\(\int \cot (c+d x) (a+b \sin ^4(c+d x))^p \, dx\) [567]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \cot (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 a d (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3308, 272, 67} \[ \int \cot (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=-\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)}{a}+1\right )}{4 a d (p+1)} \]

[In]

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

[Out]

-1/4*(Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^4)/a]*(a + b*Sin[c + d*x]^4)^(1 + p))/(a*d*(1 + 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 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*} \text {integral}& = \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 {\operatorname {Hypergeometric2F1}\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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 a d (1+p)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \cot \left (d x +c \right ) {\left (a +b \left (\sin ^{4}\left (d x +c \right )\right )\right )}^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

integrate(cot(d*x+c)*(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), x)

Sympy [F(-1)]

Timed out. \[ \int \cot (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cot(d*x+c)*(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), x)

Giac [F]

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

[In]

integrate(cot(d*x+c)*(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), x)

Mupad [F(-1)]

Timed out. \[ \int \cot (c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int \mathrm {cot}\left (c+d\,x\right )\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \]

[In]

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

[Out]

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