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

Optimal. Leaf size=114 \[ \frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sec ^2(e+f x)+a}{a+b}\right )}{2 f (p+1) (a+b)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {4139, 446, 86, 68, 65} \[ \frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sec ^2(e+f x)+a}{a+b}\right )}{2 f (p+1) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sec[e + f*x]^2)/(a + b)]*(a + b*Sec[e + f*x]^2)^(1 + p))/(2*(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 65

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

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 86

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 446

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 4139

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*} \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac {\operatorname {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 {\operatorname {Subst}\left (\int \frac {(a+b x)^p}{(-1+x) x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p}{-1+x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}-\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac {\, _2F_1\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 {\, _2F_1\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]  time = 2.13, size = 115, normalized size = 1.01 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a+b \sec ^2(e+f x)\right )^p \left ((a+b) \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)+a+b}{a}\right )-a \, _2F_1\left (1,p+1;p+2;\frac {b \tan ^2(e+f x)}{a+b}+1\right )\right )}{4 a f (p+1) (a+b)} \]

Antiderivative was successfully verified.

[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))

________________________________________________________________________________________

fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 2.03, size = 0, normalized size = 0.00 \[ \int \cot \left (f x +e \right ) \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {cot}\left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________