\(\int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1163]

Optimal result

Integrand size = 32, antiderivative size = 129 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{\left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}} \] Output:

-I*2^(1/2)*a^(1/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^ 
(1/2)/(a+I*a*tan(f*x+e))^(1/2))/(c-I*d)^(3/2)/f-2*d*(a+I*a*tan(f*x+e))^(1/ 
2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {2} a (-i c+d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{\left (c^2+d^2\right ) f} \] Input:

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4031, 3042, 4027, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4031

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4027

\(\displaystyle -\frac {2 i a^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f (c-i d)}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}\)

Input:

Int[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

((-I)*Sqrt[2]*Sqrt[a]*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/( 
Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2)*f) - (2*d*Sqr 
t[a + I*a*Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]])
 

Defintions of rubi rules used

rule 221

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027

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

rule 4031

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1290 vs. \(2 (104 ) = 208\).

Time = 0.64 (sec) , antiderivative size = 1291, normalized size of antiderivative = 10.01

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1291\)
default	\(\text {Expression too large to display}\)	\(1291\)

Input:

int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/2/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-2*2^(1/2)*d^2*(-a*(I*d-c))^(1/2 
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)-2*I*ln((3*a*c+I*a 
*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d* 
tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c*d^2*tan(f*x+e)^2+ 
2*I*2^(1/2)*c*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^( 
1/2)*tan(f*x+e)-I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1 
/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan 
(f*x+e)))*a*c^2*d*tan(f*x+e)-I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan( 
f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^ 
(1/2))/(I+tan(f*x+e)))*a*d^3*tan(f*x+e)-ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3 
*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan( 
f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c^2*d*tan(f*x+e)^2+ln((3*a*c+I*a*tan(f*x 
+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+ 
e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*d^3*tan(f*x+e)^2+2*I*2^(1/2 
)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^2+I*ln( 
(3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2 
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c^3-I*ln( 
(3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2 
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c*d^2-ln( 
(3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(...
 

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. 599 vs. \(2 (99) = 198\).

Time = 0.09 (sec) , antiderivative size = 599, normalized size of antiderivative = 4.64 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

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

Sympy [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral(sqrt(I*a*(tan(e + f*x) - I))/(c + d*tan(e + f*x))**(3/2), 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. 8488 vs. \(2 (99) = 198\).

Time = 0.67 (sec) , antiderivative size = 8488, normalized size of antiderivative = 65.80 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

-1/4*(8*((I*sqrt(2)*c^2*d + I*sqrt(2)*d^3)*cos(f*x + e) - (sqrt(2)*c^2*d + 
 sqrt(2)*d^3)*sin(f*x + e))*sqrt(a) + ((c^2 + d^2)*cos(2*f*x + 2*e)^2 + 4* 
c*d*sin(2*f*x + 2*e) + (c^2 + d^2)*sin(2*f*x + 2*e)^2 + c^2 + d^2 + 2*(c^2 
 - d^2)*cos(2*f*x + 2*e))^(1/4)*((sqrt(c^2 + d^2)*(2*(-I*sqrt(2)*cos(1/2*a 
rctan2(-d*cos(2*f*x + 2*e) + c*sin(2*f*x + 2*e) + d, c*cos(2*f*x + 2*e) + 
d*sin(2*f*x + 2*e) + c)) + sqrt(2)*sin(1/2*arctan2(-d*cos(2*f*x + 2*e) + c 
*sin(2*f*x + 2*e) + d, c*cos(2*f*x + 2*e) + d*sin(2*f*x + 2*e) + c)))*arct 
an2((sqrt(c^2 + d^2)*d*cos(f*x + e) - sqrt(c^2 + d^2)*c*sin(f*x + e) + (c^ 
2 + d^2)*(((c^2 + d^2)*cos(f*x + e)^4 + (c^2 + d^2)*sin(f*x + e)^4 + 8*c*d 
*cos(f*x + e)*sin(f*x + e) + 2*(c^2 - d^2)*cos(f*x + e)^2 + 2*((c^2 + d^2) 
*cos(f*x + e)^2 - c^2 + d^2)*sin(f*x + e)^2 + c^2 + d^2)/(c^2 + d^2))^(1/4 
)*sin(1/2*arctan2(-2*(c*d*cos(f*x + e)^2 - c*d*sin(f*x + e)^2 - (c^2 - d^2 
)*cos(f*x + e)*sin(f*x + e))/(c^2 + d^2), (4*c*d*cos(f*x + e)*sin(f*x + e) 
 + (c^2 - d^2)*cos(f*x + e)^2 - (c^2 - d^2)*sin(f*x + e)^2 + c^2 + d^2)/(c 
^2 + d^2))))/(c^2 + d^2), -(sqrt(c^2 + d^2)*c*cos(f*x + e) + sqrt(c^2 + d^ 
2)*d*sin(f*x + e) - (c^2 + d^2)*(((c^2 + d^2)*cos(f*x + e)^4 + (c^2 + d^2) 
*sin(f*x + e)^4 + 8*c*d*cos(f*x + e)*sin(f*x + e) + 2*(c^2 - d^2)*cos(f*x 
+ e)^2 + 2*((c^2 + d^2)*cos(f*x + e)^2 - c^2 + d^2)*sin(f*x + e)^2 + c^2 + 
 d^2)/(c^2 + d^2))^(1/4)*cos(1/2*arctan2(-2*(c*d*cos(f*x + e)^2 - c*d*sin( 
f*x + e)^2 - (c^2 - d^2)*cos(f*x + e)*sin(f*x + e))/(c^2 + d^2), (4*c*d...
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(3/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}}{\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}}d x \right ) \] Input:

int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x)
 

Output:

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