3.568 \(\int \cot ^3(c+d x) (a+b \sin ^4(c+d x))^p \, dx\)

Optimal. Leaf size=127 \[ \frac{\left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)}{a}+1\right )}{4 a d (p+1)}-\frac{\csc ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d} \]

[Out]

(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^4)/a]*(a + b*Sin[c + d*x]^4)^(1 + p))/(4*a*d*(1 + p))
- (Csc[c + d*x]^2*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sin[c + d*x]^4)/a)]*(a + b*Sin[c + d*x]^4)^p)/(2*d*(1
+ (b*Sin[c + d*x]^4)/a)^p)

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Rubi [A]  time = 0.101792, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3229, 764, 365, 364, 266, 65} \[ \frac{\left (a+b \sin ^4(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)}{a}+1\right )}{4 a d (p+1)}-\frac{\csc ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sin ^4(c+d x)}{a}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^4)^p,x]

[Out]

(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^4)/a]*(a + b*Sin[c + d*x]^4)^(1 + p))/(4*a*d*(1 + p))
- (Csc[c + d*x]^2*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sin[c + d*x]^4)/a)]*(a + b*Sin[c + d*x]^4)^p)/(2*d*(1
+ (b*Sin[c + d*x]^4)/a)^p)

Rule 3229

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

Rule 764

Int[(x_)^(m_.)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[f, Int[x^m*(a + c*x^2)^p, x]
, x] + Dist[g, Int[x^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, f, g, p}, x] && IntegerQ[m] &&  !IntegerQ[2
*p]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1-x) \left (a+b x^2\right )^p}{x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,\sin ^4(c+d x)\right )}{4 d}+\frac{\left (\left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac{b \sin ^4(c+d x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^p}{x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac{\, _2F_1\left (1,1+p;2+p;1+\frac{b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 a d (1+p)}-\frac{\csc ^2(c+d x) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac{b \sin ^4(c+d x)}{a}\right )^{-p}}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.649591, size = 119, normalized size = 0.94 \[ \frac{\left (a+b \sin ^4(c+d x)\right )^p \left (\frac{\left (a+b \sin ^4(c+d x)\right ) \, _2F_1\left (1,p+1;p+2;\frac{b \sin ^4(c+d x)}{a}+1\right )}{a (p+1)}-2 \csc ^2(c+d x) \left (\frac{b \sin ^4(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sin ^4(c+d x)}{a}\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^4)^p,x]

[Out]

((a + b*Sin[c + d*x]^4)^p*((Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^4)/a]*(a + b*Sin[c + d*x]^4
))/(a*(1 + p)) - (2*Csc[c + d*x]^2*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sin[c + d*x]^4)/a)])/(1 + (b*Sin[c +
d*x]^4)/a)^p))/(4*d)

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Maple [F]  time = 1.055, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( dx+c \right ) \right ) ^{3} \left ( a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^4)^p,x)

[Out]

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^4)^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 )^{4} + a\right )}^{p} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^4)^p,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^4 + a)^p*cot(d*x + c)^3, 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 )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b\right )}^{p} \cot \left (d x + c\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^4)^p,x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)^p*cot(d*x + c)^3, 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)**3*(a+b*sin(d*x+c)**4)**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 )^{4} + a\right )}^{p} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^4)^p,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^4 + a)^p*cot(d*x + c)^3, x)