\(\int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\) [790]

Optimal result

Integrand size = 45, antiderivative size = 104 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 i \sqrt {a} A \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f} \]

[Out]

-2*I*A*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))*a^(1/2)*c^(1/2)/f+B*(a+I*a*ta
n(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/f

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3669, 81, 65, 223, 209} \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i \sqrt {a} A \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \]

[In]

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

[Out]

((-2*I)*Sqrt[a]*A*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f
+ (B*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/f

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 81

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

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 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 3669

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*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

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

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 i \sqrt {a} A \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {B \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f} \]

[In]

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

[Out]

((-2*I)*Sqrt[a]*A*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f
+ (B*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/f

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(121\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (A \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(121\)
parts	\(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}+\frac {B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}}{f}\)	\(134\)

[In]

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

[Out]

1/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*(A*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e
)^2))^(1/2))/(a*c)^(1/2))*a*c+B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*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. 293 vs. \(2 (78) = 156\).

Time = 0.26 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.82 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {4 \, B \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} + \sqrt {\frac {A^{2} a c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (A e^{\left (3 i \, f x + 3 i \, e\right )} + A e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{A e^{\left (2 i \, f x + 2 i \, e\right )} + A}\right ) - \sqrt {\frac {A^{2} a c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (A e^{\left (3 i \, f x + 3 i \, e\right )} + A e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {A^{2} a c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{A e^{\left (2 i \, f x + 2 i \, e\right )} + A}\right )}{2 \, f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(I*a*(tan(e + f*x) - I))*sqrt(-I*c*(tan(e + f*x) + I))*(A + B*tan(e + f*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. 447 vs. \(2 (78) = 156\).

Time = 0.55 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.30 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {{\left (2 \, {\left (A \cos \left (2 \, f x + 2 \, e\right ) + i \, A \sin \left (2 \, f x + 2 \, e\right ) + A\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 2 \, {\left (A \cos \left (2 \, f x + 2 \, e\right ) + i \, A \sin \left (2 \, f x + 2 \, e\right ) + A\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 i \, B \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (-i \, A \cos \left (2 \, f x + 2 \, e\right ) + A \sin \left (2 \, f x + 2 \, e\right ) - i \, A\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - {\left (i \, A \cos \left (2 \, f x + 2 \, e\right ) - A \sin \left (2 \, f x + 2 \, e\right ) + i \, A\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 \, B \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{-2 \, f {\left (i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \]

[In]

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

[Out]

-(2*(A*cos(2*f*x + 2*e) + I*A*sin(2*f*x + 2*e) + A)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 2*(A*cos(2*f*x + 2*e) + I*A*sin(2*f*x + 2*e) +
A)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 1) + 4*I*B*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-I*A*cos(2*f*x + 2*e) + A*sin(2
*f*x + 2*e) - I*A)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*
e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (I*A*cos(2*f*x + 2*e)
- A*sin(2*f*x + 2*e) + I*A)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(
2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 4*B*sin(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/(f*(-2*I*cos(2*f*x + 2*e) + 2*sin(2*f*x + 2*e)
- 2*I))

Giac [F]

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

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)*sqrt(I*a*tan(f*x + e) + a)*sqrt(-I*c*tan(f*x + e) + c), x)

Mupad [B] (verification not implemented)

Time = 10.46 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.28 \[ \int \sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {A\,\sqrt {a}\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )\,4{}\mathrm {i}}{f}+\frac {\sqrt {2}\,B\,\sqrt {\frac {c}{2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}\,\sqrt {\frac {a\,\left (2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}}{f} \]

[In]

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

[Out]

(2^(1/2)*B*(c/(sin(2*e + 2*f*x)*1i + 2*cos(e + f*x)^2))^(1/2)*((a*(sin(2*e + 2*f*x)*1i + 2*cos(e + f*x)^2))/(2
*cos(e + f*x)^2))^(1/2))/f - (A*a^(1/2)*c^(1/2)*atan((c^(1/2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(a^(1
/2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))))*4i)/f