\(\int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [320]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [F(-1)]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 25, antiderivative size = 122 \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {a-i b} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (i A-B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{d} \]

[Out]

-(I*A+B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))*(a-I*b)^(1/2)/d+(I*A-B)*arctanh((a+b*tan(d*x+c))^(1/2)/
(a+I*b)^(1/2))*(a+I*b)^(1/2)/d+2*B*(a+b*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3609, 3620, 3618, 65, 214} \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {a-i b} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {\sqrt {a+i b} (-B+i A) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 B \sqrt {a+b \tan (c+d x)}}{d} \]

[In]

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

[Out]

-((Sqrt[a - I*b]*(I*A + B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d) + (Sqrt[a + I*b]*(I*A - B)*ArcT
anh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d + (2*B*Sqrt[a + b*Tan[c + d*x]])/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 214

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

Rule 3609

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

Rule 3618

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

Rule 3620

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.98 \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {-i \sqrt {a-i b} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+i \sqrt {a+i b} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 B \sqrt {a+b \tan (c+d x)}}{d} \]

[In]

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

[Out]

((-I)*Sqrt[a - I*b]*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + I*Sqrt[a + I*b]*(A + I*B)*ArcT
anh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]] + 2*B*Sqrt[a + b*Tan[c + d*x]])/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(811\) vs. \(2(102)=204\).

Time = 0.09 (sec) , antiderivative size = 812, normalized size of antiderivative = 6.66

method result size
parts \(\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {B \left (2 \sqrt {a +b \tan \left (d x +c \right )}-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right )}{4}+\frac {\left (\sqrt {a^{2}+b^{2}}-a \right ) \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{d}\) \(812\)
derivativedivides \(\frac {2 B \sqrt {a +b \tan \left (d x +c \right )}}{d}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) B \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}-\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) B \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\) \(968\)
default \(\frac {2 B \sqrt {a +b \tan \left (d x +c \right )}}{d}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}+\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}-\frac {\ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right ) B \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}-\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {b \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \sqrt {a^{2}+b^{2}}}{4 d b}-\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) A \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, a}{4 d b}+\frac {\ln \left (\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-b \tan \left (d x +c \right )-a -\sqrt {a^{2}+b^{2}}\right ) B \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{4 d}+\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B \sqrt {a^{2}+b^{2}}}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {b \arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) A}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}-\frac {\arctan \left (\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}-2 \sqrt {a +b \tan \left (d x +c \right )}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right ) B a}{d \sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\) \(968\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1199 vs. \(2 (97) = 194\).

Time = 0.33 (sec) , antiderivative size = 1199, normalized size of antiderivative = 9.83 \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.81 (sec) , antiderivative size = 845, normalized size of antiderivative = 6.93 \[ \int \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=2\,\mathrm {atanh}\left (\frac {32\,B^2\,b^4\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}-\frac {32\,a\,b^2\,\sqrt {\frac {B^2\,a}{4\,d^2}-\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {-\frac {\sqrt {-B^4\,b^2\,d^4}-B^2\,a\,d^2}{4\,d^4}}-\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}-\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {\frac {\sqrt {-A^4\,b^2\,d^4}-A^2\,a\,d^2}{d^4}}-2\,\mathrm {atanh}\left (\frac {32\,B^2\,b^4\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d^3}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d^3}}+\frac {32\,a\,b^2\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}}{4\,d^4}+\frac {B^2\,a}{4\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-B^4\,b^2\,d^4}}{\frac {16\,B\,b^4\,\sqrt {-B^4\,b^2\,d^4}}{d}+\frac {16\,B\,a^2\,b^2\,\sqrt {-B^4\,b^2\,d^4}}{d}}\right )\,\sqrt {\frac {\sqrt {-B^4\,b^2\,d^4}+B^2\,a\,d^2}{4\,d^4}}-\mathrm {atanh}\left (\frac {d^3\,\left (\frac {16\,\left (A^2\,b^4-A^2\,a^2\,b^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^2}+\frac {16\,a\,b^2\,\left (\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d^4}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}}{16\,\left (A^3\,a^2\,b^3+A^3\,b^5\right )}\right )\,\sqrt {-\frac {\sqrt {-A^4\,b^2\,d^4}+A^2\,a\,d^2}{d^4}}+\frac {2\,B\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{d} \]

[In]

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

[Out]

2*atanh((32*B^2*b^4*((B^2*a)/(4*d^2) - (-B^4*b^2*d^4)^(1/2)/(4*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16*B*
b^4*(-B^4*b^2*d^4)^(1/2))/d^3 + (16*B*a^2*b^2*(-B^4*b^2*d^4)^(1/2))/d^3) - (32*a*b^2*((B^2*a)/(4*d^2) - (-B^4*
b^2*d^4)^(1/2)/(4*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-B^4*b^2*d^4)^(1/2))/((16*B*b^4*(-B^4*b^2*d^4)^(1/2)
)/d + (16*B*a^2*b^2*(-B^4*b^2*d^4)^(1/2))/d))*(-((-B^4*b^2*d^4)^(1/2) - B^2*a*d^2)/(4*d^4))^(1/2) - atanh((d^3
*((16*(A^2*b^4 - A^2*a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2 - (16*a*b^2*((-A^4*b^2*d^4)^(1/2) - A^2*a*d^2)*(
a + b*tan(c + d*x))^(1/2))/d^4)*(((-A^4*b^2*d^4)^(1/2) - A^2*a*d^2)/d^4)^(1/2))/(16*(A^3*b^5 + A^3*a^2*b^3)))*
(((-A^4*b^2*d^4)^(1/2) - A^2*a*d^2)/d^4)^(1/2) - 2*atanh((32*B^2*b^4*((-B^4*b^2*d^4)^(1/2)/(4*d^4) + (B^2*a)/(
4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((16*B*b^4*(-B^4*b^2*d^4)^(1/2))/d^3 + (16*B*a^2*b^2*(-B^4*b^2*d^4)^
(1/2))/d^3) + (32*a*b^2*((-B^4*b^2*d^4)^(1/2)/(4*d^4) + (B^2*a)/(4*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(-B^
4*b^2*d^4)^(1/2))/((16*B*b^4*(-B^4*b^2*d^4)^(1/2))/d + (16*B*a^2*b^2*(-B^4*b^2*d^4)^(1/2))/d))*(((-B^4*b^2*d^4
)^(1/2) + B^2*a*d^2)/(4*d^4))^(1/2) - atanh((d^3*((16*(A^2*b^4 - A^2*a^2*b^2)*(a + b*tan(c + d*x))^(1/2))/d^2
+ (16*a*b^2*((-A^4*b^2*d^4)^(1/2) + A^2*a*d^2)*(a + b*tan(c + d*x))^(1/2))/d^4)*(-((-A^4*b^2*d^4)^(1/2) + A^2*
a*d^2)/d^4)^(1/2))/(16*(A^3*b^5 + A^3*a^2*b^3)))*(-((-A^4*b^2*d^4)^(1/2) + A^2*a*d^2)/d^4)^(1/2) + (2*B*(a + b
*tan(c + d*x))^(1/2))/d