\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx\) [136]

Optimal result

Integrand size = 36, antiderivative size = 236 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \]

[Out]

(1/8-1/8*I)*(A+(2-I)*B)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(1/8-1/8*I)*(A+(2-I)*B)*arctan(1+2^(1/
2)*tan(d*x+c)^(1/2))/a/d*2^(1/2)+(1/16+1/16*I)*(A-(2+I)*B)*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/
2)-(1/16+1/16*I)*(A-(2+I)*B)*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d*2^(1/2)+1/2*(I*A-B)*tan(d*x+c)^(1/2
)/d/(a+I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3676, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {(-B+i A) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} a d} \]

[In]

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

[Out]

((-1/4 + I/4)*(A + (2 - I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) + ((1/4 - I/4)*(A + (2 - I
)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*a*d) + ((1/8 + I/8)*(A - (2 + I)*B)*Log[1 - Sqrt[2]*Sqrt
[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) - ((1/8 + I/8)*(A - (2 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
] + Tan[c + d*x]])/(Sqrt[2]*a*d) + ((I*A - B)*Sqrt[Tan[c + d*x]])/(2*d*(a + I*a*Tan[c + d*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 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 3676

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

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{2 a^2} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {\frac {1}{2} a (i A-B)-\frac {1}{2} a (A-3 i B) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a^2 d} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}-\frac {\left (\left (\frac {1}{4}+\frac {i}{4}\right ) (A-(2+i) B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}--\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d} \\ & = \frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac {\left (\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}--\frac {\left (\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}--\frac {\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (A+(2-i) B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}--\frac {\left (\left (\frac {1}{8}-\frac {i}{8}\right ) (A+(2-i) B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d} \\ & = \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))}--\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d} \\ & = -\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A+(2-i) B) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-(2+i) B) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{\sqrt {2} a d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{2 d (a+i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [4]{-1} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-2 \sqrt [4]{-1} B \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\frac {(A+i B) \sqrt {\tan (c+d x)}}{-i+\tan (c+d x)}}{2 a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.53

method	result	size
derivativedivides	\(\frac {\frac {i \left (-\frac {i \left (i B +A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 B \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (\frac {A}{4}-\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(124\)
default	\(\frac {\frac {i \left (-\frac {i \left (i B +A \right ) \left (\sqrt {\tan }\left (d x +c \right )\right )}{\tan \left (d x +c \right )-i}-\frac {4 B \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}\right )}{2}+\frac {4 \left (\frac {A}{4}-\frac {i B}{4}\right ) \arctan \left (\frac {2 \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}}{d a}\)	\(124\)

[In]

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

[Out]

1/d/a*(1/2*I*(-I*(A+I*B)*tan(d*x+c)^(1/2)/(tan(d*x+c)-I)-4*B/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^
(1/2)-I*2^(1/2))))+4*(1/4*A-1/4*I*B)/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(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. 570 vs. \(2 (177) = 354\).

Time = 0.27 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.42 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {{\left (a d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} - {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 2 \, a d \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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}} \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, a d \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \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}} \sqrt {\frac {i \, B^{2}}{a^{2} d^{2}}} - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - B\right )} \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 (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

[In]

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

[Out]

-1/8*(a*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(2*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*s
qrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) + (A - I*B
)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - a*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*e^(2*I*d*
x + 2*I*c)*log(-2*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)
)*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 2
*a*d*sqrt(I*B^2/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*
c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(I*B^2/(a^2*d^2)) + I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*a*d*sqrt(I*B^2
/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(-((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) + 1))*sqrt(I*B^2/(a^2*d^2)) - I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*((I*A - B)*e^(2*I*d*x + 2*I*
c) + I*A - B)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(a*d)

Sympy [F]

\[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=- \frac {i \left (\int \frac {A \sqrt {\tan {\left (c + d x \right )}}}{\tan {\left (c + d x \right )} - i}\, dx + \int \frac {B \tan ^{\frac {3}{2}}{\left (c + d x \right )}}{\tan {\left (c + d x \right )} - i}\, dx\right )}{a} \]

[In]

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

[Out]

-I*(Integral(A*sqrt(tan(c + d*x))/(tan(c + d*x) - I), x) + Integral(B*tan(c + d*x)**(3/2)/(tan(c + d*x) - I),
x))/a

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.41 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\frac {\left (i - 1\right ) \, \sqrt {2} B \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{2 \, a d} - \frac {\left (i - 1\right ) \, \sqrt {2} {\left (A - i \, B\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{4 \, a d} + \frac {A \sqrt {\tan \left (d x + c\right )} + i \, B \sqrt {\tan \left (d x + c\right )}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.23 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx=-\mathrm {atan}\left (\frac {2\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}}{B}\right )\,\sqrt {\frac {B^2\,1{}\mathrm {i}}{4\,a^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {4\,a\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}}{B}\right )\,\sqrt {-\frac {B^2\,1{}\mathrm {i}}{16\,a^2\,d^2}}\,2{}\mathrm {i}-\frac {2\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,A\,\mathrm {atanh}\left (4\,\sqrt {\frac {1}{16}{}\mathrm {i}}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{a\,d}+\frac {A\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {B\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \]

[In]

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

[Out]

(A*tan(c + d*x)^(1/2)*1i)/(2*a*d*(tan(c + d*x)*1i + 1)) - atan((4*a*d*tan(c + d*x)^(1/2)*(-(B^2*1i)/(16*a^2*d^
2))^(1/2))/B)*(-(B^2*1i)/(16*a^2*d^2))^(1/2)*2i - (2*(1i/16)^(1/2)*A*atanh(4*(1i/16)^(1/2)*tan(c + d*x)^(1/2))
)/(a*d) - atan((2*a*d*tan(c + d*x)^(1/2)*((B^2*1i)/(4*a^2*d^2))^(1/2))/B)*((B^2*1i)/(4*a^2*d^2))^(1/2)*2i - (B
*tan(c + d*x)^(1/2))/(2*a*d*(tan(c + d*x)*1i + 1))