\(\int \cot ^3(c+d x) (a+b \sin ^4(c+d x))^p \, dx\) [499]

Optimal result

Integrand size = 23, antiderivative size = 127 \[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^{1+p}}{4 a d (1+p)}-\frac {\csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )^p \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}}{2 d} \] Output:

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)-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)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.94 \[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\frac {\left (a+b \sin ^4(c+d x)\right )^p \left (\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sin ^4(c+d x)}{a}\right ) \left (a+b \sin ^4(c+d x)\right )}{a (1+p)}-2 \csc ^2(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sin ^4(c+d x)}{a}\right ) \left (1+\frac {b \sin ^4(c+d x)}{a}\right )^{-p}\right )}{4 d} \] Input:

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

Output:

((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*Hyp 
ergeometric2F1[-1/2, -p, 1/2, -((b*Sin[c + d*x]^4)/a)])/(1 + (b*Sin[c + d* 
x]^4)/a)^p))/(4*d)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3708, 542, 243, 75, 279, 278}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin (c+d x)^4\right )^p}{\tan (c+d x)^3}dx\)

\(\Big \downarrow \) 3708

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

\(\Big \downarrow \) 542

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

\(\displaystyle \frac {\int \csc ^4(c+d x) \left (b \sin ^4(c+d x)+a\right )^pd\sin ^2(c+d x)+\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)}{a}+1\right )}{2 a (p+1)}}{2 d}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \int \csc ^4(c+d x) \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^pd\sin ^2(c+d x)+\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)}{a}+1\right )}{2 a (p+1)}}{2 d}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {\frac {\left (a+b \sin ^4(c+d x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sin ^4(c+d x)}{a}+1\right )}{2 a (p+1)}-\csc ^2(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \left (\frac {b \sin ^4(c+d x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sin ^4(c+d x)}{a}\right )}{2 d}\)

Input:

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

Output:

((Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sin[c + d*x]^4)/a]*(a + b*Sin[ 
c + d*x]^4)^(1 + p))/(2*a*(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)/(1 + (b*Sin 
[c + d*x]^4)/a)^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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 542

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[c   Int[x^m*(a + b*x^2)^p, x], x] + Simp[d   Int[x^(m + 1)*(a + b*x^2 
)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && IntegerQ[m] &&  !IntegerQ[2*p]
 

rule 3042

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

rule 3708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(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^(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]
 

Maple [F]

\[\int \cot \left (d x +c \right )^{3} \left (a +b \sin \left (d x +c \right )^{4}\right )^{p}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

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

Output:

integral((b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)^p*cot(d*x + c)^3, 
 x)
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int { {\left (b \sin \left (d x + c\right )^{4} + a\right )}^{p} \cot \left (d x + c\right )^{3} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (b\,{\sin \left (c+d\,x\right )}^4+a\right )}^p \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cot ^3(c+d x) \left (a+b \sin ^4(c+d x)\right )^p \, dx=\int \left (\sin \left (d x +c \right )^{4} b +a \right )^{p} \cot \left (d x +c \right )^{3}d x \] Input:

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

Output:

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