∫ cot3 (e+fx)___∘ a+b tan2(e+fx) dx

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

Optimal. Leaf size=116 \[ \frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}}-\frac {\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f} \]

[Out]

1/2*(2*a+b)*arctanh((a+b*tan(f*x+e)^2)^(1/2)/a^(1/2))/a^(3/2)/f-arctanh((a+b*tan(f*x+e)^2)^(1/2)/(a-b)^(1/2))/

f/(a-b)^(1/2)-1/2*cot(f*x+e)^2*(a+b*tan(f*x+e)^2)^(1/2)/a/f

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Rubi [A] time = 0.16, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3670, 446, 103, 156, 63, 208} \[ \frac {(2 a+b) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 a^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f \sqrt {a-b}}-\frac {\cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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 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 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 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]

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 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]))

Rubi steps

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

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Mathematica [A] time = 0.76, size = 135, normalized size = 1.16 \[ \frac {\left (2 a^2-a b-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a} \left ((b-a) \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}-2 a \sqrt {a-b} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )\right )}{2 a^{3/2} f (a-b)} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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fricas [A] time = 0.46, size = 697, normalized size = 6.01 \[ \left [\frac {2 \, \sqrt {a - b} a^{2} \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 ) \tan \left (f x + e\right )^{2} + {\left (2 \, a^{2} - a b - b^{2}\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 ) \tan \left (f x + e\right )^{2} - 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a^{2} - a b\right )}}{4 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{2}}, -\frac {4 \, a^{2} \sqrt {-a + b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{2} - {\left (2 \, a^{2} - a b - b^{2}\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 ) \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a^{2} - a b\right )}}{4 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{2}}, \frac {\sqrt {a - b} a^{2} \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 ) \tan \left (f x + e\right )^{2} - {\left (2 \, a^{2} - a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} - \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a^{2} - a b\right )}}{2 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{2}}, -\frac {2 \, a^{2} \sqrt {-a + b} \arctan \left (-\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a + b}}{a - b}\right ) \tan \left (f x + e\right )^{2} + {\left (2 \, a^{2} - a b - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} + \sqrt {b \tan \left (f x + e\right )^{2} + a} {\left (a^{2} - a b\right )}}{2 \, {\left (a^{3} - a^{2} b\right )} f \tan \left (f x + e\right )^{2}}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(a - b)*a^2*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)^2 + (2*a^2 - a*b - b^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)*tan(f*x + e)^2 - 2*sqrt(b*tan(f*x + e)^2 + a)*(a^2 - a*b))/((a^3 - a^2*b)*f*tan

(f*x + e)^2), -1/4*(4*a^2*sqrt(-a + b)*arctan(-sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2

- (2*a^2 - a*b - b^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)*tan(f*x + e)^2 + 2*sqrt(b*tan(f*x + e)^2 + a)*(a^2 - a*b))/((a^3 - a^2*b)*f*tan(f*x + e)^2), 1/2*(sqrt(a

- b)*a^2*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)^2 - (2*a^2 - a*b - b^2)*sqrt(-a)*arctan(sqrt(b*tan(f*x + e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^2 - sqrt

(b*tan(f*x + e)^2 + a)*(a^2 - a*b))/((a^3 - a^2*b)*f*tan(f*x + e)^2), -1/2*(2*a^2*sqrt(-a + b)*arctan(-sqrt(b*

tan(f*x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2 + (2*a^2 - a*b - b^2)*sqrt(-a)*arctan(sqrt(b*tan(f*x

+ e)^2 + a)*sqrt(-a)/a)*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*(a^2 - a*b))/((a^3 - a^2*b)*f*tan(f*x + e)

^2)]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^3/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab