3.546 \(\int \cot (c+d x) (a+b \sin ^2(c+d x))^p \, dx\)

Optimal. Leaf size=54 \[ -\frac{\left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)}{a}+1\right )}{2 a d (p+1)} \]

[Out]

-(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^2)/a]*(a + b*Sin[c + d*x]^2)^(1 + p))/(2*a*d*(1 + p))

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Rubi [A]  time = 0.0557917, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {3194, 65} \[ -\frac{\left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)}{a}+1\right )}{2 a d (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*(a + b*Sin[c + d*x]^2)^p,x]

[Out]

-(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^2)/a]*(a + b*Sin[c + d*x]^2)^(1 + p))/(2*a*d*(1 + p))

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int \cot (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac{\, _2F_1\left (1,1+p;2+p;1+\frac{b \sin ^2(c+d x)}{a}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a d (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0594202, size = 54, normalized size = 1. \[ -\frac{\left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^2(c+d x)}{a}+1\right )}{2 a d (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*(a + b*Sin[c + d*x]^2)^p,x]

[Out]

-(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^2)/a]*(a + b*Sin[c + d*x]^2)^(1 + p))/(2*a*d*(1 + p))

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Maple [F]  time = 1.125, size = 0, normalized size = 0. \begin{align*} \int \cot \left ( dx+c \right ) \left ( a+ \left ( \sin \left ( dx+c \right ) \right ) ^{2}b \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+sin(d*x+c)^2*b)^p,x)

[Out]

int(cot(d*x+c)*(a+sin(d*x+c)^2*b)^p,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c), 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 )^{2} + a + b\right )}^{p} \cot \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^2)^p,x, algorithm="fricas")

[Out]

integral((-b*cos(d*x + c)^2 + a + b)^p*cot(d*x + c), 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(cot(d*x+c)*(a+b*sin(d*x+c)**2)**p,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sin(d*x+c)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c), x)