\(\int \cot (e+f x) (a+b \sec ^2(e+f x))^p \, dx\) [445]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 21, antiderivative size = 114 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sec ^2(e+f x)}{a+b}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 (a+b) f (1+p)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)} \]

[Out]

-1/2*hypergeom([1, p+1],[2+p],(a+b*sec(f*x+e)^2)/(a+b))*(a+b*sec(f*x+e)^2)^(p+1)/(a+b)/f/(p+1)+1/2*hypergeom([
1, p+1],[2+p],1+b*sec(f*x+e)^2/a)*(a+b*sec(f*x+e)^2)^(p+1)/a/f/(p+1)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4224, 457, 88, 70, 67} \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sec ^2(e+f x)+a}{a+b}\right )}{2 f (p+1) (a+b)} \]

[In]

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

[Out]

-1/2*(Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sec[e + f*x]^2)/(a + b)]*(a + b*Sec[e + f*x]^2)^(1 + p))/((a +
 b)*f*(1 + p)) + (Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Sec[e + f*x]^2)/a]*(a + b*Sec[e + f*x]^2)^(1 + p))
/(2*a*f*(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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 88

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4224

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p/x)
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x \left (-1+x^2\right )} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b x)^p}{(-1+x) x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \frac {(a+b x)^p}{-1+x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}-\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b \sec ^2(e+f x)}{a+b}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 (a+b) f (1+p)}+\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.90 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \left ((a+b) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {a+b+b \tan ^2(e+f x)}{a}\right )-a \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \tan ^2(e+f x)}{a+b}\right )\right ) \sec ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^p}{4 a (a+b) f (1+p)} \]

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*((a + b)*Hypergeometric2F1[1, 1 + p, 2 + p, (a + b + b*Tan[e + f*x]^2)/a] - a*
Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*Tan[e + f*x]^2)/(a + b)])*Sec[e + f*x]^2*(a + b*Sec[e + f*x]^2)^p)/(
4*a*(a + b)*f*(1 + p))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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