\(\int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx\) [1272]

Optimal result

Integrand size = 27, antiderivative size = 178 \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {3+4 \tan (e+f x)}}{\sqrt {4 a-3 b}}\right )}{\sqrt {4 a-3 b} \left (a^2+b^2\right ) f}-\frac {(2 a-b) \arctan \left (\frac {2 a-b-2 (2 a-b) \tan (e+f x)}{(2 a-b) \sqrt {3+4 \tan (e+f x)}}\right )}{5 \left (a^2+b^2\right ) f}+\frac {(a+2 b) \text {arctanh}\left (\frac {2+\tan (e+f x)}{\sqrt {3+4 \tan (e+f x)}}\right )}{5 \left (a^2+b^2\right ) f} \] Output:

2*b^(3/2)*arctan(b^(1/2)*(3+4*tan(f*x+e))^(1/2)/(4*a-3*b)^(1/2))/(4*a-3*b) 
^(1/2)/(a^2+b^2)/f-1/5*(2*a-b)*arctan((2*a-b-2*(2*a-b)*tan(f*x+e))/(2*a-b) 
/(3+4*tan(f*x+e))^(1/2))/(a^2+b^2)/f+1/5*(a+2*b)*arctanh((2+tan(f*x+e))/(3 
+4*tan(f*x+e))^(1/2))/(a^2+b^2)/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\frac {(2-i) \sqrt {4 a-3 b} (a-i b) \arctan \left (\left (\frac {1}{5}+\frac {2 i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )+10 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {3+4 \tan (e+f x)}}{\sqrt {4 a-3 b}}\right )+(2+i) \sqrt {4 a-3 b} (-i a+b) \text {arctanh}\left (\left (\frac {2}{5}+\frac {i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )}{5 \sqrt {4 a-3 b} \left (a^2+b^2\right ) f} \] Input:

Integrate[1/(Sqrt[3 + 4*Tan[e + f*x]]*(a + b*Tan[e + f*x])),x]
 

Output:

((2 - I)*Sqrt[4*a - 3*b]*(a - I*b)*ArcTan[(1/5 + (2*I)/5)*Sqrt[3 + 4*Tan[e 
 + f*x]]] + 10*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[3 + 4*Tan[e + f*x]])/Sqrt[4*a 
- 3*b]] + (2 + I)*Sqrt[4*a - 3*b]*((-I)*a + b)*ArcTanh[(2/5 + I/5)*Sqrt[3 
+ 4*Tan[e + f*x]]])/(5*Sqrt[4*a - 3*b]*(a^2 + b^2)*f)
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3042, 4057, 3042, 4019, 27, 3042, 4018, 216, 220, 4117, 73, 218}

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.

\(\displaystyle \int \frac {1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx\)

\(\Big \downarrow \) 4057

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4019

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {b^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}+\frac {\frac {1}{5} \int \frac {\tan (e+f x) (2 a-b)+2 (2 a-b)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {1}{5} \int \frac {a+2 b-2 (a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}+\frac {\frac {1}{5} \int \frac {\tan (e+f x) (2 a-b)+2 (2 a-b)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {1}{5} \int \frac {a+2 b-2 (a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}\)

\(\Big \downarrow \) 4018

\(\displaystyle \frac {b^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}+\frac {-\frac {2 (2 a-b)^2 \int \frac {1}{4 (2 a-b)^2+\frac {4 (2 a-b-2 (2 a-b) \tan (e+f x))^2}{4 \tan (e+f x)+3}}d\frac {2 (2 a-b-2 (2 a-b) \tan (e+f x))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}-\frac {8 (a+2 b)^2 \int \frac {1}{\frac {64 (\tan (e+f x) (a+2 b)+2 (a+2 b))^2}{4 \tan (e+f x)+3}-64 (a+2 b)^2}d\frac {8 (\tan (e+f x) (a+2 b)+2 (a+2 b))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}}{a^2+b^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {8 (a+2 b)^2 \int \frac {1}{\frac {64 (\tan (e+f x) (a+2 b)+2 (a+2 b))^2}{4 \tan (e+f x)+3}-64 (a+2 b)^2}d\frac {8 (\tan (e+f x) (a+2 b)+2 (a+2 b))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}-\frac {(2 a-b) \arctan \left (\frac {-2 (2 a-b) \tan (e+f x)+2 a-b}{(2 a-b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}}{a^2+b^2}+\frac {b^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {b^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}+\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {(a+2 b) \tan (e+f x)+2 (a+2 b)}{(a+2 b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}-\frac {(2 a-b) \arctan \left (\frac {-2 (2 a-b) \tan (e+f x)+2 a-b}{(2 a-b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}}{a^2+b^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {b^2 \int \frac {1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}d\tan (e+f x)}{f \left (a^2+b^2\right )}+\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {(a+2 b) \tan (e+f x)+2 (a+2 b)}{(a+2 b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}-\frac {(2 a-b) \arctan \left (\frac {-2 (2 a-b) \tan (e+f x)+2 a-b}{(2 a-b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}}{a^2+b^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {b^2 \int \frac {1}{\frac {1}{4} (4 a-3 b)+\frac {1}{4} b (4 \tan (e+f x)+3)}d\sqrt {4 \tan (e+f x)+3}}{2 f \left (a^2+b^2\right )}+\frac {\frac {(a+2 b) \text {arctanh}\left (\frac {(a+2 b) \tan (e+f x)+2 (a+2 b)}{(a+2 b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}-\frac {(2 a-b) \arctan \left (\frac {-2 (2 a-b) \tan (e+f x)+2 a-b}{(2 a-b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}}{a^2+b^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {(a+2 b) \text {arctanh}\left (\frac {(a+2 b) \tan (e+f x)+2 (a+2 b)}{(a+2 b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}-\frac {(2 a-b) \arctan \left (\frac {-2 (2 a-b) \tan (e+f x)+2 a-b}{(2 a-b) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f}}{a^2+b^2}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {4 \tan (e+f x)+3}}{\sqrt {4 a-3 b}}\right )}{f \sqrt {4 a-3 b} \left (a^2+b^2\right )}\)

Input:

Int[1/(Sqrt[3 + 4*Tan[e + f*x]]*(a + b*Tan[e + f*x])),x]
 

Output:

(2*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[3 + 4*Tan[e + f*x]])/Sqrt[4*a - 3*b]])/(Sq 
rt[4*a - 3*b]*(a^2 + b^2)*f) + (-1/5*((2*a - b)*ArcTan[(2*a - b - 2*(2*a - 
 b)*Tan[e + f*x])/((2*a - b)*Sqrt[3 + 4*Tan[e + f*x]])])/f + ((a + 2*b)*Ar 
cTanh[(2*(a + 2*b) + (a + 2*b)*Tan[e + f*x])/((a + 2*b)*Sqrt[3 + 4*Tan[e + 
 f*x]])])/(5*f))/(a^2 + b^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 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 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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 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 4057

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)/((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[(a + b*Tan[e + f*x])^m 
*(c - d*Tan[e + f*x]), x], x] + Simp[d^2/(c^2 + d^2)   Int[(a + b*Tan[e + f 
*x])^m*((1 + Tan[e + f*x]^2)/(c + d*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 4117

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {32 b^{2} \arctan \left (\frac {\sqrt {3+4 \tan \left (f x +e \right )}\, b}{\sqrt {b \left (4 a -3 b \right )}}\right )}{\left (16 a^{2}+16 b^{2}\right ) \sqrt {b \left (4 a -3 b \right )}}+\frac {16 \left (-a -2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -b \right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{160 a^{2}+160 b^{2}}+\frac {16 \left (a +2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -b \right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{160 a^{2}+160 b^{2}}}{f}\)	\(205\)
default	\(\frac {\frac {32 b^{2} \arctan \left (\frac {\sqrt {3+4 \tan \left (f x +e \right )}\, b}{\sqrt {b \left (4 a -3 b \right )}}\right )}{\left (16 a^{2}+16 b^{2}\right ) \sqrt {b \left (4 a -3 b \right )}}+\frac {16 \left (-a -2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -b \right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{160 a^{2}+160 b^{2}}+\frac {16 \left (a +2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -b \right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{160 a^{2}+160 b^{2}}}{f}\)	\(205\)

Input:

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

Output:

1/f*(32*b^2/(16*a^2+16*b^2)/(b*(4*a-3*b))^(1/2)*arctan((3+4*tan(f*x+e))^(1 
/2)*b/(b*(4*a-3*b))^(1/2))+32/(160*a^2+160*b^2)*(1/2*(-a-2*b)*ln(8+4*tan(f 
*x+e)-4*(3+4*tan(f*x+e))^(1/2))+(2*a-b)*arctan(-2+(3+4*tan(f*x+e))^(1/2))) 
+32/(160*a^2+160*b^2)*(1/2*(a+2*b)*ln(8+4*tan(f*x+e)+4*(3+4*tan(f*x+e))^(1 
/2))+(2*a-b)*arctan(2+(3+4*tan(f*x+e))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.05 \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\left [\frac {10 \, b \sqrt {-\frac {b}{4 \, a - 3 \, b}} \log \left (\frac {{\left (4 \, a - 3 \, b\right )} \sqrt {-\frac {b}{4 \, a - 3 \, b}} \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2 \, b \tan \left (f x + e\right ) - 2 \, a + 3 \, b}{b \tan \left (f x + e\right ) + a}\right ) + 2 \, {\left (2 \, a - b\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 2 \, {\left (2 \, a - b\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) + {\left (a + 2 \, b\right )} \log \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) - {\left (a + 2 \, b\right )} \log \left (-\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right )}{10 \, {\left (a^{2} + b^{2}\right )} f}, \frac {20 \, b \sqrt {\frac {b}{4 \, a - 3 \, b}} \arctan \left (\sqrt {\frac {b}{4 \, a - 3 \, b}} \sqrt {4 \, \tan \left (f x + e\right ) + 3}\right ) + 2 \, {\left (2 \, a - b\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 2 \, {\left (2 \, a - b\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) + {\left (a + 2 \, b\right )} \log \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) - {\left (a + 2 \, b\right )} \log \left (-\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right )}{10 \, {\left (a^{2} + b^{2}\right )} f}\right ] \] Input:

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

Output:

[1/10*(10*b*sqrt(-b/(4*a - 3*b))*log(((4*a - 3*b)*sqrt(-b/(4*a - 3*b))*sqr 
t(4*tan(f*x + e) + 3) + 2*b*tan(f*x + e) - 2*a + 3*b)/(b*tan(f*x + e) + a) 
) + 2*(2*a - b)*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 2*(2*a - b)*arctan( 
sqrt(4*tan(f*x + e) + 3) - 2) + (a + 2*b)*log(sqrt(4*tan(f*x + e) + 3) + t 
an(f*x + e) + 2) - (a + 2*b)*log(-sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) 
+ 2))/((a^2 + b^2)*f), 1/10*(20*b*sqrt(b/(4*a - 3*b))*arctan(sqrt(b/(4*a - 
 3*b))*sqrt(4*tan(f*x + e) + 3)) + 2*(2*a - b)*arctan(sqrt(4*tan(f*x + e) 
+ 3) + 2) + 2*(2*a - b)*arctan(sqrt(4*tan(f*x + e) + 3) - 2) + (a + 2*b)*l 
og(sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) - (a + 2*b)*log(-sqrt(4*ta 
n(f*x + e) + 3) + tan(f*x + e) + 2))/((a^2 + b^2)*f)]
 

Sympy [F]

\[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\int \frac {1}{\left (a + b \tan {\left (e + f x \right )}\right ) \sqrt {4 \tan {\left (e + f x \right )} + 3}}\, dx \] Input:

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

Output:

Integral(1/((a + b*tan(e + f*x))*sqrt(4*tan(e + f*x) + 3)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(3*b-4*a>0)', see `assume?` for m 
ore detail
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 5361, normalized size of antiderivative = 30.12 \[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((((((2097152*(96*a*b^7*f^2 - 364*b^8*f^2 - 664*a^2*b^6*f^2 + 19 
2*a^3*b^5*f^2 - 236*a^4*b^4*f^2 + 96*a^5*b^3*f^2 + 64*a^6*b^2*f^2))/f^3 - 
((4*tan(e + f*x) + 3)^(1/2)*(472*b^9*f^4 + 96*a*b^8*f^4 + 616*a^2*b^7*f^4 
+ 288*a^3*b^6*f^4 - 184*a^4*b^5*f^4 + 288*a^5*b^4*f^4 - 328*a^6*b^3*f^4 + 
96*a^7*b^2*f^4)*(1048576/5 - 2097152i/5))/(f^4*(a*f - b*f*1i)))*(1/10 - 1i 
/5))/(a*f - b*f*1i) + (2097152*(4*tan(e + f*x) + 3)^(1/2)*(54*b^7*f^2 + 12 
0*a*b^6*f^2 - 36*a^2*b^5*f^2 - 16*a^3*b^4*f^2 + 6*a^4*b^3*f^2 - 8*a^5*b^2* 
f^2))/f^4)*(1/10 - 1i/5))/(a*f - b*f*1i) + (2097152*(20*a*b^5 + 9*b^6 - 3* 
a^2*b^4 + 4*a^3*b^3))/f^3)*(1/10 - 1i/5))/(a*f - b*f*1i) - (6291456*b^5*(4 
*tan(e + f*x) + 3)^(1/2))/f^4)*(1/5 + 1i/10))/(a*f - b*f*1i) - (((((((((20 
97152*(96*a*b^7*f^2 - 364*b^8*f^2 - 664*a^2*b^6*f^2 + 192*a^3*b^5*f^2 - 23 
6*a^4*b^4*f^2 + 96*a^5*b^3*f^2 + 64*a^6*b^2*f^2))/f^3 + ((4*tan(e + f*x) + 
 3)^(1/2)*(472*b^9*f^4 + 96*a*b^8*f^4 + 616*a^2*b^7*f^4 + 288*a^3*b^6*f^4 
- 184*a^4*b^5*f^4 + 288*a^5*b^4*f^4 - 328*a^6*b^3*f^4 + 96*a^7*b^2*f^4)*(1 
048576/5 - 2097152i/5))/(f^4*(a*f - b*f*1i)))*(1/10 - 1i/5))/(a*f - b*f*1i 
) - (2097152*(4*tan(e + f*x) + 3)^(1/2)*(54*b^7*f^2 + 120*a*b^6*f^2 - 36*a 
^2*b^5*f^2 - 16*a^3*b^4*f^2 + 6*a^4*b^3*f^2 - 8*a^5*b^2*f^2))/f^4)*(1/10 - 
 1i/5))/(a*f - b*f*1i) + (2097152*(20*a*b^5 + 9*b^6 - 3*a^2*b^4 + 4*a^3*b^ 
3))/f^3)*(1/10 - 1i/5))/(a*f - b*f*1i) + (6291456*b^5*(4*tan(e + f*x) + 3) 
^(1/2))/f^4)*(1/5 + 1i/10))/(a*f - b*f*1i))/((((((((((2097152*(96*a*b^7...
 

Reduce [F]

\[ \int \frac {1}{\sqrt {3+4 \tan (e+f x)} (a+b \tan (e+f x))} \, dx=\frac {\sqrt {4 \tan \left (f x +e \right )+3}-2 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{3}}{4 \tan \left (f x +e \right )^{2} b +4 \tan \left (f x +e \right ) a +3 \tan \left (f x +e \right ) b +3 a}d x \right ) b f -2 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}}{4 \tan \left (f x +e \right )^{2} b +4 \tan \left (f x +e \right ) a +3 \tan \left (f x +e \right ) b +3 a}d x \right ) a f -2 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )}{4 \tan \left (f x +e \right )^{2} b +4 \tan \left (f x +e \right ) a +3 \tan \left (f x +e \right ) b +3 a}d x \right ) b f}{2 a f} \] Input:

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

Output:

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