\(\int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx\) [828]

Optimal result

Integrand size = 23, antiderivative size = 318 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2} \left (a^2+b^2\right )^2 d}+\frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^2 d} \]

[Out]

-b^(3/2)*(5*a^2+b^2)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/a^(3/2)/(a^2+b^2)^2/d-1/2*(a^2-2*a*b-b^2)*arctan
(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)-1/2*(a^2-2*a*b-b^2)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^
2+b^2)^2/d*2^(1/2)-1/4*(a^2+2*a*b-b^2)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+1/4*(a^
2+2*a*b-b^2)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^2/d*2^(1/2)+b^2*cot(d*x+c)^(1/2)/a/(a^2+b^2)/
d/(b+a*cot(d*x+c))

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3754, 3646, 3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^2}+\frac {b^2 \sqrt {\cot (c+d x)}}{a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^2}-\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2} d \left (a^2+b^2\right )^2} \]

[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]*(a^2 + b^2)^2*d) - ((a^2 - 2*a*b - b^2)*
ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^2*d) - (b^(3/2)*(5*a^2 + b^2)*ArcTan[(Sqrt[a]*Sqr
t[Cot[c + d*x]])/Sqrt[b]])/(a^(3/2)*(a^2 + b^2)^2*d) + (b^2*Sqrt[Cot[c + d*x]])/(a*(a^2 + b^2)*d*(b + a*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]*(a^2 + b^2)^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]*(a^2 + b^2)^2*d)

Rule 65

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 210

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

Rule 211

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

Rule 631

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 642

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]

Rule 1176

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 1179

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 1182

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 3615

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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3715

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]

Rule 3734

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0] &&  !LeQ[n, -
1]

Rule 3754

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(b+a \cot (c+d x))^2} \, dx \\ & = \frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\int \frac {-\frac {b^2}{2}+a b \cot (c+d x)-\frac {1}{2} \left (2 a^2+b^2\right ) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a \left (a^2+b^2\right )} \\ & = \frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\int \frac {2 a^2 b-a \left (a^2-b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{a \left (a^2+b^2\right )^2}+\frac {\left (b^2 \left (5 a^2+b^2\right )\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 a \left (a^2+b^2\right )^2} \\ & = \frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {2 \text {Subst}\left (\int \frac {-2 a^2 b+a \left (a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a \left (a^2+b^2\right )^2 d}+\frac {\left (b^2 \left (5 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{2 a \left (a^2+b^2\right )^2 d} \\ & = \frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left (b^2 \left (5 a^2+b^2\right )\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a \left (a^2+b^2\right )^2 d} \\ & = -\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2} \left (a^2+b^2\right )^2 d}+\frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \cot (c+d x))}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \text {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} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2+2 a b-b^2\right ) \text {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} \left (a^2+b^2\right )^2 d} \\ & = -\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2} \left (a^2+b^2\right )^2 d}+\frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \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} \left (a^2+b^2\right )^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} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d} \\ & = \frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2-2 a b-b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {b^{3/2} \left (5 a^2+b^2\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2} \left (a^2+b^2\right )^2 d}+\frac {b^2 \sqrt {\cot (c+d x)}}{a \left (a^2+b^2\right ) d (b+a \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} \left (a^2+b^2\right )^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} \left (a^2+b^2\right )^2 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.95 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\frac {-28 a^{3/2} b^2 \left (a^2-b^2\right ) \cot ^{\frac {3}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\cot ^2(c+d x)\right )-12 a^{7/2} \left (a^2+b^2\right ) \cot ^{\frac {7}{2}}(c+d x) \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},-\frac {a \cot (c+d x)}{b}\right )+7 b^2 \left (-6 \sqrt {2} a^{5/2} b \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+6 \sqrt {2} a^{5/2} b \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )+24 b^{7/2} \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )-24 a^{5/2} b \sqrt {\cot (c+d x)}-24 \sqrt {a} b^3 \sqrt {\cot (c+d x)}+4 a^{7/2} \cot ^{\frac {3}{2}}(c+d x)+4 a^{3/2} b^2 \cot ^{\frac {3}{2}}(c+d x)-3 \sqrt {2} a^{5/2} b \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )+3 \sqrt {2} a^{5/2} b \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )}{42 a^{3/2} b^2 \left (a^2+b^2\right )^2 d} \]

[In]

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

[Out]

(-28*a^(3/2)*b^2*(a^2 - b^2)*Cot[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2] - 12*a^(7/2)*(
a^2 + b^2)*Cot[c + d*x]^(7/2)*Hypergeometric2F1[2, 7/2, 9/2, -((a*Cot[c + d*x])/b)] + 7*b^2*(-6*Sqrt[2]*a^(5/2
)*b*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] + 6*Sqrt[2]*a^(5/2)*b*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + 24*b
^(7/2)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]] - 24*a^(5/2)*b*Sqrt[Cot[c + d*x]] - 24*Sqrt[a]*b^3*Sqrt[Co
t[c + d*x]] + 4*a^(7/2)*Cot[c + d*x]^(3/2) + 4*a^(3/2)*b^2*Cot[c + d*x]^(3/2) - 3*Sqrt[2]*a^(5/2)*b*Log[1 - Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] + 3*Sqrt[2]*a^(5/2)*b*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*
x]]))/(42*a^(3/2)*b^2*(a^2 + b^2)^2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(970\) vs. \(2(278)=556\).

Time = 1.10 (sec) , antiderivative size = 971, normalized size of antiderivative = 3.05

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(971\)
default	\(\text {Expression too large to display}\)	\(971\)

[In]

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

[Out]

1/4/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(
1/2)-tan(d*x+c)-1))*2^(1/2)*(a*b)^(1/2)*a^3*b*tan(d*x+c)-ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*
tan(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*(a*b)^(1/2)*a*b^3*tan(d*x+c)+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1
/2)*(a*b)^(1/2)*a^3*b*tan(d*x+c)-4*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^2*b^2*tan(d*x+c)-2
*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a*b^3*tan(d*x+c)+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))
*2^(1/2)*(a*b)^(1/2)*a^3*b*tan(d*x+c)-4*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^2*b^2*tan(d*
x+c)-2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a*b^3*tan(d*x+c)-2*ln(-(2^(1/2)*tan(d*x+c)^(1/2
)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*2^(1/2)*(a*b)^(1/2)*a^2*b^2*tan(d*x+c)+ln(-(1+2^(1/2)
*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*(a*b)^(1/2)*a^4-ln(-(1+2^(1/2)*
tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*(a*b)^(1/2)*a^2*b^2+2*arctan(1+2
^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^4-4*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^3*
b-2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^2*b^2+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/
2)*(a*b)^(1/2)*a^4-4*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^3*b-2*arctan(-1+2^(1/2)*tan(d*x
+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a^2*b^2-2*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/
2)+tan(d*x+c)))*2^(1/2)*(a*b)^(1/2)*a^3*b+20*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*a^2*b^3*tan(d*x+c)+4*arcta
n(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*b^5*tan(d*x+c)+20*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*a^3*b^2+4*arctan(b*
tan(d*x+c)^(1/2)/(a*b)^(1/2))*a*b^4+4*(a*b)^(1/2)*tan(d*x+c)^(1/2)*a^2*b^2+4*(a*b)^(1/2)*tan(d*x+c)^(1/2)*b^4)
/(a^2+b^2)^2/a/(a+b*tan(d*x+c))/(a*b)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2654 vs. \(2 (278) = 556\).

Time = 0.55 (sec) , antiderivative size = 5337, normalized size of antiderivative = 16.78 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

\[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 283, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (5 \, a^{2} b^{2} + b^{4}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \sqrt {a b}} - \frac {4 \, b^{2}}{{\left (a^{3} b + a b^{3} + \frac {a^{4} + a^{2} b^{2}}{\tan \left (d x + c\right )}\right )} \sqrt {\tan \left (d x + c\right )}} + \frac {2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}}}{4 \, d} \]

[In]

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

[Out]

-1/4*(4*(5*a^2*b^2 + b^4)*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^5 + 2*a^3*b^2 + a*b^4)*sqrt(a*b)) - 4*b
^2/((a^3*b + a*b^3 + (a^4 + a^2*b^2)/tan(d*x + c))*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)))) + 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)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sq
rt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^4 + 2*a^2*b^2 + b^4))/d

Giac [F]

\[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\int { \frac {\sqrt {\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^2} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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