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

Optimal. Leaf size=74 \[ \frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.100548, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3670, 446, 83, 63, 208} \[ \frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[a]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/Sqrt[a]])/f) + (Sqrt[a - b]*ArcTanh[Sqrt[a + b*Tan[e + f*x]^2]/S
qrt[a - b]])/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 83

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*
x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

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

Mathematica [A]  time = 0.0612342, size = 72, normalized size = 0.97 \[ \frac{\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.256, size = 615, normalized size = 8.3 \begin{align*} \text{result 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)^(1/2),x)

[Out]

-1/4/f/(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*4^(1/2)*cos(f*x+e)*(cos(f*x+e)-1)*(l
n(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos
(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*a^(
1/2)*(a-b)^(1/2)-a^(1/2)*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+
cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1)
)*(a-b)^(1/2)+2*ln(4*cos(f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f
*x+e)*a-4*b*cos(f*x+e)+4*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a-2*ln(4*cos(
f*x+e)*(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)+4*cos(f*x+e)*a-4*b*cos(f*x+e)+4*
(a-b)^(1/2)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*b)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f
*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72163, size = 953, normalized size = 12.88 \begin{align*} \left [\frac{\sqrt{a - b} \log \left (\frac{b \tan \left (f x + e\right )^{2} + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right ) + \sqrt{a} \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac{2 \, \sqrt{-a + b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{a - b}\right ) + \sqrt{a} \log \left (\frac{b \tan \left (f x + e\right )^{2} - 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a} + 2 \, a}{\tan \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac{2 \, \sqrt{-a} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a}}{a}\right ) + \sqrt{a - b} \log \left (\frac{b \tan \left (f x + e\right )^{2} + 2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 2 \, a - b}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, f}, \frac{\sqrt{-a} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a}}{a}\right ) + \sqrt{-a + b} \arctan \left (-\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{a - b}\right )}{f}\right ] \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)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(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)) + 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))/f, 1/2*(2
*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b)) + sqrt(a)*log((b*tan(f*x + e)^2 - 2*sqr
t(b*tan(f*x + e)^2 + a)*sqrt(a) + 2*a)/tan(f*x + e)^2))/f, 1/2*(2*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*s
qrt(-a)/a) + 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)))/f, (sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a) + sqrt(-a + b)*arctan(-sqrt(b*tan(f*x
+ e)^2 + a)*sqrt(-a + b)/(a - b)))/f]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan ^{2}{\left (e + f x \right )}} \cot{\left (e + f x \right )}\, 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)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (f x + e\right )^{2} + a} \cot \left (f x + e\right )\,{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)^(1/2),x, algorithm="giac")

[Out]

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