\(\int \cot ^3(c+d x) (a+b \sin ^2(c+d x))^p \, dx\) [478]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=-\frac {\csc ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a d}+\frac {(a-b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^2(c+d x)}{a}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a^2 d (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=-\frac {\left (a \csc ^2(c+d x)+\frac {(-a+b p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^2(c+d x)}{a}\right )}{1+p}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 a^2 d} \] Input:

Integrate[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^2)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 87, 75}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (c+d x)^2\right )^p}{\tan (c+d x)^3}dx\)

\(\Big \downarrow \) 3673

\(\displaystyle \frac {\int \csc ^4(c+d x) \left (1-\sin ^2(c+d x)\right ) \left (b \sin ^2(c+d x)+a\right )^pd\sin ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {-\frac {(a-b p) \int \csc ^2(c+d x) \left (b \sin ^2(c+d x)+a\right )^pd\sin ^2(c+d x)}{a}-\frac {\csc ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{a}}{2 d}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {(a-b p) \left (a+b \sin ^2(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^2(c+d x)}{a}+1\right )}{a^2 (p+1)}-\frac {\csc ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^{p+1}}{a}}{2 d}\)

Input:

Int[Cot[c + d*x]^3*(a + b*Sin[c + d*x]^2)^p,x]
 

Output:

(-((Csc[c + d*x]^2*(a + b*Sin[c + d*x]^2)^(1 + p))/a) + ((a - b*p)*Hyperge 
ometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^2)/a]*(a + b*Sin[c + d*x]^ 
2)^(1 + p))/(a^2*(1 + p)))/(2*d)
 

Defintions of rubi rules used

rule 75

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] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3673

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

Maple [F]

\[\int \cot \left (d x +c \right )^{3} \left (a +b \sin \left (d x +c \right )^{2}\right )^{p}d x\]

Input:

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x)
 

Output:

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x)
 

Fricas [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="fricas")
 

Output:

integral((-b*cos(d*x + c)^2 + a + b)^p*cot(d*x + c)^3, x)
 

Sympy [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int \left (a + b \sin ^{2}{\left (c + d x \right )}\right )^{p} \cot ^{3}{\left (c + d x \right )}\, dx \] Input:

integrate(cot(d*x+c)**3*(a+b*sin(d*x+c)**2)**p,x)
 

Output:

Integral((a + b*sin(c + d*x)**2)**p*cot(c + d*x)**3, x)
 

Maxima [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c)^3, x)
 

Giac [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{2} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

integrate(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sin(d*x + c)^2 + a)^p*cot(d*x + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^2+a\right )}^p \,d x \] Input:

int(cot(c + d*x)^3*(a + b*sin(c + d*x)^2)^p,x)
 

Output:

int(cot(c + d*x)^3*(a + b*sin(c + d*x)^2)^p, x)
 

Reduce [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \, dx=\int \left (\sin \left (d x +c \right )^{2} b +a \right )^{p} \cot \left (d x +c \right )^{3}d x \] Input:

int(cot(d*x+c)^3*(a+b*sin(d*x+c)^2)^p,x)
 

Output:

int((sin(c + d*x)**2*b + a)**p*cot(c + d*x)**3,x)