\(\int \cot ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [451]

Optimal result

Integrand size = 23, antiderivative size = 90 \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},1,-p,-\frac {1}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right ) \cot ^3(e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (\frac {a+b+b \tan ^2(e+f x)}{a+b}\right )^{-p}}{3 f} \] Output:

-1/3*AppellF1(-3/2,1,-p,-1/2,-tan(f*x+e)^2,-b*tan(f*x+e)^2/(a+b))*cot(f*x+ 
e)^3*(a+b+b*tan(f*x+e)^2)^p/f/(((a+b+b*tan(f*x+e)^2)/(a+b))^p)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3033\) vs. \(2(90)=180\).

Time = 16.48 (sec) , antiderivative size = 3033, normalized size of antiderivative = 33.70 \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \] Input:

Integrate[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^7*(Sec[e + f*x]^2)^p*(a + b 
*Sec[e + f*x]^2)^p*((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x] 
^2)/(a + b)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(3*(a + b)*A 
ppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 
 2*(b*p*AppellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e 
 + f*x]^2] - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b 
)), -Tan[e + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, 
 -((b*Tan[e + f*x]^2)/(a + b))] - 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b* 
Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a + b)) 
^p))/(3*f*((2*p*(a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^2*(Sec[e + f 
*x]^2)^p*((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b 
)), -Tan[e + f*x]^2]*Sin[e + f*x]^2*Tan[e + f*x]^2)/(3*(a + b)*AppellF1[1/ 
2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] + 2*(b*p*Ap 
pellF1[3/2, 1 - p, 1, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e + f*x]^2] 
 - (a + b)*AppellF1[3/2, -p, 2, 5/2, -((b*Tan[e + f*x]^2)/(a + b)), -Tan[e 
 + f*x]^2])*Tan[e + f*x]^2) - (Hypergeometric2F1[-3/2, -p, -1/2, -((b*Tan[ 
e + f*x]^2)/(a + b))] - 3*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Tan[e + f* 
x]^2)/(a + b))]*Tan[e + f*x]^2)/(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/3 - ( 
a + 2*b + a*Cos[2*(e + f*x)])^p*Cot[e + f*x]^2*Csc[e + f*x]^2*(Sec[e + f*x 
]^2)^p*((9*(a + b)*AppellF1[1/2, -p, 1, 3/2, -((b*Tan[e + f*x]^2)/(a + ...
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4629, 2075, 395, 394}

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.

\(\displaystyle \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^p}{\tan (e+f x)^4}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\cot ^4(e+f x) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\cot ^4(e+f x) \left (b \tan ^2(e+f x)+a+b\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 395

\(\displaystyle \frac {\left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \int \frac {\cot ^4(e+f x) \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^p}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 394

\(\displaystyle -\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {3}{2},1,-p,-\frac {1}{2},-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a+b}\right )}{3 f}\)

Input:

Int[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]
 

Output:

-1/3*(AppellF1[-3/2, 1, -p, -1/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a 
 + b))]*Cot[e + f*x]^3*(a + b + b*Tan[e + f*x]^2)^p)/(f*(1 + (b*Tan[e + f* 
x]^2)/(a + b))^p)
 

Defintions of rubi rules used

rule 394

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/2 
, -p, -q, 1 + (m + 1)/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; FreeQ[{a, b, c, 
 d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 1] && (Int 
egerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 395

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^ 
FracPart[p])   Int[(e*x)^m*(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ 
[{a, b, c, d, e, m, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 
1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

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

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [F]

\[\int \cot \left (f x +e \right )^{4} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]

Input:

int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Fricas [F]

\[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
 

Output:

integral((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^4, x)
 

Sympy [F(-1)]

Timed out. \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \] Input:

integrate(cot(f*x+e)**4*(a+b*sec(f*x+e)**2)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^4, x)
 

Giac [F]

\[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{4} \,d x } \] Input:

integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
 

Output:

integrate((b*sec(f*x + e)^2 + a)^p*cot(f*x + e)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^4\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \] Input:

int(cot(e + f*x)^4*(a + b/cos(e + f*x)^2)^p,x)
 

Output:

int(cot(e + f*x)^4*(a + b/cos(e + f*x)^2)^p, x)
 

Reduce [F]

\[ \int \cot ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \cot \left (f x +e \right )^{4} \left (\sec \left (f x +e \right )^{2} b +a \right )^{p}d x \] Input:

int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)
 

Output:

int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)