\(\int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [70]

Optimal result

Integrand size = 28, antiderivative size = 75 \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {2} \sqrt {a} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 B \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3608, 3561, 212} \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2 B \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\sqrt {2} \sqrt {a} (B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \]

[In]

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

[Out]

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

Rule 212

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

Rule 3561

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

Rule 3608

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] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 0]

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {-i \sqrt {2} \sqrt {a} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 B \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {2 i \left (-i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d}\)	\(63\)
default	\(\frac {2 i \left (-i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}\right )}{d}\)	\(63\)
parts	\(-\frac {i A \sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d}+\frac {B \left (2 \sqrt {a +i a \tan \left (d x +c \right )}-\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(91\)

[In]

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

[Out]

2*I/d*(-I*B*(a+I*a*tan(d*x+c))^(1/2)-1/2*a^(1/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/
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. 272 vs. \(2 (59) = 118\).

Time = 0.26 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.63 \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {4 \, \sqrt {2} B \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} d \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} d \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((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)), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16 \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {i \, {\left (\sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 4 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a\right )}}{2 \, a d} \]

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 7.49 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,A\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d}-\frac {\sqrt {2}\,B\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{d} \]

[In]

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

[Out]

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