\(\int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [591]

Optimal result

Integrand size = 33, antiderivative size = 278 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {(b (A-B)-a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(b (A-B)-a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3662, 3693, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {(b (A-B)-a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )}+\frac {(b (A-B)-a (A+B)) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )}-\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}-\frac {(a (A-B)+b (A+B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )}+\frac {(a (A-B)+b (A+B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )} \]

[In]

Int[(Sqrt[Cot[c + d*x]]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]

[Out]

-(((b*(A - B) - a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d)) + ((b*(A - B) - a*
(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)*d) - (2*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt
[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(Sqrt[a]*(a^2 + b^2)*d) - ((a*(A - B) + b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[
c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)*d) + ((a*(A - B) + b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d
*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^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 3662

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

Rule 3693

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

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]

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \left (-2 \sqrt {2} (b (-A+B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\frac {8 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\sqrt {2} (a (A-B)+b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right ) d} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*(-2*Sqrt[2]*(b*(-A + B) + a*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 +
Sqrt[2]*Sqrt[Tan[c + d*x]]]) + (8*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] -
Sqrt[2]*(a*(A - B) + b*(A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan
[c + d*x]] + Tan[c + d*x]]))*Sqrt[Tan[c + d*x]])/(4*(a^2 + b^2)*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(530\) vs. \(2(240)=480\).

Time = 0.40 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.91

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b -A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +8 A \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) b^{2}-8 B \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) a b \right )}{4 d \left (a^{2}+b^{2}\right ) \sqrt {a b}}\)	\(531\)
default	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b -A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +8 A \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) b^{2}-8 B \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) a b \right )}{4 d \left (a^{2}+b^{2}\right ) \sqrt {a b}}\)	\(531\)

[In]

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

[Out]

1/4/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(A*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*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a-2*A*a
rctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^
(1/2)*a-2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b-A*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)*b+B*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)*b+2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(
1/2)*(a*b)^(1/2)*a+2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+2*B*arctan(-1+2^(1/2)*tan(d*x+
c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a+2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+B*ln(-(2^(1/2)*t
an(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+8*A*arctan(b*tan(
d*x+c)^(1/2)/(a*b)^(1/2))*b^2-8*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*a*b)/(a^2+b^2)/(a*b)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2995 vs. \(2 (240) = 480\).

Time = 4.33 (sec) , antiderivative size = 6020, normalized size of antiderivative = 21.65 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {8 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}}}{4 \, d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(cot(d*x + c))/(b*tan(d*x + c) + a), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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