3.338 \(\int \frac{\cot ^5(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=215 \[ \frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 f (a-b) \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (8 a^2+12 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}} \]

[Out]

-((8*a^2 + 12*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(8*a^(7/2)*f) + ArcTanh[Sqrt[a + b*Ta
n[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(3/2)*f) + (b*(4*a^2 + 3*a*b - 15*b^2))/(8*a^3*(a - b)*f*Sqrt[a + b*Tan[e
+ f*x]^2]) + ((4*a + 5*b)*Cot[e + f*x]^2)/(8*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^4/(4*a*f*Sqrt[a
+ b*Tan[e + f*x]^2])

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Rubi [A]  time = 0.34645, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3670, 446, 103, 151, 152, 156, 63, 208} \[ \frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 f (a-b) \sqrt{a+b \tan ^2(e+f x)}}-\frac{\left (8 a^2+12 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

-((8*a^2 + 12*a*b + 15*b^2)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/(8*a^(7/2)*f) + ArcTanh[Sqrt[a + b*Ta
n[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(3/2)*f) + (b*(4*a^2 + 3*a*b - 15*b^2))/(8*a^3*(a - b)*f*Sqrt[a + b*Tan[e
+ f*x]^2]) + ((4*a + 5*b)*Cot[e + f*x]^2)/(8*a^2*f*Sqrt[a + b*Tan[e + f*x]^2]) - Cot[e + f*x]^4/(4*a*f*Sqrt[a
+ b*Tan[e + f*x]^2])

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 152

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] && IntegersQ[2*m, 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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^5 \left (1+x^2\right ) \left (a+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 (1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (4 a+5 b)+\frac{5 b x}{2}}{x^2 (1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{4 a f}\\ &=\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (8 a^2+12 a b+15 b^2\right )+\frac{3}{4} b (4 a+5 b) x}{x (1+x) (a+b x)^{3/2}} \, dx,x,\tan ^2(e+f x)\right )}{4 a^2 f}\\ &=\frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{8} (a-b) \left (8 a^2+12 a b+15 b^2\right )-\frac{1}{8} b \left (4 a^2+3 a b-15 b^2\right ) x}{x (1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 a^3 (a-b) f}\\ &=\frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 (a-b) f}+\frac{\left (8 a^2+12 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{16 a^3 f}\\ &=\frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^2(e+f x)}\right )}{(a-b) b f}+\frac{\left (8 a^2+12 a b+15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^2(e+f x)}\right )}{8 a^3 b f}\\ &=-\frac{\left (8 a^2+12 a b+15 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2} f}+\frac{b \left (4 a^2+3 a b-15 b^2\right )}{8 a^3 (a-b) f \sqrt{a+b \tan ^2(e+f x)}}+\frac{(4 a+5 b) \cot ^2(e+f x)}{8 a^2 f \sqrt{a+b \tan ^2(e+f x)}}-\frac{\cot ^4(e+f x)}{4 a f \sqrt{a+b \tan ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.19724, size = 142, normalized size = 0.66 \[ \frac{(a-b) \left (a \cot ^2(e+f x) \left (2 a \cot ^2(e+f x)-4 a-5 b\right )-\left (8 a^2+12 a b+15 b^2\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{b \tan ^2(e+f x)}{a}+1\right )\right )+8 a^3 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{8 a^3 f (b-a) \sqrt{a+b \tan ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5/(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(8*a^3*Hypergeometric2F1[-1/2, 1, 1/2, (a + b*Tan[e + f*x]^2)/(a - b)] + (a - b)*(a*Cot[e + f*x]^2*(-4*a - 5*b
+ 2*a*Cot[e + f*x]^2) - (8*a^2 + 12*a*b + 15*b^2)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Tan[e + f*x]^2)/a]))
/(8*a^3*(-a + b)*f*Sqrt[a + b*Tan[e + f*x]^2])

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Maple [B]  time = 2.861, size = 79934, normalized size = 371.8 \begin{align*} \text{output too large to display} \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)^(3/2),x)

[Out]

result too large to display

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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)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.69519, size = 3421, normalized size = 15.91 \begin{align*} \text{result too large to display} \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(8*(a^4*b*tan(f*x + e)^6 + a^5*tan(f*x + e)^4)*sqrt(a - b)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e
)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e)^2 + 1)) - ((8*a^4*b - 4*a^3*b^2 - a^2*b^3 - 18*a*b^4 + 15*b^5)*t
an(f*x + e)^6 + (8*a^5 - 4*a^4*b - a^3*b^2 - 18*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^4)*sqrt(a)*log((b*tan(f*x + e
)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) + 2*(2*a^5 - 4*a^4*b + 2*a^3*b^2 - (4*a^4*b
- a^3*b^2 - 18*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^4 - (4*a^5 - 3*a^4*b - 6*a^3*b^2 + 5*a^2*b^3)*tan(f*x + e)^2)*
sqrt(b*tan(f*x + e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^6 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(
f*x + e)^4), 1/16*(16*(a^4*b*tan(f*x + e)^6 + a^5*tan(f*x + e)^4)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 +
a)*sqrt(-a + b)/(a - b)) + ((8*a^4*b - 4*a^3*b^2 - a^2*b^3 - 18*a*b^4 + 15*b^5)*tan(f*x + e)^6 + (8*a^5 - 4*a
^4*b - a^3*b^2 - 18*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^4)*sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^
2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2) - 2*(2*a^5 - 4*a^4*b + 2*a^3*b^2 - (4*a^4*b - a^3*b^2 - 18*a^2*b^3 + 15*
a*b^4)*tan(f*x + e)^4 - (4*a^5 - 3*a^4*b - 6*a^3*b^2 + 5*a^2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/
((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^6 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^4), 1/8*(((8*a^4*b
- 4*a^3*b^2 - a^2*b^3 - 18*a*b^4 + 15*b^5)*tan(f*x + e)^6 + (8*a^5 - 4*a^4*b - a^3*b^2 - 18*a^2*b^3 + 15*a*b^4
)*tan(f*x + e)^4)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) - 4*(a^4*b*tan(f*x + e)^6 + a^5*tan(f
*x + e)^4)*sqrt(a - b)*log((b*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x +
e)^2 + 1)) - (2*a^5 - 4*a^4*b + 2*a^3*b^2 - (4*a^4*b - a^3*b^2 - 18*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^4 - (4*a^
5 - 3*a^4*b - 6*a^3*b^2 + 5*a^2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3
)*f*tan(f*x + e)^6 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^4), 1/8*(((8*a^4*b - 4*a^3*b^2 - a^2*b^3 - 18*a*
b^4 + 15*b^5)*tan(f*x + e)^6 + (8*a^5 - 4*a^4*b - a^3*b^2 - 18*a^2*b^3 + 15*a*b^4)*tan(f*x + e)^4)*sqrt(-a)*ar
ctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + 8*(a^4*b*tan(f*x + e)^6 + a^5*tan(f*x + e)^4)*sqrt(-a + b)*arcta
n(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) - (2*a^5 - 4*a^4*b + 2*a^3*b^2 - (4*a^4*b - a^3*b^2 - 18*a
^2*b^3 + 15*a*b^4)*tan(f*x + e)^4 - (4*a^5 - 3*a^4*b - 6*a^3*b^2 + 5*a^2*b^3)*tan(f*x + e)^2)*sqrt(b*tan(f*x +
e)^2 + a))/((a^6*b - 2*a^5*b^2 + a^4*b^3)*f*tan(f*x + e)^6 + (a^7 - 2*a^6*b + a^5*b^2)*f*tan(f*x + e)^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{5}{\left (e + f x \right )}}{\left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral(cot(e + f*x)**5/(a + b*tan(e + f*x)**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{5}}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^5/(b*tan(f*x + e)^2 + a)^(3/2), x)