3.446 \(\int \cot ^3(e+f x) (a+b \sec ^2(e+f x))^p \, dx\)

Optimal. Leaf size=157 \[ \frac{(a-b p+b) \left (a+b \sec ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \sec ^2(e+f x)}{a+b}\right )}{2 f (p+1) (a+b)^2}-\frac{\left (a+b \sec ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac{\cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 f (a+b)} \]

[Out]

-(Cot[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(1 + p))/(2*(a + b)*f) + ((a + b - b*p)*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)^2*f*(1 + p)) - (Hypergeometric
2F1[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))

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Rubi [A]  time = 0.174266, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4139, 446, 103, 156, 68, 65} \[ \frac{(a-b p+b) \left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sec ^2(e+f x)+a}{a+b}\right )}{2 f (p+1) (a+b)^2}-\frac{\left (a+b \sec ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)}-\frac{\cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 f (a+b)} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

-(Cot[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(1 + p))/(2*(a + b)*f) + ((a + b - b*p)*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)^2*f*(1 + p)) - (Hypergeometric
2F1[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 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])

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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

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

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 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])

Rubi steps

\begin{align*} \int \cot ^3(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 )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{(-1+x)^2 x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p (a+b-b p x)}{(-1+x) x} \, dx,x,\sec ^2(e+f x)\right )}{2 (a+b) f}\\ &=-\frac{\cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f}-\frac{(a+b-b p) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{-1+x} \, dx,x,\sec ^2(e+f x)\right )}{2 (a+b) f}\\ &=-\frac{\cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 (a+b) f}+\frac{(a+b-b p) \, _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)^2 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 = 3.53619, size = 139, normalized size = 0.89 \[ -\frac{\tan ^2(e+f x) \left ((a+b) \cot ^2(e+f x)+b\right ) \left (a+b \sec ^2(e+f x)\right )^p \left ((a+b)^2 \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \tan ^2(e+f x)+b}{a}\right )-a (a-b p+b) \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \tan ^2(e+f x)}{a+b}+1\right )+a (p+1) (a+b) \cot ^2(e+f x)\right )}{2 a f (p+1) (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]  time = 0.354, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{3} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x)

[Out]

int(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(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)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(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)^3, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**3*(a+b*sec(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3*(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)^3, x)