\(\int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [115]

Optimal result

Integrand size = 34, antiderivative size = 55 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 i a B \sqrt {\tan (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3673, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 i a B \sqrt {\tan (c+d x)}}{d}-\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d} \]

[In]

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

[Out]

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

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 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 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 a B \sqrt {\tan (c+d x)}}{d}+\int \frac {a (A-i B)+a (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 i a B \sqrt {\tan (c+d x)}}{d}+\frac {\left (2 a^2 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{a (A-i B)-a (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 i a B \sqrt {\tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {2 \sqrt [4]{-1} a (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}+\frac {2 i a B \sqrt {\tan (c+d x)}}{d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (45 ) = 90\).

Time = 0.03 (sec) , antiderivative size = 201, normalized size of antiderivative = 3.65

method	result	size
derivativedivides	\(\frac {a \left (2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (i A +B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(201\)
default	\(\frac {a \left (2 i B \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-i B +A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (i A +B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(201\)
parts	\(\frac {\left (i a A +B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {a A \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {i a B \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(293\)

[In]

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

[Out]

1/d*a*(2*I*B*tan(d*x+c)^(1/2)+1/4*(A-I*B)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d
*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(I*A+
B)*2^(1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(
1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(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. 314 vs. \(2 (43) = 86\).

Time = 0.25 (sec) , antiderivative size = 314, normalized size of antiderivative = 5.71 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {-4 i \, B a \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (43) = 86\).

Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.75 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {-8 i \, B a \sqrt {\tan \left (d x + c\right )} + {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a}{4 \, d} \]

[In]

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

[Out]

-1/4*(-8*I*B*a*sqrt(tan(d*x + c)) + (2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(t
an(d*x + c)))) + 2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + sq
rt(2)*((I - 1)*A + (I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1
)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {\left (i - 1\right ) \, \sqrt {2} {\left (-i \, A a - B a\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} + \frac {2 i \, B a \sqrt {\tan \left (d x + c\right )}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {2\,{\left (-1\right )}^{1/4}\,B\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {B\,a\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,2{}\mathrm {i}}{d}+\frac {\sqrt {2}\,A\,a\,\mathrm {atan}\left (\sqrt {2}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (1+1{}\mathrm {i}\right )}{d} \]

[In]

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

[Out]

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