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

Optimal. Leaf size=97 \[ -\frac{\sqrt{\cos ^2(c+d x)} \csc (c+d x) \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (\frac{b \sin ^2(c+d x)}{a}+1\right )^{-p} F_1\left (-\frac{1}{2};-\frac{1}{2},-p;\frac{1}{2};\sin ^2(c+d x),-\frac{b \sin ^2(c+d x)}{a}\right )}{d} \]

[Out]

-((AppellF1[-1/2, -1/2, -p, 1/2, Sin[c + d*x]^2, -((b*Sin[c + d*x]^2)/a)]*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]*Se
c[c + d*x]*(a + b*Sin[c + d*x]^2)^p)/(d*(1 + (b*Sin[c + d*x]^2)/a)^p))

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Rubi [A]  time = 0.100471, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3196, 511, 510} \[ -\frac{\sqrt{\cos ^2(c+d x)} \csc (c+d x) \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (\frac{b \sin ^2(c+d x)}{a}+1\right )^{-p} F_1\left (-\frac{1}{2};-\frac{1}{2},-p;\frac{1}{2};\sin ^2(c+d x),-\frac{b \sin ^2(c+d x)}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((AppellF1[-1/2, -1/2, -p, 1/2, Sin[c + d*x]^2, -((b*Sin[c + d*x]^2)/a)]*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]*Se
c[c + d*x]*(a + b*Sin[c + d*x]^2)^p)/(d*(1 + (b*Sin[c + d*x]^2)/a)^p))

Rule 3196

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

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \cot ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx &=\frac{\left (\sqrt{\cos ^2(c+d x)} \sec (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2} \left (a+b x^2\right )^p}{x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac{\left (\sqrt{\cos ^2(c+d x)} \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (1+\frac{b \sin ^2(c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2} \left (1+\frac{b x^2}{a}\right )^p}{x^2} \, dx,x,\sin (c+d x)\right )}{d}\\ &=-\frac{F_1\left (-\frac{1}{2};-\frac{1}{2},-p;\frac{1}{2};\sin ^2(c+d x),-\frac{b \sin ^2(c+d x)}{a}\right ) \sqrt{\cos ^2(c+d x)} \csc (c+d x) \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (1+\frac{b \sin ^2(c+d x)}{a}\right )^{-p}}{d}\\ \end{align*}

Mathematica [A]  time = 0.197504, size = 98, normalized size = 1.01 \[ -\frac{\sqrt{\cos ^2(c+d x)} \csc (c+d x) \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (\frac{a+b \sin ^2(c+d x)}{a}\right )^{-p} F_1\left (-\frac{1}{2};-\frac{1}{2},-p;\frac{1}{2};\sin ^2(c+d x),-\frac{b \sin ^2(c+d x)}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((AppellF1[-1/2, -1/2, -p, 1/2, Sin[c + d*x]^2, -((b*Sin[c + d*x]^2)/a)]*Sqrt[Cos[c + d*x]^2]*Csc[c + d*x]*Se
c[c + d*x]*(a + b*Sin[c + d*x]^2)^p)/(d*((a + b*Sin[c + d*x]^2)/a)^p))

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Maple [F]  time = 0.686, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( dx+c \right ) \right ) ^{2} \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)^2*(a+sin(d*x+c)^2*b)^p,x)

[Out]

int(cot(d*x+c)^2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(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)^2, 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 )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(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)^2, 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)**2*(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 )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(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)^2, x)