\(\int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx\) [1243]

Optimal result

Integrand size = 27, antiderivative size = 171 \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=-\frac {2 \left (a^2-a b-b^2\right ) \arctan \left (\frac {a^2-a b-b^2-2 \left (a^2-a b-b^2\right ) \tan (e+f x)}{\left (a^2-a b-b^2\right ) \sqrt {3+4 \tan (e+f x)}}\right )}{f}-\frac {\left (a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {2+\tan (e+f x)}{\sqrt {3+4 \tan (e+f x)}}\right )}{f}+\frac {4 a b \sqrt {3+4 \tan (e+f x)}}{f}+\frac {b^2 (3+4 \tan (e+f x))^{3/2}}{6 f} \] Output:

-2*(a^2-a*b-b^2)*arctan((a^2-a*b-b^2-2*(a^2-a*b-b^2)*tan(f*x+e))/(a^2-a*b- 
b^2)/(3+4*tan(f*x+e))^(1/2))/f-(a^2+4*a*b-b^2)*arctanh((2+tan(f*x+e))/(3+4 
*tan(f*x+e))^(1/2))/f+4*a*b*(3+4*tan(f*x+e))^(1/2)/f+1/6*b^2*(3+4*tan(f*x+ 
e))^(3/2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.66 \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\frac {(12+6 i) (a+i b)^2 \arctan \left (\left (\frac {1}{5}+\frac {2 i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )-(6+12 i) (a-i b)^2 \text {arctanh}\left (\left (\frac {2}{5}+\frac {i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )+b \sqrt {3+4 \tan (e+f x)} (3 (8 a+b)+4 b \tan (e+f x))}{6 f} \] Input:

Integrate[Sqrt[3 + 4*Tan[e + f*x]]*(a + b*Tan[e + f*x])^2,x]
 

Output:

((12 + 6*I)*(a + I*b)^2*ArcTan[(1/5 + (2*I)/5)*Sqrt[3 + 4*Tan[e + f*x]]] - 
 (6 + 12*I)*(a - I*b)^2*ArcTanh[(2/5 + I/5)*Sqrt[3 + 4*Tan[e + f*x]]] + b* 
Sqrt[3 + 4*Tan[e + f*x]]*(3*(8*a + b) + 4*b*Tan[e + f*x]))/(6*f)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.25, 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, 4019, 27, 3042, 4018, 216, 220}

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.

\(\displaystyle \int \sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))^2dx\)

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4011

\(\displaystyle \int \frac {(a-3 b) (3 a+b)+2 (2 a-b) (a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a-3 b) (3 a+b)+2 (2 a-b) (a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {1}{10} \int \frac {20 \left (\tan (e+f x) \left (a^2-b a-b^2\right )+2 \left (a^2-b a-b^2\right )\right )}{\sqrt {4 \tan (e+f x)+3}}dx-\frac {1}{10} \int \frac {10 \left (a^2+4 b a-b^2-2 \left (a^2+4 b a-b^2\right ) \tan (e+f x)\right )}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4018

\(\displaystyle -\frac {4 \left (a^2-a b-b^2\right )^2 \int \frac {1}{4 \left (a^2-b a-b^2\right )^2+\frac {4 \left (a^2-b a-b^2-2 \left (a^2-b a-b^2\right ) \tan (e+f x)\right )^2}{4 \tan (e+f x)+3}}d\frac {2 \left (a^2-b a-b^2-2 \left (a^2-b a-b^2\right ) \tan (e+f x)\right )}{\sqrt {4 \tan (e+f x)+3}}}{f}+\frac {8 \left (a^2+4 a b-b^2\right )^2 \int \frac {1}{\frac {64 \left (\tan (e+f x) \left (a^2+4 b a-b^2\right )+2 \left (a^2+4 b a-b^2\right )\right )^2}{4 \tan (e+f x)+3}-64 \left (a^2+4 b a-b^2\right )^2}d\frac {8 \left (\tan (e+f x) \left (a^2+4 b a-b^2\right )+2 \left (a^2+4 b a-b^2\right )\right )}{\sqrt {4 \tan (e+f x)+3}}}{f}+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {8 \left (a^2+4 a b-b^2\right )^2 \int \frac {1}{\frac {64 \left (\tan (e+f x) \left (a^2+4 b a-b^2\right )+2 \left (a^2+4 b a-b^2\right )\right )^2}{4 \tan (e+f x)+3}-64 \left (a^2+4 b a-b^2\right )^2}d\frac {8 \left (\tan (e+f x) \left (a^2+4 b a-b^2\right )+2 \left (a^2+4 b a-b^2\right )\right )}{\sqrt {4 \tan (e+f x)+3}}}{f}-\frac {2 \left (a^2-a b-b^2\right ) \arctan \left (\frac {-2 \left (a^2-a b-b^2\right ) \tan (e+f x)+a^2-a b-b^2}{\left (a^2-a b-b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{f}+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

\(\Big \downarrow \) 220

\(\displaystyle -\frac {2 \left (a^2-a b-b^2\right ) \arctan \left (\frac {-2 \left (a^2-a b-b^2\right ) \tan (e+f x)+a^2-a b-b^2}{\left (a^2-a b-b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{f}-\frac {\left (a^2+4 a b-b^2\right ) \text {arctanh}\left (\frac {\left (a^2+4 a b-b^2\right ) \tan (e+f x)+2 \left (a^2+4 a b-b^2\right )}{\left (a^2+4 a b-b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{f}+\frac {4 a b \sqrt {4 \tan (e+f x)+3}}{f}+\frac {b^2 (4 \tan (e+f x)+3)^{3/2}}{6 f}\)

Input:

Int[Sqrt[3 + 4*Tan[e + f*x]]*(a + b*Tan[e + f*x])^2,x]
 

Output:

(-2*(a^2 - a*b - b^2)*ArcTan[(a^2 - a*b - b^2 - 2*(a^2 - a*b - b^2)*Tan[e 
+ f*x])/((a^2 - a*b - b^2)*Sqrt[3 + 4*Tan[e + f*x]])])/f - ((a^2 + 4*a*b - 
 b^2)*ArcTanh[(2*(a^2 + 4*a*b - b^2) + (a^2 + 4*a*b - b^2)*Tan[e + f*x])/( 
(a^2 + 4*a*b - b^2)*Sqrt[3 + 4*Tan[e + f*x]])])/f + (4*a*b*Sqrt[3 + 4*Tan[ 
e + f*x]])/f + (b^2*(3 + 4*Tan[e + f*x])^(3/2))/(6*f)
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

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 4018

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*b*c*d - 4*a*d^2 
+ x^2), x], x, (b*c - 2*a*d - b*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] && NeQ[a^2 + b^2, 0 
] && NeQ[c^2 + d^2, 0] && EqQ[2*a*c*d - b*(c^2 - d^2), 0]
 

rule 4019

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*tan[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> With[{q = Rt[a^2 + b^2, 2]}, Simp[1/(2*q)   Int[( 
a*c + b*d + c*q + (b*c - a*d + d*q)*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]], 
 x], x] - Simp[1/(2*q)   Int[(a*c + b*d - c*q + (b*c - a*d - d*q)*Tan[e + f 
*x])/Sqrt[a + b*Tan[e + f*x]], x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && N 
eQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && NeQ[2*a*c*d - 
 b*(c^2 - d^2), 0] && NiceSqrtQ[a^2 + b^2]
 

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 [A] (verified)

Time = 0.27 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {b^{2} \left (3+4 \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+4 \sqrt {3+4 \tan \left (f x +e \right )}\, a b +\frac {\left (-2 a^{2}-8 a b +2 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{4}+\frac {\left (4 a^{2}-4 a b -4 b^{2}\right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+\frac {\left (2 a^{2}+8 a b -2 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{4}+\frac {\left (4 a^{2}-4 a b -4 b^{2}\right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}}{f}\)	\(187\)
default	\(\frac {\frac {b^{2} \left (3+4 \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+4 \sqrt {3+4 \tan \left (f x +e \right )}\, a b +\frac {\left (-2 a^{2}-8 a b +2 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{4}+\frac {\left (4 a^{2}-4 a b -4 b^{2}\right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+\frac {\left (2 a^{2}+8 a b -2 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{4}+\frac {\left (4 a^{2}-4 a b -4 b^{2}\right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}}{f}\)	\(187\)
parts	\(\frac {a^{2} \left (\frac {\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )-\frac {\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}+2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )\right )}{f}+\frac {b^{2} \left (\frac {\left (3+4 \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}-\frac {\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}-2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )+\frac {\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{2}-2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )\right )}{f}+\frac {2 a b \left (2 \sqrt {3+4 \tan \left (f x +e \right )}+\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )-\arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )-\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )-\arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )\right )}{f}\)	\(316\)

Input:

int((3+4*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/6*b^2*(3+4*tan(f*x+e))^(3/2)+4*(3+4*tan(f*x+e))^(1/2)*a*b+1/4*(-2*a 
^2-8*a*b+2*b^2)*ln(8+4*tan(f*x+e)+4*(3+4*tan(f*x+e))^(1/2))+1/2*(4*a^2-4*a 
*b-4*b^2)*arctan(2+(3+4*tan(f*x+e))^(1/2))+1/4*(2*a^2+8*a*b-2*b^2)*ln(8+4* 
tan(f*x+e)-4*(3+4*tan(f*x+e))^(1/2))+1/2*(4*a^2-4*a*b-4*b^2)*arctan(-2+(3+ 
4*tan(f*x+e))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.02 \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\frac {12 \, {\left (a^{2} - a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 12 \, {\left (a^{2} - a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) - 3 \, {\left (a^{2} + 4 \, a b - b^{2}\right )} \log \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) + 3 \, {\left (a^{2} + 4 \, a b - b^{2}\right )} \log \left (-\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) + {\left (4 \, b^{2} \tan \left (f x + e\right ) + 24 \, a b + 3 \, b^{2}\right )} \sqrt {4 \, \tan \left (f x + e\right ) + 3}}{6 \, f} \] Input:

integrate((3+4*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

1/6*(12*(a^2 - a*b - b^2)*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 12*(a^2 - 
 a*b - b^2)*arctan(sqrt(4*tan(f*x + e) + 3) - 2) - 3*(a^2 + 4*a*b - b^2)*l 
og(sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + 3*(a^2 + 4*a*b - b^2)*lo 
g(-sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + (4*b^2*tan(f*x + e) + 24 
*a*b + 3*b^2)*sqrt(4*tan(f*x + e) + 3))/f
 

Sympy [F]

\[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \sqrt {4 \tan {\left (e + f x \right )} + 3}\, dx \] Input:

integrate((3+4*tan(f*x+e))**(1/2)*(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((a + b*tan(e + f*x))**2*sqrt(4*tan(e + f*x) + 3), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.04 \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\frac {b^{2} {\left (4 \, \tan \left (f x + e\right ) + 3\right )}^{\frac {3}{2}} + 24 \, a b \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 12 \, {\left (a^{2} - a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 12 \, {\left (a^{2} - a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) - 3 \, {\left (a^{2} + 4 \, a b - b^{2}\right )} \log \left (4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right ) + 3 \, {\left (a^{2} + 4 \, a b - b^{2}\right )} \log \left (-4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right )}{6 \, f} \] Input:

integrate((3+4*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

1/6*(b^2*(4*tan(f*x + e) + 3)^(3/2) + 24*a*b*sqrt(4*tan(f*x + e) + 3) + 12 
*(a^2 - a*b - b^2)*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 12*(a^2 - a*b - 
b^2)*arctan(sqrt(4*tan(f*x + e) + 3) - 2) - 3*(a^2 + 4*a*b - b^2)*log(4*sq 
rt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8) + 3*(a^2 + 4*a*b - b^2)*log(- 
4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8))/f
 

Giac [F(-2)]

Exception generated. \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\text {Exception raised: TypeError} \] Input:

integrate((3+4*tan(f*x+e))^(1/2)*(a+b*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:Unable to divide, perhaps due to ro 
unding error%%%{%%%{64,[14]%%%}+%%%{384,[12]%%%}+%%%{960,[10]%%%}+%%%{1280 
,[8]%%%}+
 

Mupad [B] (verification not implemented)

Time = 1.84 (sec) , antiderivative size = 1403, normalized size of antiderivative = 8.20 \[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\text {Too large to display} \] Input:

int((4*tan(e + f*x) + 3)^(1/2)*(a + b*tan(e + f*x))^2,x)
 

Output:

(4*tan(e + f*x) + 3)^(1/2)*((3*b^2)/f + (b*(4*a - 3*b))/f) + (b^2*(4*tan(e 
 + f*x) + 3)^(3/2))/(6*f) + (atan(((((256*(4*tan(e + f*x) + 3)^(1/2)*(96*a 
^3*b - 96*a*b^3 + 7*a^4 + 7*b^4 - 42*a^2*b^2))/f^2 - (((3072*(4*tan(e + f* 
x) + 3)^(1/2)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i)))/f - (25600 
*a*b)/f)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i)))/f)*(a*b*(2 - 1i 
) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i))*1i)/f + (((256*(4*tan(e + f*x) + 3)^( 
1/2)*(96*a^3*b - 96*a*b^3 + 7*a^4 + 7*b^4 - 42*a^2*b^2))/f^2 - (((3072*(4* 
tan(e + f*x) + 3)^(1/2)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i)))/ 
f + (25600*a*b)/f)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i)))/f)*(a 
*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i))*1i)/f)/((12800*(3*a*b^5 + 3 
*a^5*b + 2*a^6 - 2*b^6 - 2*a^2*b^4 + 6*a^3*b^3 + 2*a^4*b^2))/f^3 - (((256* 
(4*tan(e + f*x) + 3)^(1/2)*(96*a^3*b - 96*a*b^3 + 7*a^4 + 7*b^4 - 42*a^2*b 
^2))/f^2 - (((3072*(4*tan(e + f*x) + 3)^(1/2)*(a*b*(2 - 1i) + a^2*(1/2 + 1 
i) - b^2*(1/2 + 1i)))/f - (25600*a*b)/f)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - 
b^2*(1/2 + 1i)))/f)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i)))/f + 
(((256*(4*tan(e + f*x) + 3)^(1/2)*(96*a^3*b - 96*a*b^3 + 7*a^4 + 7*b^4 - 4 
2*a^2*b^2))/f^2 - (((3072*(4*tan(e + f*x) + 3)^(1/2)*(a*b*(2 - 1i) + a^2*( 
1/2 + 1i) - b^2*(1/2 + 1i)))/f + (25600*a*b)/f)*(a*b*(2 - 1i) + a^2*(1/2 + 
 1i) - b^2*(1/2 + 1i)))/f)*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i) 
))/f))*(a*b*(2 - 1i) + a^2*(1/2 + 1i) - b^2*(1/2 + 1i))*2i)/f + (atan((...
 

Reduce [F]

\[ \int \sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))^2 \, dx=\left (\int \sqrt {4 \tan \left (f x +e \right )+3}d x \right ) a^{2}+\left (\int \sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}d x \right ) b^{2}+2 \left (\int \sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )d x \right ) a b \] Input:

int((3+4*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^2,x)
 

Output:

int(sqrt(4*tan(e + f*x) + 3),x)*a**2 + int(sqrt(4*tan(e + f*x) + 3)*tan(e 
+ f*x)**2,x)*b**2 + 2*int(sqrt(4*tan(e + f*x) + 3)*tan(e + f*x),x)*a*b