\(\int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [976]

Optimal result

Integrand size = 31, antiderivative size = 25 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

-2*I*a/f/(c-I*c*tan(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 32} \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {\sec ^2(e+f x)}{(c-i c \tan (e+f x))^{3/2}} \, dx \\ & = \frac {(i a) \text {Subst}\left (\int \frac {1}{(c+x)^{3/2}} \, dx,x,-i c \tan (e+f x)\right )}{f} \\ & = -\frac {2 i a}{f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {2 i a}{f \sqrt {c -i c \tan \left (f x +e \right )}}\)	\(22\)
default	\(-\frac {2 i a}{f \sqrt {c -i c \tan \left (f x +e \right )}}\)	\(22\)
risch	\(-\frac {i a \sqrt {2}}{\sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(28\)
parts	\(\frac {2 i a c \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}+\frac {i a \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f}\)	\(116\)

[In]

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

[Out]

-2*I*a/f/(c-I*c*tan(f*x+e))^(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. 43 vs. \(2 (19) = 38\).

Time = 0.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.72 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {2} {\left (-i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \]

[In]

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

[Out]

sqrt(2)*(-I*a*e^(2*I*f*x + 2*I*e) - I*a)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f)

Sympy [A] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=\begin {cases} - \frac {2 i a}{f \sqrt {- i c \tan {\left (e + f x \right )} + c}} & \text {for}\: f \neq 0 \\\frac {x \left (i a \tan {\left (e \right )} + a\right )}{\sqrt {- i c \tan {\left (e \right )} + c}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*I*a/(f*sqrt(-I*c*tan(e + f*x) + c)), Ne(f, 0)), (x*(I*a*tan(e) + a)/sqrt(-I*c*tan(e) + c), True)
)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, a}{\sqrt {-i \, c \tan \left (f x + e\right ) + c} f} \]

[In]

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

[Out]

-2*I*a/(sqrt(-I*c*tan(f*x + e) + c)*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.60 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {a\,\left (\sin \left (2\,e+2\,f\,x\right )-{\cos \left (e+f\,x\right )}^2\,2{}\mathrm {i}\right )\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}}{c\,f} \]

[In]

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

[Out]

(a*(sin(2*e + 2*f*x) - cos(e + f*x)^2*2i)*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(
1/2))/(c*f)