (a \sin (c+d x)+a)^{-n-2}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^n*(a + a*Sin[c + d*x])^(-2 - n)*(-1 - n - (-2 - n)*Sin[c + d*x]),x]

[Out]

-((Cos[c + d*x]*Sin[c + d*x]^(1 + n)*(a + a*Sin[c + d*x])^(-2 - n))/d)

Rule 2974

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

Rubi steps

\begin{align*} \int \sin ^n(c+d x) (a+a \sin (c+d x))^{-2-n} (-1-n-(-2-n) \sin (c+d x)) \, dx &=-\frac{\cos (c+d x) \sin ^{1+n}(c+d x) (a+a \sin (c+d x))^{-2-n}}{d}\\ \end{align*}

Mathematica [B]  time = 1.5116, size = 107, normalized size = 2.89 \[ -\frac{2^n \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) (-\sin (c+d x)+\cos (c+d x)+1) (a (\sin (c+d x)+1))^{-n-2} \left (\left (\sin \left (\frac{3}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )\right ) \cos \left (\frac{1}{4} (c+d x)\right )\right )^n}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^n*(a + a*Sin[c + d*x])^(-2 - n)*(-1 - n - (-2 - n)*Sin[c + d*x]),x]

[Out]

-((2^n*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(Cos[(c + d*x)/4]*(-Sin[(c + d*x)/4] + Sin[(3*(c
+ d*x))/4]))^n*(1 + Cos[c + d*x] - Sin[c + d*x])*(a*(1 + Sin[c + d*x]))^(-2 - n))/d)

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Maple [F]  time = 0.546, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{-2-n} \left ( -1-n- \left ( -2-n \right ) \sin \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^n*(a+a*sin(d*x+c))^(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x)

[Out]

int(sin(d*x+c)^n*(a+a*sin(d*x+c))^(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left ({\left (n + 2\right )} \sin \left (d x + c\right ) - n - 1\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{-n - 2} \sin \left (d x + c\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^n*(a+a*sin(d*x+c))^(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x, algorithm="maxima")

[Out]

integrate(((n + 2)*sin(d*x + c) - n - 1)*(a*sin(d*x + c) + a)^(-n - 2)*sin(d*x + c)^n, x)

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Fricas [A]  time = 1.5027, size = 101, normalized size = 2.73 \begin{align*} -\frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{-n - 2} \sin \left (d x + c\right )^{n} \cos \left (d x + c\right ) \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^n*(a+a*sin(d*x+c))^(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x, algorithm="fricas")

[Out]

-(a*sin(d*x + c) + a)^(-n - 2)*sin(d*x + c)^n*cos(d*x + c)*sin(d*x + c)/d

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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(sin(d*x+c)**n*(a+a*sin(d*x+c))**(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left ({\left (n + 2\right )} \sin \left (d x + c\right ) - n - 1\right )}{\left (a \sin \left (d x + c\right ) + a\right )}^{-n - 2} \sin \left (d x + c\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^n*(a+a*sin(d*x+c))^(-2-n)*(-1-n-(-2-n)*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate(((n + 2)*sin(d*x + c) - n - 1)*(a*sin(d*x + c) + a)^(-n - 2)*sin(d*x + c)^n, x)