\(\int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx\) [978]

Optimal result

Integrand size = 33, antiderivative size = 167 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {15 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^2 \sqrt {c} f}-\frac {15 i}{32 a^2 f \sqrt {c-i c \tan (e+f x)}}+\frac {i}{4 a^2 f (1+i \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}+\frac {5 i}{16 a^2 f (1+i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Output:

15/64*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a^2/ 
c^(1/2)/f-15/32*I/a^2/f/(c-I*c*tan(f*x+e))^(1/2)+1/4*I/a^2/f/(1+I*tan(f*x+ 
e))^2/(c-I*c*tan(f*x+e))^(1/2)+5/16*I/a^2/f/(1+I*tan(f*x+e))/(c-I*c*tan(f* 
x+e))^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.55 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.32 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},3,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{4 a^2 f \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.42 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4005, 3042, 3968, 52, 52, 61, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \cos ^4(e+f x) (c-i c \tan (e+f x))^{3/2}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{3/2}}{\sec (e+f x)^4}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^3 \left (\frac {5 \int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {1}{4 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^3 \left (\frac {5 \left (\frac {3 \int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^3 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {i c^3 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {i c^3 \left (\frac {5 \left (\frac {3 \left (-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}\right )}{4 c}+\frac {1}{2 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c \sqrt {c-i c \tan (e+f x)} (c+i c \tan (e+f x))^2}\right )}{a^2 f}\)

Input:

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

Output:

(I*c^3*(1/(4*c*Sqrt[c - I*c*Tan[e + f*x]]*(c + I*c*Tan[e + f*x])^2) + (5*( 
1/(2*c*Sqrt[c - I*c*Tan[e + f*x]]*(c + I*c*Tan[e + f*x])) + (3*(((-I)*ArcT 
an[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2)) - 1/(c*Sqrt[c - I*c* 
Tan[e + f*x]])))/(4*c)))/(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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {\frac {-\frac {7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {9 \sqrt {c -i c \tan \left (f x +e \right )}\, c}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{8 c^{3}}-\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,a^{2}}\)	\(120\)
default	\(\frac {2 i c^{3} \left (\frac {\frac {-\frac {7 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{8}+\frac {9 \sqrt {c -i c \tan \left (f x +e \right )}\, c}{4}}{\left (c +i c \tan \left (f x +e \right )\right )^{2}}+\frac {15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{16 \sqrt {c}}}{8 c^{3}}-\frac {1}{8 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,a^{2}}\)	\(120\)

Input:

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

Output:

2*I/f/a^2*c^3*(1/8/c^3*(4*(-7/32*(c-I*c*tan(f*x+e))^(3/2)+9/16*(c-I*c*tan( 
f*x+e))^(1/2)*c)/(c+I*c*tan(f*x+e))^2+15/16*2^(1/2)/c^(1/2)*arctanh(1/2*(c 
-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/8/c^3/(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. 295 vs. \(2 (124) = 248\).

Time = 0.08 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {\frac {1}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{2} c f \sqrt {\frac {1}{a^{4} c f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {15 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{4} c f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-8 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 11 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{64 \, a^{2} c f} \] Input:

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

Output:

1/64*(-15*I*sqrt(1/2)*a^2*c*f*sqrt(1/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log( 
-15/16*(sqrt(2)*sqrt(1/2)*(I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(c/( 
e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c*f^2)) - I)*e^(-I*f*x - I*e)/(a^2*f 
)) + 15*I*sqrt(1/2)*a^2*c*f*sqrt(1/(a^4*c*f^2))*e^(4*I*f*x + 4*I*e)*log(-1 
5/16*(sqrt(2)*sqrt(1/2)*(-I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt(c/(e 
^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^4*c*f^2)) - I)*e^(-I*f*x - I*e)/(a^2*f) 
) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-8*I*e^(6*I*f*x + 6*I*e) + 
I*e^(4*I*f*x + 4*I*e) + 11*I*e^(2*I*f*x + 2*I*e) + 2*I))*e^(-4*I*f*x - 4*I 
*e)/(a^2*c*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.01 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c - 50 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} + 32 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{2} c + 4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} a^{2} c^{2}} + \frac {15 \, \sqrt {2} \sqrt {c} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}}\right )}}{128 \, c f} \] Input:

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

Output:

-1/128*I*(4*(15*(-I*c*tan(f*x + e) + c)^2*c - 50*(-I*c*tan(f*x + e) + c)*c 
^2 + 32*c^3)/((-I*c*tan(f*x + e) + c)^(5/2)*a^2 - 4*(-I*c*tan(f*x + e) + c 
)^(3/2)*a^2*c + 4*sqrt(-I*c*tan(f*x + e) + c)*a^2*c^2) + 15*sqrt(2)*sqrt(c 
)*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)/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

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 = 2.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(a+i a \tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,15{}\mathrm {i}}{32\,a^2\,f}+\frac {c^2\,1{}\mathrm {i}}{a^2\,f}-\frac {c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{16\,a^2\,f}}{-4\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+4\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,15{}\mathrm {i}}{64\,a^2\,\sqrt {-c}\,f} \] Input:

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

Output:

- (((c - c*tan(e + f*x)*1i)^2*15i)/(32*a^2*f) + (c^2*1i)/(a^2*f) - (c*(c - 
 c*tan(e + f*x)*1i)*25i)/(16*a^2*f))/((c - c*tan(e + f*x)*1i)^(5/2) - 4*c* 
(c - c*tan(e + f*x)*1i)^(3/2) + 4*c^2*(c - c*tan(e + f*x)*1i)^(1/2)) - (2^ 
(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*15i)/(6 
4*a^2*(-c)^(1/2)*f)
 

Reduce [F]

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

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

Output:

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