3.6.85 \(\int \cot ^3(c+d x) (a+b \sin ^n(c+d x))^p \, dx\) [585]

Optimal. Leaf size=136 \[ \frac {\, _2F_1\left (1,1+p;2+p;1+\frac {b \sin ^n(c+d x)}{a}\right ) \left (a+b \sin ^n(c+d x)\right )^{1+p}}{a d n (1+p)}-\frac {\csc ^2(c+d x) \, _2F_1\left (-\frac {2}{n},-p;-\frac {2-n}{n};-\frac {b \sin ^n(c+d x)}{a}\right ) \left (a+b \sin ^n(c+d x)\right )^p \left (1+\frac {b \sin ^n(c+d x)}{a}\right )^{-p}}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3309, 1858, 372, 371, 272, 67} \begin {gather*} \frac {\left (a+b \sin ^n(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sin ^n(c+d x)}{a}+1\right )}{a d n (p+1)}-\frac {\csc ^2(c+d x) \left (a+b \sin ^n(c+d x)\right )^p \left (\frac {b \sin ^n(c+d x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {2}{n},-p;-\frac {2-n}{n};-\frac {b \sin ^n(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]^n)^p,x]

[Out]

(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^n)/a]*(a + b*Sin[c + d*x]^n)^(1 + p))/(a*d*n*(1 + p))
- (Csc[c + d*x]^2*Hypergeometric2F1[-2/n, -p, -((2 - n)/n), -((b*Sin[c + d*x]^n)/a)]*(a + b*Sin[c + d*x]^n)^p)
/(2*d*(1 + (b*Sin[c + d*x]^n)/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 1858

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rule 3309

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + b*(c*ff*x)^n)^p/(1 - ff^2*x^2)^((
m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.73, size = 129, normalized size = 0.95 \begin {gather*} \frac {\left (a+b \sin ^n(c+d x)\right )^p \left (\frac {2 \, _2F_1\left (1,1+p;2+p;1+\frac {b \sin ^n(c+d x)}{a}\right ) \left (a+b \sin ^n(c+d x)\right )}{a n (1+p)}-\csc ^2(c+d x) \, _2F_1\left (-\frac {2}{n},-p;\frac {-2+n}{n};-\frac {b \sin ^n(c+d x)}{a}\right ) \left (1+\frac {b \sin ^n(c+d x)}{a}\right )^{-p}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.39, size = 0, normalized size = 0.00 \[\int \left (\cot ^{3}\left (d x +c \right )\right ) \left (a +b \left (\sin ^{n}\left (d x +c \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.41, size = 25, normalized size = 0.18 \begin {gather*} {\rm integral}\left ({\left (b \sin \left (d x + c\right )^{n} + a\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)^n)^p,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c)^n + a)^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)**n)**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)^n)^p,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^n + 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 (a+b\,{\sin \left (c+d\,x\right )}^n\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)^n)^p,x)

[Out]

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

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