\(\int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx\) [1230]

Optimal result

Integrand size = 27, antiderivative size = 157 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=-\frac {i (a-i b)^2 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}+\frac {i (a+i b)^2 \sqrt {c+i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {4 a b \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 b^2 (c+d \tan (e+f x))^{3/2}}{3 d f} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.95 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\frac {-3 i (a-i b)^2 \sqrt {c-i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+3 i (a+i b)^2 \sqrt {c+i d} d \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+2 b \sqrt {c+d \tan (e+f x)} (b c+6 a d+b d \tan (e+f x))}{3 d f} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.86, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {3042, 4026, 3042, 4011, 3042, 4022, 3042, 4020, 25, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)}dx\)

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 4022

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

rule 4026

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1313\) vs. \(2(131)=262\).

Time = 0.41 (sec) , antiderivative size = 1314, normalized size of antiderivative = 8.37

method	result	size
parts	\(\text {Expression too large to display}\)	\(1314\)
derivativedivides	\(\text {Expression too large to display}\)	\(1495\)
default	\(\text {Expression too large to display}\)	\(1495\)

Input:

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

Output:

-1/4/f/d*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))*(c^2+d^2)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+1 
/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^ 
2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*a^2+1/4/f/d*ln(d*t 
an(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)*a^2*c+1/4/f/d*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))*(c^2+d^2 
)^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*a^2+1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1 
/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^ 
2+d^2)^(1/2)-2*c)^(1/2))*a^2-1/4/f/d*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)*a^2*c+b^2*(2/3/d/f*(c+d*tan(f*x+e))^(3/2)+1/4/f/d*(2*(c^2+d^2)^(1/2 
)+2*c)^(1/2)*(c^2+d^2)^(1/2)*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))-1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^( 
1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c 
^2+d^2)^(1/2)-2*c)^(1/2))-1/4/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*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))-1/4/f/d*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*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))- 
1/f*d/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2203 vs. \(2 (123) = 246\).

Time = 0.17 (sec) , antiderivative size = 2203, normalized size of antiderivative = 14.03 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/6*(3*d*f*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*( 
a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^ 
4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a^3*b - a 
*b^3)*d)/f^2)*log((4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c + (a^8 - 4*a^6* 
b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) + ((a^2 - 
b^2)*f^3*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7*b - 7*a^5* 
b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 
 + b^8)*d^2)/f^4) + 2*(4*(a^4*b^2 - a^2*b^4)*c + (a^5*b - 6*a^3*b^3 + a*b^ 
5)*d)*f)*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7 
*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 
 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a^3*b - a*b^ 
3)*d)/f^2)) - 3*d*f*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c 
^2 + 8*(a^7*b - 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 3 
8*a^4*b^4 - 12*a^2*b^6 + b^8)*d^2)/f^4) + (a^4 - 6*a^2*b^2 + b^4)*c - 4*(a 
^3*b - a*b^3)*d)/f^2)*log((4*(a^7*b + a^5*b^3 - a^3*b^5 - a*b^7)*c + (a^8 
- 4*a^6*b^2 - 10*a^4*b^4 - 4*a^2*b^6 + b^8)*d)*sqrt(d*tan(f*x + e) + c) - 
((a^2 - b^2)*f^3*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c^2 + 8*(a^7*b 
- 7*a^5*b^3 + 7*a^3*b^5 - a*b^7)*c*d + (a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12 
*a^2*b^6 + b^8)*d^2)/f^4) + 2*(4*(a^4*b^2 - a^2*b^4)*c + (a^5*b - 6*a^3*b^ 
3 + a*b^5)*d)*f)*sqrt(-(f^2*sqrt(-(16*(a^6*b^2 - 2*a^4*b^4 + a^2*b^6)*c...
 

Sympy [F]

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

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

Output:

Integral((a + b*tan(e + f*x))**2*sqrt(c + d*tan(e + f*x)), x)
 

Maxima [F]

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

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

Output:

integrate((b*tan(f*x + e) + a)^2*sqrt(d*tan(f*x + e) + c), x)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
 

Mupad [B] (verification not implemented)

Time = 7.09 (sec) , antiderivative size = 3722, normalized size of antiderivative = 23.71 \[ \int (a+b \tan (e+f x))^2 \sqrt {c+d \tan (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*b^2*(c + d*tan(e + f*x))^(3/2))/(3*d*f) - atan(((((8*(8*a*b*d^4*f^2 + 8 
*a*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a^4*c - a^ 
4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3* 
b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2))*(-(a^4*c - a^4*d*1i + b^4* 
c - b^4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a* 
b^3*d - 4*a^3*b*d)/(4*f^2))^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^ 
4 + b^4*d^4 - 6*a^2*b^2*d^4 - a^4*c^2*d^2 - b^4*c^2*d^2 + 6*a^2*b^2*c^2*d^ 
2 - 8*a*b^3*c*d^3 + 8*a^3*b*c*d^3))/f^2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4 
*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 
 4*a^3*b*d)/(4*f^2))^(1/2)*1i - (((8*(8*a*b*d^4*f^2 + 8*a*b*c^2*d^2*f^2))/ 
f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(a^4*c - a^4*d*1i + b^4*c - b^ 
4*d*1i - 6*a^2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d 
- 4*a^3*b*d)/(4*f^2))^(1/2))*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^ 
2*b^2*c + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/ 
(4*f^2))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^4*d^4 + b^4*d^4 - 6*a^2 
*b^2*d^4 - a^4*c^2*d^2 - b^4*c^2*d^2 + 6*a^2*b^2*c^2*d^2 - 8*a*b^3*c*d^3 + 
 8*a^3*b*c*d^3))/f^2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b^2*c 
 + a^2*b^2*d*6i + a*b^3*c*4i - a^3*b*c*4i + 4*a*b^3*d - 4*a^3*b*d)/(4*f^2) 
)^(1/2)*1i)/((((8*(8*a*b*d^4*f^2 + 8*a*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + 
 d*tan(e + f*x))^(1/2)*(-(a^4*c - a^4*d*1i + b^4*c - b^4*d*1i - 6*a^2*b...
 

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x)*d + c),x)*a**2 + int(sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)**2,x)*b**2 + 2*int(sqrt(tan(e + f*x)*d + c)*tan(e + f*x),x)*a*b