3.6.68 \(\int \cot ^3(c+d x) (a+b \sin ^4(c+d x))^p \, dx\) [568]

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

[Out]

1/4*hypergeom([1, 1+p],[2+p],1+b*sin(d*x+c)^4/a)*(a+b*sin(d*x+c)^4)^(1+p)/a/d/(1+p)-1/2*csc(d*x+c)^2*hypergeom
([-1/2, -p],[1/2],-b*sin(d*x+c)^4/a)*(a+b*sin(d*x+c)^4)^p/d/((1+b*sin(d*x+c)^4/a)^p)

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Rubi [A]
time = 0.07, antiderivative size = 127, normalized size of antiderivative = 1.00, 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 = {3308, 778, 372, 371, 272, 67} \begin {gather*} \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} \end {gather*}

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 67

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

Rule 272

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 778

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 3308

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]

Rubi steps

\begin {align*} \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx &=\frac {\text {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 {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\sin ^2(c+d x)\right )}{2 d}-\frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac {\text {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 ) \text {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*}

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Mathematica [A]
time = 0.46, size = 119, normalized size = 0.94 \begin {gather*} \frac {\left (a+b \sin ^4(c+d x)\right )^p \left (\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 )}{a (1+p)}-2 \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 (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}\right )}{4 d} \end {gather*}

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 = 1.63, size = 0, normalized size = 0.00 \[\int \left (\cot ^{3}\left (d x +c \right )\right ) \left (a +b \left (\sin ^{4}\left (d x +c \right )\right )\right )^{p}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.48, size = 37, normalized size = 0.29 \begin {gather*} {\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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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