\(\int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [155]

Optimal result

Integrand size = 38, antiderivative size = 152 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {(-1)^{3/4} \sqrt {a} (2 A-i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

-(-1)^(3/4)*(2*A-I*B)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)/d-(1+I)*(A-
I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^(1/2)/d+B*tan(d*x+c)^(1/2)*(a+I*a*tan(
d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3678, 3682, 3625, 211, 3680, 65, 223, 209} \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {(-1)^{3/4} \sqrt {a} (2 A-i B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

-(((-1)^(3/4)*Sqrt[a]*(2*A - I*B)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/
d) - ((1 + I)*Sqrt[a]*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d +
(B*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3625

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

Rule 3678

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

Rule 3680

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

Rule 3682

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

Rubi steps \begin{align*} \text {integral}& = \frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {a B}{2}+\frac {1}{2} a (2 A-i B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{a} \\ & = \frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+(-i A-B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx+\frac {(2 i A+B) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a} \\ & = \frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\left (2 a^2 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(a (2 i A+B)) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d} \\ & = -\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {(a (2 i A+B)) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {(a (2 i A+B)) \text {Subst}\left (\int \frac {1}{1-i a x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \\ & = -\frac {\sqrt [4]{-1} \sqrt {a} (2 i A+B) \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1+i) \sqrt {a} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {B \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.55 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt [4]{-1} a (2 A-i B) \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}+\frac {a B (1+i \tan (c+d x)) \tan (c+d x)-\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}}}{d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((-1)^(1/4)*a*(2*A - I*B)*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[1 + I*Tan[c + d*x]] + (a*B*(1 + I*Tan[c
+ d*x])*Tan[c + d*x] - Sqrt[2]*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*
Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 576 vs. \(2 (123 ) = 246\).

Time = 0.16 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.80

method	result	size
parts	\(-\frac {A \left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )-\sqrt {i a}\, \sqrt {2}\, \tan \left (d x +c \right ) \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+2 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \tan \left (d x +c \right )+2 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )}+\frac {B \left (i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}+\ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right ) \left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(577\)
derivativedivides	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+2 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+i A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -2 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +2 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(709\)
default	\(-\frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+2 i A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+i A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a -A \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a -2 i B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )+B \sqrt {i a}\, \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +2 B \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+2 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \left (-\tan \left (d x +c \right )+i\right ) \sqrt {-i a}}\)	\(709\)

[In]

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

[Out]

-1/2*A/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))-(I*a)^(1/2)*2^(1/2)*tan(d*x+c)*ln((2*2^(1
/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))+2*I*(-I*a)^(1/2)*ln
(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*tan(d*x+c)+2*(-I*a)
^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2)))/(a*tan(d*
x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)+1/2*B/d*(I*2^(1/2)*ln((2*2^(1/2)*(-I*a)^
(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*a+2*(a*tan(d*x+c)*
(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)+ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a)*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/(a*tan(d*x+c)*(1+I
*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 673 vs. \(2 (116) = 232\).

Time = 0.27 (sec) , antiderivative size = 673, normalized size of antiderivative = 4.43 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2 \, \sqrt {2} B \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} d \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} d \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} d \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} \log \left (-\frac {{\left (\sqrt {2} d \sqrt {-\frac {{\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 2 \, d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, A + B}\right ) - d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (2 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - 2 \, d \sqrt {\frac {{\left (4 i \, A^{2} + 4 \, A B - i \, B^{2}\right )} a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 i \, A + B}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*sqrt(tan(d*x + c)), x)

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0]Warning, replacing 0 by 32, a substitution variable should perhaps be pur
ged.Warning

Mupad [B] (verification not implemented)

Time = 28.49 (sec) , antiderivative size = 2225, normalized size of antiderivative = 14.64 \[ \int \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

- ((B*tan(c + d*x)^(3/2)*2i)/(d*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^3) + (2*B*tan(c + d*x)^(1/2))/(a*d*(
(a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))))/(tan(c + d*x)^2/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^4 - 1/a^2
+ (tan(c + d*x)*2i)/(a*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - ((-a)^(1/2)*atan((A^4*(-a)^(21/2)*tan(
c + d*x)^(1/2)*(7168 - 7168i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 40
96*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)
^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2
*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(
c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2
)) + (B^4*(-a)^(21/2)*tan(c + d*x)^(1/2)*(1024 - 1024i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*
3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x)
)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a
^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c
+ d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d
*x)*1i)^(1/2) - a^(1/2))^2)) + (A*B^3*(-a)^(21/2)*tan(c + d*x)^(1/2)*(8192 + 8192i))/(((a + a*tan(c + d*x)*1i)
^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i -
(3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(
1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*
x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - (A^3*B*(-a)^(21/2)*tan(c + d*x)^(1/2)*(20480 + 2048
0i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512i - 4096*A*B^3*a^10 + 10240*A^3*
B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (512
*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A^2*B^2*a^11*tan(c + d*x))/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/
2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2)) - (A^2*B^2*(-a)^(21/2)*
tan(c + d*x)^(1/2)*(20480 - 20480i))/(((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*(A^4*a^10*3584i + B^4*a^10*512
i - 4096*A*B^3*a^10 + 10240*A^3*B*a^10 - A^2*B^2*a^10*10240i - (3584*A^4*a^11*tan(c + d*x))/((a + a*tan(c + d*
x)*1i)^(1/2) - a^(1/2))^2 - (512*B^4*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (10240*A
^2*B^2*a^11*tan(c + d*x))/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 - (A*B^3*a^11*tan(c + d*x)*4096i)/((a +
a*tan(c + d*x)*1i)^(1/2) - a^(1/2))^2 + (A^3*B*a^11*tan(c + d*x)*10240i)/((a + a*tan(c + d*x)*1i)^(1/2) - a^(1
/2))^2)))*(A*1i + B)*(1 - 1i))/d - ((-1)^(1/4)*a^(1/2)*atan(((-1)^(1/4)*A^5*tan(c + d*x)^(1/2)*25690112i)/(a^(
15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8
- (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) + (262144*(-1)^(1/4)*B^5*tan(c +
d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (367
0016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) + ((-1)^(1/4)*A*B
^4*tan(c + d*x)^(1/2)*3670016i)/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5
*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8
)) + (56885248*(-1)^(1/4)*A^4*B*tan(c + d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((2569
0112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (4
8234496*A^3*B^2)/a^8)) - (19398656*(-1)^(1/4)*A^2*B^3*tan(c + d*x)^(1/2))/(a^(15/2)*((a + a*tan(c + d*x)*1i)^(
1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*56885248i)/a^8 + (A^2*B
^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)) - ((-1)^(1/4)*A^3*B^2*tan(c + d*x)^(1/2)*48234496i)/(a^(15/2)*((a
+ a*tan(c + d*x)*1i)^(1/2) - a^(1/2))*((25690112*A^5)/a^8 - (B^5*262144i)/a^8 + (3670016*A*B^4)/a^8 - (A^4*B*
56885248i)/a^8 + (A^2*B^3*19398656i)/a^8 - (48234496*A^3*B^2)/a^8)))*(2*A - B*1i)*2i)/d