3.366 \(\int \cot ^5(e+f x) (a+b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=217 \[ -\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \left (a+b \tan ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \tan ^2(e+f x)}{a}+1\right )}{4 a^3 f (p+1)}+\frac{\left (a+b \tan ^2(e+f x)\right )^{p+1} \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{2 f (p+1) (a-b)}+\frac{(2 a-b p+b) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a f} \]

[Out]

((2*a + b - b*p)*Cot[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^2*f) - (Cot[e + f*x]^4*(a + b*Tan[e + f*x
]^2)^(1 + p))/(4*a*f) + (Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Tan[e + f*x]^2)/(a - b)]*(a + b*Tan[e + f*x
]^2)^(1 + p))/(2*(a - b)*f*(1 + p)) - ((2*a^2 - 2*a*b*p - b^2*(1 - p)*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1
+ (b*Tan[e + f*x]^2)/a]*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^3*f*(1 + p))

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Rubi [A]  time = 0.253299, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3670, 446, 103, 151, 156, 65, 68} \[ -\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \tan ^2(e+f x)}{a}+1\right )}{4 a^3 f (p+1)}+\frac{(2 a-b p+b) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a^2 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b \tan ^2(e+f x)+a}{a-b}\right )}{2 f (p+1) (a-b)}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{p+1}}{4 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

((2*a + b - b*p)*Cot[e + f*x]^2*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^2*f) - (Cot[e + f*x]^4*(a + b*Tan[e + f*x
]^2)^(1 + p))/(4*a*f) + (Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Tan[e + f*x]^2)/(a - b)]*(a + b*Tan[e + f*x
]^2)^(1 + p))/(2*(a - b)*f*(1 + p)) - ((2*a^2 - 2*a*b*p - b^2*(1 - p)*p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1
+ (b*Tan[e + f*x]^2)/a]*(a + b*Tan[e + f*x]^2)^(1 + p))/(4*a^3*f*(1 + p))

Rule 3670

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

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 151

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

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 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]

Rubi steps

\begin{align*} \int \cot ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x^5 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{x^3 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p (2 a+b-b p+b (1-p) x)}{x^2 (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p \left (2 a^2-2 a b p-b^2 (1-p) p-b p (2 a+b-b p) x\right )}{x (1+x)} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^p}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}+\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^p}{x} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac{(2 a+b-b p) \cot ^2(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^2 f}-\frac{\cot ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a f}+\frac{\, _2F_1\left (1,1+p;2+p;\frac{a+b \tan ^2(e+f x)}{a-b}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{2 (a-b) f (1+p)}-\frac{\left (2 a^2-2 a b p-b^2 (1-p) p\right ) \, _2F_1\left (1,1+p;2+p;1+\frac{b \tan ^2(e+f x)}{a}\right ) \left (a+b \tan ^2(e+f x)\right )^{1+p}}{4 a^3 f (1+p)}\\ \end{align*}

Mathematica [A]  time = 2.54215, size = 172, normalized size = 0.79 \[ -\frac{\tan ^2(e+f x) \left (a \cot ^2(e+f x)+b\right ) \left (a+b \tan ^2(e+f x)\right )^p \left ((a-b) \left (\left (2 a^2-2 a b p+b^2 (p-1) p\right ) \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{b \tan ^2(e+f x)}{a}+1\right )+a (p+1) \cot ^2(e+f x) \left (a \cot ^2(e+f x)-2 a+b (p-1)\right )\right )-2 a^3 \text{Hypergeometric2F1}\left (1,p+1,p+2,\frac{a+b \tan ^2(e+f x)}{a-b}\right )\right )}{4 a^3 f (p+1) (a-b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

-((b + a*Cot[e + f*x]^2)*(-2*a^3*Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Tan[e + f*x]^2)/(a - b)] + (a - b)*
(a*(1 + p)*Cot[e + f*x]^2*(-2*a + b*(-1 + p) + a*Cot[e + f*x]^2) + (2*a^2 - 2*a*b*p + b^2*(-1 + p)*p)*Hypergeo
metric2F1[1, 1 + p, 2 + p, 1 + (b*Tan[e + f*x]^2)/a]))*Tan[e + f*x]^2*(a + b*Tan[e + f*x]^2)^p)/(4*a^3*(a - b)
*f*(1 + p))

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Maple [F]  time = 0.301, size = 0, normalized size = 0. \begin{align*} \int \left ( \cot \left ( fx+e \right ) \right ) ^{5} \left ( a+b \left ( \tan \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)^5*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^p,x)

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Maxima [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)^5*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2 + a)^p*cot(f*x + e)^5, 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)**5*(a+b*tan(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 \tan \left (f x + e\right )^{2} + a\right )}^{p} \cot \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*cot(f*x + e)^5, x)