\(\int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [960]

Optimal result

Integrand size = 33, antiderivative size = 138 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {i \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {3 i \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))} \] Output:

3/32*I*c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/ 
2)/a^2/f+1/4*I*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))^2+3/16*I*(c 
-I*c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.49 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.38 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f} \] Input:

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

Output:

((I/4)*Hypergeometric2F1[1/2, 3, 3/2, (-1/2*I)*(I + Tan[e + f*x])]*Sqrt[c 
- I*c*Tan[e + f*x]])/(a^2*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3042, 4005, 3042, 3968, 52, 52, 73, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{5/2}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{5/2}}{\sec (e+f x)^4}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i c^3 \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^3 \left (\frac {3 \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{4 c (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^3 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )}{8 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{4 c (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {i c^3 \left (\frac {3 \left (\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{2 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}\right )}{8 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{4 c (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {i c^3 \left (\frac {3 \left (\frac {\sqrt {c-i c \tan (e+f x)}}{2 c (c+i c \tan (e+f x))}-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{2 \sqrt {2} c^{3/2}}\right )}{8 c}+\frac {\sqrt {c-i c \tan (e+f x)}}{4 c (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

Input:

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

Output:

(I*c^3*(Sqrt[c - I*c*Tan[e + f*x]]/(4*c*(c + I*c*Tan[e + f*x])^2) + (3*((( 
-1/2*I)*ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) + Sqrt[c 
 - I*c*Tan[e + f*x]]/(2*c*(c + I*c*Tan[e + f*x]))))/(8*c)))/(a^2*f)
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[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 4005

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]))
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\frac {3 \sqrt {c -i c \tan \left (f x +e \right )}}{32 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}}{c}\right )}{f \,a^{2}}\)	\(117\)
default	\(\frac {2 i c^{3} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{8 c \left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {\frac {3 \sqrt {c -i c \tan \left (f x +e \right )}}{32 c \left (c +i c \tan \left (f x +e \right )\right )}+\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}}{c}\right )}{f \,a^{2}}\)	\(117\)

Input:

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

Output:

2*I/f/a^2*c^3*(1/8*(c-I*c*tan(f*x+e))^(1/2)/c/(c+I*c*tan(f*x+e))^2+3/8/c*( 
1/4*(c-I*c*tan(f*x+e))^(1/2)/c/(c+I*c*tan(f*x+e))+1/8/c^(3/2)*2^(1/2)*arct 
anh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/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. 275 vs. \(2 (103) = 206\).

Time = 0.08 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.99 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{4} f^{2}}} + i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c}{a^{4} f^{2}}} - i \, c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (5 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 7 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} f} \] Input:

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

Output:

1/32*(3*sqrt(1/2)*a^2*f*sqrt(-c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(3/8*(sq 
rt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2 
*I*e) + 1))*sqrt(-c/(a^4*f^2)) + I*c)*e^(-I*f*x - I*e)/(a^2*f)) - 3*sqrt(1 
/2)*a^2*f*sqrt(-c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-3/8*(sqrt(2)*sqrt(1/ 
2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*s 
qrt(-c/(a^4*f^2)) - I*c)*e^(-I*f*x - I*e)/(a^2*f)) + sqrt(2)*sqrt(c/(e^(2* 
I*f*x + 2*I*e) + 1))*(5*I*e^(4*I*f*x + 4*I*e) + 7*I*e^(2*I*f*x + 2*I*e) + 
2*I))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c^{2} - 10 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{64 \, c f} \] Input:

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

Output:

-1/64*I*(3*sqrt(2)*c^(3/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) 
+ c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/a^2 + 4*(3*(-I*c*ta 
n(f*x + e) + c)^(3/2)*c^2 - 10*sqrt(-I*c*tan(f*x + e) + c)*c^3)/((-I*c*tan 
(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 2.01 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{32\,a^2\,f} \] Input:

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

Output:

((c^2*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(8*a^2*f) - (c*(c - c*tan(e + f*x) 
*1i)^(3/2)*3i)/(16*a^2*f))/((c - c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + 
 f*x)*1i) + 4*c^2) + (2^(1/2)*(-c)^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x) 
*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(32*a^2*f)
 

Reduce [F]

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

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

Output:

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