\(\int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx\) [1101]

Optimal result

Integrand size = 30, antiderivative size = 150 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {8 i a^3 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {8 i a^3 \sqrt {c+d \tan (e+f x)}}{f}+\frac {4 a^3 (i c-6 d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f} \] Output:

-8*I*a^3*(c-I*d)^(1/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f+8*I 
*a^3*(c+d*tan(f*x+e))^(1/2)/f+4/15*a^3*(I*c-6*d)*(c+d*tan(f*x+e))^(3/2)/d^ 
2/f-2/5*(a^3+I*a^3*tan(f*x+e))*(c+d*tan(f*x+e))^(3/2)/d/f
 

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.80 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {i a^2 \left (-8 a \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+8 a \sqrt {c+d \tan (e+f x)}+\frac {2 a (c+3 i d) (c+d \tan (e+f x))^{3/2}}{3 d^2}-\frac {2 a (c+d \tan (e+f x))^{5/2}}{5 d^2}\right )}{f} \] Input:

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

Output:

(I*a^2*(-8*a*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] 
 + 8*a*Sqrt[c + d*Tan[e + f*x]] + (2*a*(c + (3*I)*d)*(c + d*Tan[e + f*x])^ 
(3/2))/(3*d^2) - (2*a*(c + d*Tan[e + f*x])^(5/2))/(5*d^2)))/f
 

Rubi [A] (warning: unable to verify)

Time = 0.92 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 4039, 3042, 4075, 3042, 4011, 3042, 4020, 27, 73, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

\(\displaystyle \frac {2 a \int (i \tan (e+f x) a+a) (a (i c+4 d)+a (c+6 i d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}dx}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \int (i \tan (e+f x) a+a) (a (i c+4 d)+a (c+6 i d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}dx}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {2 a \left (\int \sqrt {c+d \tan (e+f x)} \left (10 d a^2+10 i d \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int \sqrt {c+d \tan (e+f x)} \left (10 d a^2+10 i d \tan (e+f x) a^2\right )dx+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 a \left (\int \frac {10 (c-i d) d a^2+10 d (i c+d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\int \frac {10 (c-i d) d a^2+10 d (i c+d) \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 a \left (\frac {100 i a^4 d^2 (c-i d)^2 \int \frac {1}{\sqrt {10} a^2 d \sqrt {10 c+10 d \tan (e+f x)} \left (10 d (i c+d)^2 a^2+10 (c-i d) d (i c+d) \tan (e+f x) a^2\right )}d\left (10 a^2 d (i c+d) \tan (e+f x)\right )}{f}+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a \left (\frac {10 i \sqrt {10} a^2 d (c-i d)^2 \int \frac {1}{\sqrt {10 c+10 d \tan (e+f x)} \left (10 d (i c+d)^2 a^2+10 (c-i d) d (i c+d) \tan (e+f x) a^2\right )}d\left (10 a^2 d (i c+d) \tan (e+f x)\right )}{f}+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 a \left (\frac {20 i \sqrt {10} a^4 d (c-i d)^2 (d+i c) \int \frac {1}{100 i (c-i d)^2 d^2 (i c+d)^2 \tan ^2(e+f x) a^6+10 (i c+d)^3 a^2}d\sqrt {10 c+10 d \tan (e+f x)}}{f}+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a \left (-\frac {20 a^2 d (d+i c) \text {arctanh}\left (\frac {\sqrt {10} a^2 d (d+i c) \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}+\frac {2 a^2 (-6 d+i c) (c+d \tan (e+f x))^{3/2}}{3 d f}+\frac {20 i a^2 d \sqrt {c+d \tan (e+f x)}}{f}\right )}{5 d}-\frac {2 \left (a^3+i a^3 \tan (e+f x)\right ) (c+d \tan (e+f x))^{3/2}}{5 d f}\)

Input:

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

Output:

(-2*(a^3 + I*a^3*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2))/(5*d*f) + (2*a* 
((-20*a^2*d*(I*c + d)*ArcTanh[(Sqrt[10]*a^2*d*(I*c + d)*Tan[e + f*x])/Sqrt 
[c - I*d]])/(Sqrt[c - I*d]*f) + ((20*I)*a^2*d*Sqrt[c + d*Tan[e + f*x]])/f 
+ (2*a^2*(I*c - 6*d)*(c + d*Tan[e + f*x])^(3/2))/(3*d*f)))/(5*d)
 

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 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 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 4011

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

rule 4020

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

rule 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x 
] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4075

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B 
*d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* 
x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[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. 1277 vs. \(2 (128 ) = 256\).

Time = 0.66 (sec) , antiderivative size = 1278, normalized size of antiderivative = 8.52

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1278\)
default	\(\text {Expression too large to display}\)	\(1278\)
parts	\(\text {Expression too large to display}\)	\(1689\)

Input:

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

Output:

-2/5*I/f*a^3/d^2*(c+d*tan(f*x+e))^(5/2)-4*I/f*a^3/(4*(c^2+d^2)^(1/2)+4*c)* 
ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^ 
2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c+8*I*a^3*(c+d*tan(f*x+e))^(1/ 
2)/f-2/f*a^3/d*(c+d*tan(f*x+e))^(3/2)+2/3*I/f*a^3/d^2*c*(c+d*tan(f*x+e))^( 
3/2)+4/f*a^3*d/(4*(c^2+d^2)^(1/2)+4*c)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d 
^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2* 
c)^(1/2)+16*I/f*a^3*d^2/(4*(c^2+d^2)^(1/2)+4*c)/(2*(c^2+d^2)^(1/2)-2*c)^(1 
/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^ 
2+d^2)^(1/2)-2*c)^(1/2))-16/f*a^3*d/(4*(c^2+d^2)^(1/2)+4*c)/(2*(c^2+d^2)^( 
1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^( 
1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*(c^2+d^2)^(1/2)-16/f*a^3*d/(4*(c^2+d^ 
2)^(1/2)+4*c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c 
)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-4*I/f*a 
^3/(4*(c^2+d^2)^(1/2)+4*c)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/2)*(2*(c^ 
2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^ 
2+d^2)^(1/2)-4/f*a^3*d/(4*(c^2+d^2)^(1/2)+4*c)*ln(d*tan(f*x+e)+c+(c+d*tan( 
f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^ 
(1/2)+2*c)^(1/2)+4*I/f*a^3/(4*(c^2+d^2)^(1/2)+4*c)*ln((c+d*tan(f*x+e))^(1/ 
2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*(2*(c^2+d 
^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)+4*I/f*a^3/(4*(c^2+d^2)^(1/2)+4*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. 512 vs. \(2 (124) = 248\).

Time = 0.16 (sec) , antiderivative size = 512, normalized size of antiderivative = 3.41 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=\frac {2 \, {\left (15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c + {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} + {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3}}\right ) - 15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} \log \left (\frac {2 \, {\left (a^{3} c + {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {a^{6} c - i \, a^{6} d}{f^{2}}} + {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a^{3}}\right ) - 2 \, {\left (-i \, a^{3} c^{2} + 7 \, a^{3} c d - 24 i \, a^{3} d^{2} + {\left (-i \, a^{3} c^{2} + 8 \, a^{3} c d - 39 i \, a^{3} d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (-2 i \, a^{3} c^{2} + 15 \, a^{3} c d - 57 i \, a^{3} d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )}}{15 \, {\left (d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, d^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{2} f\right )}} \] Input:

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

Output:

2/15*(15*(d^2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f) 
*sqrt(-(a^6*c - I*a^6*d)/f^2)*log(2*(a^3*c + (I*f*e^(2*I*f*x + 2*I*e) + I* 
f)*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^6*c - I*a^6*d)/f^2) + (a^3*c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e 
^(-2*I*f*x - 2*I*e)/a^3) - 15*(d^2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I* 
f*x + 2*I*e) + d^2*f)*sqrt(-(a^6*c - I*a^6*d)/f^2)*log(2*(a^3*c + (-I*f*e^ 
(2*I*f*x + 2*I*e) - I*f)*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^6*c - I*a^6*d)/f^2) + (a^3*c - I*a^3*d)* 
e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - 2*(-I*a^3*c^2 + 7*a^3*c*d 
 - 24*I*a^3*d^2 + (-I*a^3*c^2 + 8*a^3*c*d - 39*I*a^3*d^2)*e^(4*I*f*x + 4*I 
*e) + (-2*I*a^3*c^2 + 15*a^3*c*d - 57*I*a^3*d^2)*e^(2*I*f*x + 2*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)))/(d^ 
2*f*e^(4*I*f*x + 4*I*e) + 2*d^2*f*e^(2*I*f*x + 2*I*e) + d^2*f)
 

Sympy [F]

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

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

Output:

-I*a**3*(Integral(I*sqrt(c + d*tan(e + f*x)), x) + Integral(-3*sqrt(c + d* 
tan(e + f*x))*tan(e + f*x), x) + Integral(sqrt(c + d*tan(e + f*x))*tan(e + 
 f*x)**3, x) + Integral(-3*I*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2, x))
 

Maxima [F]

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

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

Output:

integrate((I*a*tan(f*x + e) + a)^3*sqrt(d*tan(f*x + e) + c), x)
 

Giac [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.64 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {2 \, a^{3} {\left (\frac {60 \, \sqrt {2} {\left (-i \, c - d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{\sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {3 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} d^{8} - 5 i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c d^{8} + 15 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d^{9} - 60 i \, \sqrt {d \tan \left (f x + e\right ) + c} d^{10}}{d^{10}}\right )}}{15 \, f} \] Input:

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

Output:

-2/15*a^3*(60*sqrt(2)*(-I*c - d)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sq 
rt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^ 
2)) - I*sqrt(2)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqr 
t(-c + sqrt(c^2 + d^2))))/(sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 
+ d^2)) + 1)) + (3*I*(d*tan(f*x + e) + c)^(5/2)*d^8 - 5*I*(d*tan(f*x + e) 
+ c)^(3/2)*c*d^8 + 15*(d*tan(f*x + e) + c)^(3/2)*d^9 - 60*I*sqrt(d*tan(f*x 
 + e) + c)*d^10)/d^10)/f
 

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.33 \[ \int (a+i a \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx=-\left (\left (c-d\,1{}\mathrm {i}\right )\,\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d^2\,f}\right )+\frac {a^3\,{\left (c+d\,1{}\mathrm {i}\right )}^2\,2{}\mathrm {i}}{d^2\,f}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}-\left (\frac {a^3\,\left (c-d\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{3\,d^2\,f}-\frac {a^3\,\left (c+d\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{3\,d^2\,f}\right )\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}-\frac {a^3\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,2{}\mathrm {i}}{5\,d^2\,f}+\frac {\sqrt {16{}\mathrm {i}}\,a^3\,\mathrm {atan}\left (\frac {\sqrt {16{}\mathrm {i}}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{4\,\sqrt {d+c\,1{}\mathrm {i}}}\right )\,\sqrt {d+c\,1{}\mathrm {i}}\,2{}\mathrm {i}}{f} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

a**3*(int(sqrt(tan(e + f*x)*d + c),x) - int(sqrt(tan(e + f*x)*d + c)*tan(e 
 + f*x)**3,x)*i - 3*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2,x) + 3*in 
t(sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*i)