\(\int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [576]

Optimal result

Integrand size = 31, antiderivative size = 205 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b B}{d \sqrt {\cot (c+d x)}}-\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3662, 3672, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {(a (A+B)+b (A-B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b B}{d \sqrt {\cot (c+d x)}} \]

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

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 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 3672

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.87 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {\cot (c+d x)} \left (2 \sqrt {2} (b (A-B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\sqrt {2} (a (A-B)-b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )-8 b B \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}}{4 d} \]

[In]

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

[Out]

-1/4*(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]]]) + 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]]) - 8*b*B*Sqrt[Tan[c + d*x]])*Sqrt[Tan[c + d*x]])/
d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(175)=350\).

Time = 0.36 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.08

method	result	size
derivativedivides	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, b -B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a -2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a -2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +8 B b \sqrt {\tan \left (d x +c \right )}\right )}{4 d}\)	\(426\)
default	\(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, b -B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a -2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a -2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, a +8 B b \sqrt {\tan \left (d x +c \right )}\right )}{4 d}\)	\(426\)

[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+2*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a+2*A*arctan(1+2^(1/2)*tan(d*x+
c)^(1/2))*2^(1/2)*b+2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a+2*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*
2^(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)*b-B*l
n(-(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)*b+2*B*arctan(1+2^(
1/2)*tan(d*x+c)^(1/2))*2^(1/2)*a-2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b+2*B*arctan(-1+2^(1/2)*tan(d*
x+c)^(1/2))*2^(1/2)*a-2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b+B*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+8*B*b*tan(d*x+c)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2216 vs. \(2 (175) = 350\).

Time = 0.71 (sec) , antiderivative size = 2216, normalized size of antiderivative = 10.81 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/2*(d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B -
A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4
))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^
2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*A^2*B
- B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b
+ d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(
A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b -
4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c))) - d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2
)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2
+ 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*
B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4
- 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B -
B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^
3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^
4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(t
an(d*x + c))) - d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 -
8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^
4)*b^4)/d^4))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4
- 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^3 - A*B^2)*a^3
- (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2
- B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^
2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^
3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c))) + d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2
*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^
4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)*d^3*sqrt(-((A
^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a
*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2
+ (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*
a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^
2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*
b^4)*sqrt(tan(d*x + c))) + 4*B*b*sqrt(tan(d*x + c)))/d

Sympy [F]

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

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.87 \[ \int \sqrt {\cot (c+d x)} (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, B b \sqrt {\tan \left (d x + c\right )}}{4 \, d} \]

[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*(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) - 8*B*b*sqrt(tan(d*x + c)))/d

Giac [F]

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

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

Mupad [F(-1)]

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

[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)