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

Optimal. Leaf size=147 \[ -\frac{b (2 a-b)}{a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{b}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.213703, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3670, 446, 85, 152, 156, 63, 208} \[ -\frac{b (2 a-b)}{a^2 f (a-b)^2 \sqrt{a+b \tan ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{a^{5/2} f}-\frac{b}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]]/(a^(5/2)*f)) + ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a - b]]/(
(a - b)^(5/2)*f) - b/(3*a*(a - b)*f*(a + b*Tan[e + f*x]^2)^(3/2)) - ((2*a - b)*b)/(a^2*(a - b)^2*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 85

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

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

Mathematica [C]  time = 0.370316, size = 94, normalized size = 0.64 \[ \frac{(a-b) \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{b \tan ^2(e+f x)}{a}+1\right )-a \text{Hypergeometric2F1}\left (-\frac{3}{2},1,-\frac{1}{2},\frac{a+b \tan ^2(e+f x)}{a-b}\right )}{3 a f (a-b) \left (a+b \tan ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 15.647, size = 331597, normalized size = 2255.8 \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)/(a+b*tan(f*x+e)^2)^(5/2),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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.18106, size = 3623, normalized size = 24.65 \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)/(a+b*tan(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^3*b^2*tan(f*x + e)^4 + 2*a^4*b*tan(f*x + e)^2 + a^5)*sqrt(a - b)*log((b*tan(f*x + e)^2 + 2*sqrt(b*t
an(f*x + e)^2 + a)*sqrt(a - b) + 2*a - b)/(tan(f*x + e)^2 + 1)) + 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^
3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*
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*(7*a^4*b - 11*
a^3*b^2 + 4*a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2
- 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x +
e)^2 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f), 1/6*(6*(a^3*b^2*tan(f*x + e)^4 + 2*a^4*b*tan(f*x + e)^2 + a^5
)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) + 3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b
^3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x
+ e)^2)*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*(7*a^4
*b - 11*a^3*b^2 + 4*a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((
a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*ta
n(f*x + e)^2 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f), 1/6*(6*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3 + (a^3*b^
2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x + e)^2)*sqrt
(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + 3*(a^3*b^2*tan(f*x + e)^4 + 2*a^4*b*tan(f*x + e)^2 + a^5)
*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*(7*a^4*b - 11*a^3*b^2 + 4*a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)
^2 + a))/((a^6*b^2 - 3*a^5*b^3 + 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^
4*b^4)*f*tan(f*x + e)^2 + (a^8 - 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f), 1/3*(3*(a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^
3 + (a^3*b^2 - 3*a^2*b^3 + 3*a*b^4 - b^5)*tan(f*x + e)^4 + 2*(a^4*b - 3*a^3*b^2 + 3*a^2*b^3 - a*b^4)*tan(f*x +
e)^2)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + 3*(a^3*b^2*tan(f*x + e)^4 + 2*a^4*b*tan(f*x +
e)^2 + a^5)*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) - (7*a^4*b - 11*a^3*b^2 + 4*
a^2*b^3 + 3*(2*a^3*b^2 - 3*a^2*b^3 + a*b^4)*tan(f*x + e)^2)*sqrt(b*tan(f*x + e)^2 + a))/((a^6*b^2 - 3*a^5*b^3
+ 3*a^4*b^4 - a^3*b^5)*f*tan(f*x + e)^4 + 2*(a^7*b - 3*a^6*b^2 + 3*a^5*b^3 - a^4*b^4)*f*tan(f*x + e)^2 + (a^8
- 3*a^7*b + 3*a^6*b^2 - a^5*b^3)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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