∫ cot3(e + fx)(a + b tan2(e + fx))3∕2 dx

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

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

[Out]

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

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

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Rubi [A] time = 0.17, 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, 98, 156, 63, 208} \[ \frac {\sqrt {a} (2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )}{f}-\frac {a \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

n[e + f*x]^2]/Sqrt[a - b]])/f - (a*Cot[e + f*x]^2*Sqrt[a + b*Tan[e + f*x]^2])/(2*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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -

a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*

f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d

*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;

FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p

] || IntegersQ[p, m + n])

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

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Mathematica [A] time = 0.34, size = 109, normalized size = 0.94 \[ \frac {\sqrt {a} (2 a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a}}\right )-2 (a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan ^2(e+f x)}}{\sqrt {a-b}}\right )-a \cot ^2(e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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fricas [A] time = 0.48, size = 584, normalized size = 5.03 \[ \left [-\frac {2 \, {\left (a - b\right )}^{\frac {3}{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 - 3 \, 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 ) \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, f \tan \left (f x + e\right )^{2}}, -\frac {4 \, {\left (a - b\right )} \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 - 3 \, 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 ) \tan \left (f x + e\right )^{2} + 2 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{4 \, f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - 3 \, b\right )} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} + {\left (a - b\right )}^{\frac {3}{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} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, f \tan \left (f x + e\right )^{2}}, -\frac {\sqrt {-a} {\left (2 \, a - 3 \, b\right )} \arctan \left (\frac {\sqrt {b \tan \left (f x + e\right )^{2} + a} \sqrt {-a}}{a}\right ) \tan \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \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} + \sqrt {b \tan \left (f x + e\right )^{2} + a} a}{2 \, 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)^(3/2),x, algorithm="fricas")

[Out]

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

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

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

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

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

x + e)^2 + a)*sqrt(-a + b)/(a - b))*tan(f*x + e)^2 + sqrt(b*tan(f*x + e)^2 + a)*a)/(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)^(3/2),x, algorithm="giac")

[Out]

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