\(\int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1169]

Optimal result

Integrand size = 32, antiderivative size = 188 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {2 (5 c+i d) d \sqrt {a+i a \tan (e+f x)}}{3 \left (c^2+d^2\right )^2 f \sqrt {c+d \tan (e+f x)}} \] 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)^(5/2)/f-2/3*d*(a+I*a*tan(f*x+e))^( 
1/2)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)-2/3*(5*c+I*d)*d*(a+I*a*tan(f*x+e)) 
^(1/2)/(c^2+d^2)^2/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.65 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {3 i \sqrt {2} \sqrt {-a (c-i d)} (c+i d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{(c-i d)^2}-\frac {2 d \sqrt {a+i a \tan (e+f x)} \left (6 c^2+i c d+d^2+(5 c+i d) d \tan (e+f x)\right )}{\left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{3 \left (c^2+d^2\right ) f} \] Input:

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(((3*I)*Sqrt[2]*Sqrt[-(a*(c - I*d))]*(c + I*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]])])/(c - 
I*d)^2 - (2*d*Sqrt[a + I*a*Tan[e + f*x]]*(6*c^2 + I*c*d + d^2 + (5*c + I*d 
)*d*Tan[e + f*x]))/((c^2 + d^2)*(c + d*Tan[e + f*x])^(3/2)))/(3*(c^2 + d^2 
)*f)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4044, 27, 3042, 4081, 27, 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.

\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

\(\displaystyle \frac {-\frac {6 i a^3 (c+i d)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}-\frac {2 a d (5 c+i d) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}-\frac {2 d \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 221

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

Input:

Int[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*d*Sqrt[a + I*a*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3 
/2)) + (((-3*I)*Sqrt[2]*a^(3/2)*(c + I*d)^2*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[ 
c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c 
- I*d]*(c^2 + d^2)*f) - (2*a*(5*c + I*d)*d*Sqrt[a + I*a*Tan[e + f*x]])/((c 
^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(3*a*(c^2 + d^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 4044

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*Tan[e 
 + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(c^2 + d^2)*(n + 
1))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - 
a*c*(n + 1) + a*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d 
, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 
0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])
 

rule 4081

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2447 vs. \(2 (153 ) = 306\).

Time = 0.65 (sec) , antiderivative size = 2448, normalized size of antiderivative = 13.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2448\)
default	\(\text {Expression too large to display}\)	\(2448\)

Input:

int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/6/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(-4*2^(1/2)*d^4*(-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)-3*ln((3*a*c+I*a*t 
an(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*ta 
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*a*c^3*d^2*tan(f*x+e)^3+ 
9*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^ 
4*tan(f*x+e)^3-6*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^4*d*tan(f*x+e)^2+9*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^3*tan(f*x+e)^2-9*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*d^2*tan(f*x+e)+6*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^4*tan(f*x+e) 
+12*2^(1/2)*c^3*d*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))) 
^(1/2)+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*d^5*tan(f*x+e)^3+2*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*2^(1 
/2)*(-a*(I*d-c))^(1/2)*d^4-9*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*t...
 

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. 968 vs. \(2 (146) = 292\).

Time = 0.12 (sec) , antiderivative size = 968, normalized size of antiderivative = 5.15 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fr 
icas")
 

Output:

-1/6*(8*sqrt(2)*((3*c^2*d - 2*I*c*d^2 + d^3)*e^(5*I*f*x + 5*I*e) + (6*c^2* 
d + I*c*d^2 + d^3)*e^(3*I*f*x + 3*I*e) + 3*(c^2*d + I*c*d^2)*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)) - 3*((c^6 - 2*I*c^5*d + c^4*d^2 - 4* 
I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^(4*I*f*x + 4*I*e) + 2*(c^6 + 3* 
c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2*I*e) + (c^6 + 2*I*c^5*d + c^4* 
d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - d^6)*f)*sqrt(-2*I*a/((I*c^5 + 5* 
c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(((I*c^3 + 3 
*c^2*d - 3*I*c*d^2 - d^3)*f*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 
 10*c^2*d^3 + 5*I*c*d^4 + d^5)*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)) + 3*(( 
c^6 - 2*I*c^5*d + c^4*d^2 - 4*I*c^3*d^3 - c^2*d^4 - 2*I*c*d^5 - d^6)*f*e^( 
4*I*f*x + 4*I*e) + 2*(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*f*e^(2*I*f*x + 2* 
I*e) + (c^6 + 2*I*c^5*d + c^4*d^2 + 4*I*c^3*d^3 - c^2*d^4 + 2*I*c*d^5 - d^ 
6)*f)*sqrt(-2*I*a/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^ 
4 + d^5)*f^2))*log(((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f*sqrt(-2*I*a/((I 
*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*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...
 

Sympy [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(5/2),x)
 

Output:

Integral(sqrt(I*a*(tan(e + f*x) - I))/(c + d*tan(e + f*x))**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="ma 
xima")
 

Output:

Timed out
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="gi 
ac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(5/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(1/2)/(c + d*tan(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{5/2}} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \] Input:

int((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(5/2),x)
 

Output:

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