3.809 \(\int \sqrt{\cot (c+d x)} (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=204 \[ -\frac{\left (a^2-2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b^2}{d \sqrt{\cot (c+d x)}} \]

[Out]

((a^2 + 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*ArcTan[1 + Sqr
t[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b^2)/(d*Sqrt[Cot[c + d*x]]) - ((a^2 - 2*a*b - b^2)*Log[1 - Sqrt[2]*
Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^2 - 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] +
Cot[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.175722, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3673, 3542, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (a^2-2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b^2}{d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^2,x]

[Out]

((a^2 + 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*ArcTan[1 + Sqr
t[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b^2)/(d*Sqrt[Cot[c + d*x]]) - ((a^2 - 2*a*b - b^2)*Log[1 - Sqrt[2]*
Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^2 - 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] +
Cot[c + d*x]])/(2*Sqrt[2]*d)

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3542

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Ta
n[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Tan[e + f*x], x], x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [C]  time = 0.716336, size = 173, normalized size = 0.85 \[ -\frac{\frac{2 \left (a^2-b^2\right ) \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};-\cot ^2(c+d x)\right )}{\sqrt{\cot (c+d x)}}-\frac{2 a^2}{\sqrt{\cot (c+d x)}}+\frac{a b \left (\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )}{\sqrt{2}}-\sqrt{2} a b \left (\tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-\tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^2,x]

[Out]

-((-(Sqrt[2]*a*b*(ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])) - (2*a^2)/
Sqrt[Cot[c + d*x]] + (2*(a^2 - b^2)*Hypergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2])/Sqrt[Cot[c + d*x]] + (a*
b*(-Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))/
Sqrt[2])/d)

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Maple [C]  time = 0.291, size = 1757, normalized size = 8.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x)

[Out]

1/2/d*2^(1/2)*(cos(d*x+c)/sin(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(-I*b^2*((cos(d*x+c)-1)/sin(d*x+c)
)^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(
(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+I*b^2*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticP
i((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)-2*I*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticP
i((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)*a*b-I*a^2*((cos(d*x+c)-1)/si
n(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*Ell
ipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)+I*a^2*((cos(d*x+c)-1)/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*E
llipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+2*I*((cos(d*x+c)-1)/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*E
llipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*a*b-sin(d*x+c)*((cos
(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+
c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a^2+2*sin(d*x+c)*((c
os(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a*b+sin(d*x+c)*((c
os(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b^2-sin(d*x+c)*((c
os(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a^2+2*sin(d*x+c)*(
(cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(
d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a*b+sin(d*x+c)*(
(cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(
d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*b^2+2*sin(d*x+c)
*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2-2*sin(d*x+c)*((cos(d*
x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))
^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+2*2^(1/2)*cos(d*x+c)*b^2-2*2^(
1/2)*b^2)/cos(d*x+c)/sin(d*x+c)^3

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Maxima [A]  time = 1.6242, size = 239, normalized size = 1.17 \begin{align*} -\frac{2 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, b^{2} \sqrt{\tan \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a^2 + 2*
a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)/sqr
t(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*(a^2 - 2*a*b - b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*
x + c) + 1) - 8*b^2*sqrt(tan(d*x + c)))/d

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Fricas [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(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(1/2)*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sqrt(cot(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^2*sqrt(cot(d*x + c)), x)