\(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [92]

Optimal result

Integrand size = 34, antiderivative size = 109 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i B \sqrt {a+i a \tan (c+d x)}}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3673, 3607, 3561, 212} \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i B \sqrt {a+i a \tan (c+d x)}}{a d} \]

[In]

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

[Out]

-(((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d)) - (A + I*B)/(d*Sqrt[a
+ I*a*Tan[c + d*x]]) - ((2*I)*B*Sqrt[a + I*a*Tan[c + d*x]])/(a*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 3607

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

Rule 3673

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

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

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {A+i B}{d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i B \sqrt {a+i a \tan (c+d x)}}{a d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {-2 i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}}{a d}\)	\(88\)
default	\(\frac {-2 i B \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a \left (i B +A \right )}{\sqrt {a +i a \tan \left (d x +c \right )}}-\frac {\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2}}{a d}\)	\(88\)
parts	\(\frac {A \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {1}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d}+\frac {2 i B \left (-\sqrt {a +i a \tan \left (d x +c \right )}+\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d a}\)	\(130\)

[In]

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

[Out]

2/d/a*(-I*B*(a+I*a*tan(d*x+c))^(1/2)-1/2*a*(A+I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/4*a^(1/2)*(A-I*B)*2^(1/2)*arctan
h(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. 327 vs. \(2 (84) = 168\).

Time = 0.25 (sec) , antiderivative size = 327, normalized size of antiderivative = 3.00 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (\sqrt {2} a d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} a d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, \sqrt {2} {\left ({\left (A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

[In]

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

[Out]

1/4*(sqrt(2)*a*d*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*e^(I*d*x + I*c)*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) + (I
*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I
*d*x - I*c)/(I*A + B)) - sqrt(2)*a*d*sqrt((A^2 - 2*I*A*B - B^2)/(a*d^2))*e^(I*d*x + I*c)*log(-4*((-I*A - B)*a*
e^(I*d*x + I*c) + (-I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B -
B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A + B)) - 2*sqrt(2)*((A + 5*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 8 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a - \frac {4 \, {\left (A + i \, B\right )} a^{2}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{4 \, a^{2} d} \]

[In]

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

[Out]

1/4*(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))) - 8*I*sqrt(I*a*tan(d*x + c) + a)*B*a - 4*(A + I*B)*a^2/sqrt(I*a*tan(d*x + c) + a))/(a^2*d
)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.29 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {A}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,d}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,A\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{2\,\sqrt {a}\,d} \]

[In]

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

[Out]

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