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

Optimal. Leaf size=166 \[ -\frac{\left (8 a^2+4 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f}+\frac{(4 a+3 b) \cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 a^2 f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f \sqrt{a-b}}-\frac{\cot ^4(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.213681, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3670, 446, 103, 151, 156, 63, 208} \[ -\frac{\left (8 a^2+4 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f}+\frac{(4 a+3 b) \cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 a^2 f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f \sqrt{a-b}}-\frac{\cot ^4(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.87309, size = 162, normalized size = 0.98 \[ \frac{\left (4 a^2 b-8 a^3+a b^2+3 b^3\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )+\sqrt{a} \left (8 a^2 \sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )+(b-a) \cot ^2(e+f x) \sqrt{a+b \tan ^2(e+f x)} \left (2 a \cot ^2(e+f x)-4 a-3 b\right )\right )}{8 a^{5/2} f (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^3 + 4*a^2*b + a*b^2 + 3*b^3)*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]] + Sqrt[a]*(8*a^2*Sqrt[a - b]*A
rcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]] + (-a + b)*Cot[e + f*x]^2*(-4*a - 3*b + 2*a*Cot[e + f*x]^2)*Sqr
t[a + b*Tan[e + f*x]^2]))/(8*a^(5/2)*(a - b)*f)

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Maple [B]  time = 0.266, size = 7641, normalized size = 46. \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)^(1/2),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)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.94811, size = 1994, normalized size = 12.01 \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)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(8*sqrt(a - b)*a^3*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))*tan(f*x + e)^4 + (8*a^3 - 4*a^2*b - a*b^2 - 3*b^3)*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)*tan(f*x + e)^4 - 2*(2*a^3 - 2*a^2*b - (4*a^3 - a^2*b - 3*a*b^2)*
tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^4), 1/16*(16*a^3*sqrt(-a + b)*arctan
(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^4 + (8*a^3 - 4*a^2*b - a*b^2 - 3*b^3)*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)*tan(f*x + e)^4 - 2*(2*a^3
- 2*a^2*b - (4*a^3 - a^2*b - 3*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e
)^4), 1/8*(4*sqrt(a - b)*a^3*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))*tan(f*x + e)^4 + (8*a^3 - 4*a^2*b - a*b^2 - 3*b^3)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)
*sqrt(-a)/a)*tan(f*x + e)^4 - (2*a^3 - 2*a^2*b - (4*a^3 - a^2*b - 3*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)
^2 + a))/((a^4 - a^3*b)*f*tan(f*x + e)^4), 1/8*(8*a^3*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a
+ b)/(a - b))*tan(f*x + e)^4 + (8*a^3 - 4*a^2*b - a*b^2 - 3*b^3)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sq
rt(-a)/a)*tan(f*x + e)^4 - (2*a^3 - 2*a^2*b - (4*a^3 - a^2*b - 3*a*b^2)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2
+ a))/((a^4 - a^3*b)*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 )}}{\sqrt{a + b \tan ^{2}{\left (e + f x \right )}}}\, 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)**(1/2),x)

[Out]

Integral(cot(e + f*x)**5/sqrt(a + b*tan(e + f*x)**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}}{\sqrt{b \tan \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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