\(\int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx\) [1278]

Optimal result

Integrand size = 27, antiderivative size = 281 \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx=-\frac {2 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {3+4 \tan (e+f x)}}{\sqrt {4 a-3 b}}\right )}{(4 a-3 b)^{3/2} \left (a^2+b^2\right ) f}-\frac {(2 a-11 b) \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {3+4 \tan (e+f x)}}\right )}{125 \left (a^2+b^2\right ) f}+\frac {(11 a+2 b) \text {arctanh}\left (\frac {2 (4 a-3 b) (11 a+2 b)+(4 a-3 b) (11 a+2 b) \tan (e+f x)}{\left (44 a^2-25 a b-6 b^2\right ) \sqrt {3+4 \tan (e+f x)}}\right )}{125 \left (a^2+b^2\right ) f}-\frac {32}{25 (4 a-3 b) f \sqrt {3+4 \tan (e+f x)}} \] Output:

-2*b^(5/2)*arctan(b^(1/2)*(3+4*tan(f*x+e))^(1/2)/(4*a-3*b)^(1/2))/(4*a-3*b 
)^(3/2)/(a^2+b^2)/f-1/125*(2*a-11*b)*arctan(((2*a-11*b)*(4*a-3*b)-2*(2*a-1 
1*b)*(4*a-3*b)*tan(f*x+e))/(8*a^2-50*a*b+33*b^2)/(3+4*tan(f*x+e))^(1/2))/( 
a^2+b^2)/f+1/125*(11*a+2*b)*arctanh((2*(4*a-3*b)*(11*a+2*b)+(4*a-3*b)*(11* 
a+2*b)*tan(f*x+e))/(44*a^2-25*a*b-6*b^2)/(3+4*tan(f*x+e))^(1/2))/(a^2+b^2) 
/f-32/25/(4*a-3*b)/f/(3+4*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.93 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx=-\frac {-\frac {(2-11 i) \arctan \left (\left (\frac {1}{5}+\frac {2 i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )}{a+i b}+\frac {250 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {3+4 \tan (e+f x)}}{\sqrt {4 a-3 b}}\right )}{(4 a-3 b)^{3/2} \left (a^2+b^2\right )}-\frac {(11-2 i) \text {arctanh}\left (\left (\frac {2}{5}+\frac {i}{5}\right ) \sqrt {3+4 \tan (e+f x)}\right )}{a-i b}+\frac {160}{(4 a-3 b) \sqrt {3+4 \tan (e+f x)}}}{125 f} \] Input:

Integrate[1/((3 + 4*Tan[e + f*x])^(3/2)*(a + b*Tan[e + f*x])),x]
 

Output:

-1/125*(((-2 + 11*I)*ArcTan[(1/5 + (2*I)/5)*Sqrt[3 + 4*Tan[e + f*x]]])/(a 
+ I*b) + (250*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[3 + 4*Tan[e + f*x]])/Sqrt[4*a - 
 3*b]])/((4*a - 3*b)^(3/2)*(a^2 + b^2)) - ((11 - 2*I)*ArcTanh[(2/5 + I/5)* 
Sqrt[3 + 4*Tan[e + f*x]]])/(a - I*b) + 160/((4*a - 3*b)*Sqrt[3 + 4*Tan[e + 
 f*x]]))/f
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.22, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4052, 27, 3042, 4136, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

\(\displaystyle -\frac {2 \int -\frac {-16 b \tan ^2(e+f x)-4 (4 a-3 b) \tan (e+f x)+12 a-25 b}{2 \sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-16 b \tan ^2(e+f x)-4 (4 a-3 b) \tan (e+f x)+12 a-25 b}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {-16 b \tan (e+f x)^2-4 (4 a-3 b) \tan (e+f x)+12 a-25 b}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {\int \frac {12 a^2-25 b a+12 b^2-(4 a-3 b) (4 a+3 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan ^2(e+f x)+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {12 a^2-25 b a+12 b^2-(4 a-3 b) (4 a+3 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {\frac {\frac {1}{10} \int \frac {2 (2 (2 a-11 b) (4 a-3 b)+(2 a-11 b) \tan (e+f x) (4 a-3 b))}{\sqrt {4 \tan (e+f x)+3}}dx-\frac {1}{10} \int -\frac {2 \left (44 a^2-25 b a-6 b^2-2 (4 a-3 b) (11 a+2 b) \tan (e+f x)\right )}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{5} \int \frac {44 a^2-25 b a-6 b^2-2 (4 a-3 b) (11 a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {1}{5} \int \frac {2 (2 a-11 b) (4 a-3 b)+(2 a-11 b) \tan (e+f x) (4 a-3 b)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {1}{5} \int \frac {44 a^2-25 b a-6 b^2-2 (4 a-3 b) (11 a+2 b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {1}{5} \int \frac {2 (2 a-11 b) (4 a-3 b)+(2 a-11 b) \tan (e+f x) (4 a-3 b)}{\sqrt {4 \tan (e+f x)+3}}dx}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4018

\(\displaystyle \frac {\frac {-\frac {8 (11 a+2 b)^2 (4 a-3 b)^2 \int \frac {1}{\frac {64 (2 (4 a-3 b) (11 a+2 b)+(4 a-3 b) \tan (e+f x) (11 a+2 b))^2}{4 \tan (e+f x)+3}-64 \left (44 a^2-25 b a-6 b^2\right )^2}d\frac {8 (2 (4 a-3 b) (11 a+2 b)+(4 a-3 b) \tan (e+f x) (11 a+2 b))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}-\frac {2 (2 a-11 b)^2 (4 a-3 b)^2 \int \frac {1}{4 (2 a-11 b)^2 (4 a-3 b)^2+\frac {4 ((2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x))^2}{4 \tan (e+f x)+3}}d\frac {2 ((2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {-\frac {8 (11 a+2 b)^2 (4 a-3 b)^2 \int \frac {1}{\frac {64 (2 (4 a-3 b) (11 a+2 b)+(4 a-3 b) \tan (e+f x) (11 a+2 b))^2}{4 \tan (e+f x)+3}-64 \left (44 a^2-25 b a-6 b^2\right )^2}d\frac {8 (2 (4 a-3 b) (11 a+2 b)+(4 a-3 b) \tan (e+f x) (11 a+2 b))}{\sqrt {4 \tan (e+f x)+3}}}{5 f}-\frac {(2 a-11 b)^2 (4 a-3 b)^2 \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (8 a^2-50 a b+33 b^2\right )}}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\frac {\frac {(4 a-3 b)^2 (11 a+2 b)^2 \text {arctanh}\left (\frac {(4 a-3 b) (11 a+2 b) \tan (e+f x)+2 (4 a-3 b) (11 a+2 b)}{\left (44 a^2-25 a b-6 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (44 a^2-25 a b-6 b^2\right )}-\frac {(2 a-11 b)^2 (4 a-3 b)^2 \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (8 a^2-50 a b+33 b^2\right )}}{a^2+b^2}-\frac {25 b^3 \int \frac {\tan (e+f x)^2+1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}dx}{a^2+b^2}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {(4 a-3 b)^2 (11 a+2 b)^2 \text {arctanh}\left (\frac {(4 a-3 b) (11 a+2 b) \tan (e+f x)+2 (4 a-3 b) (11 a+2 b)}{\left (44 a^2-25 a b-6 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (44 a^2-25 a b-6 b^2\right )}-\frac {(2 a-11 b)^2 (4 a-3 b)^2 \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (8 a^2-50 a b+33 b^2\right )}}{a^2+b^2}-\frac {25 b^3 \int \frac {1}{\sqrt {4 \tan (e+f x)+3} (a+b \tan (e+f x))}d\tan (e+f x)}{f \left (a^2+b^2\right )}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {(4 a-3 b)^2 (11 a+2 b)^2 \text {arctanh}\left (\frac {(4 a-3 b) (11 a+2 b) \tan (e+f x)+2 (4 a-3 b) (11 a+2 b)}{\left (44 a^2-25 a b-6 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (44 a^2-25 a b-6 b^2\right )}-\frac {(2 a-11 b)^2 (4 a-3 b)^2 \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (8 a^2-50 a b+33 b^2\right )}}{a^2+b^2}-\frac {25 b^3 \int \frac {1}{\frac {1}{4} (4 a-3 b)+\frac {1}{4} b (4 \tan (e+f x)+3)}d\sqrt {4 \tan (e+f x)+3}}{2 f \left (a^2+b^2\right )}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {(4 a-3 b)^2 (11 a+2 b)^2 \text {arctanh}\left (\frac {(4 a-3 b) (11 a+2 b) \tan (e+f x)+2 (4 a-3 b) (11 a+2 b)}{\left (44 a^2-25 a b-6 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (44 a^2-25 a b-6 b^2\right )}-\frac {(2 a-11 b)^2 (4 a-3 b)^2 \arctan \left (\frac {(2 a-11 b) (4 a-3 b)-2 (2 a-11 b) (4 a-3 b) \tan (e+f x)}{\left (8 a^2-50 a b+33 b^2\right ) \sqrt {4 \tan (e+f x)+3}}\right )}{5 f \left (8 a^2-50 a b+33 b^2\right )}}{a^2+b^2}-\frac {50 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {4 \tan (e+f x)+3}}{\sqrt {4 a-3 b}}\right )}{f \sqrt {4 a-3 b} \left (a^2+b^2\right )}}{25 (4 a-3 b)}-\frac {32}{25 f (4 a-3 b) \sqrt {4 \tan (e+f x)+3}}\)

Input:

Int[1/((3 + 4*Tan[e + f*x])^(3/2)*(a + b*Tan[e + f*x])),x]
 

Output:

((-50*b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[3 + 4*Tan[e + f*x]])/Sqrt[4*a - 3*b]])/ 
(Sqrt[4*a - 3*b]*(a^2 + b^2)*f) + (-1/5*((2*a - 11*b)^2*(4*a - 3*b)^2*ArcT 
an[((2*a - 11*b)*(4*a - 3*b) - 2*(2*a - 11*b)*(4*a - 3*b)*Tan[e + f*x])/(( 
8*a^2 - 50*a*b + 33*b^2)*Sqrt[3 + 4*Tan[e + f*x]])])/((8*a^2 - 50*a*b + 33 
*b^2)*f) + ((4*a - 3*b)^2*(11*a + 2*b)^2*ArcTanh[(2*(4*a - 3*b)*(11*a + 2* 
b) + (4*a - 3*b)*(11*a + 2*b)*Tan[e + f*x])/((44*a^2 - 25*a*b - 6*b^2)*Sqr 
t[3 + 4*Tan[e + f*x]])])/(5*(44*a^2 - 25*a*b - 6*b^2)*f))/(a^2 + b^2))/(25 
*(4*a - 3*b)) - 32/(25*(4*a - 3*b)*f*Sqrt[3 + 4*Tan[e + f*x]])
 

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 4052

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

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]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\frac {16 \left (11 a +2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -11 b \right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{4000 a^{2}+4000 b^{2}}-\frac {2 b^{3} \arctan \left (\frac {\sqrt {3+4 \tan \left (f x +e \right )}\, b}{\sqrt {b \left (4 a -3 b \right )}}\right )}{\left (4 a -3 b \right ) \left (a^{2}+b^{2}\right ) \sqrt {b \left (4 a -3 b \right )}}-\frac {32}{\left (100 a -75 b \right ) \sqrt {3+4 \tan \left (f x +e \right )}}+\frac {16 \left (-11 a -2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -11 b \right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{4000 a^{2}+4000 b^{2}}}{f}\)	\(235\)
default	\(\frac {\frac {16 \left (11 a +2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -11 b \right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{4000 a^{2}+4000 b^{2}}-\frac {2 b^{3} \arctan \left (\frac {\sqrt {3+4 \tan \left (f x +e \right )}\, b}{\sqrt {b \left (4 a -3 b \right )}}\right )}{\left (4 a -3 b \right ) \left (a^{2}+b^{2}\right ) \sqrt {b \left (4 a -3 b \right )}}-\frac {32}{\left (100 a -75 b \right ) \sqrt {3+4 \tan \left (f x +e \right )}}+\frac {16 \left (-11 a -2 b \right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )+32 \left (2 a -11 b \right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{4000 a^{2}+4000 b^{2}}}{f}\)	\(235\)

Input:

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

Output:

1/f*(32/(4000*a^2+4000*b^2)*(1/2*(11*a+2*b)*ln(8+4*tan(f*x+e)+4*(3+4*tan(f 
*x+e))^(1/2))+(2*a-11*b)*arctan(2+(3+4*tan(f*x+e))^(1/2)))-2*b^3/(4*a-3*b) 
/(a^2+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/(100*a-75*b)/(3+4*tan(f*x+e))^(1/2)+32/(4000*a^2+4000*b^2)*(1 
/2*(-11*a-2*b)*ln(8+4*tan(f*x+e)-4*(3+4*tan(f*x+e))^(1/2))+(2*a-11*b)*arct 
an(-2+(3+4*tan(f*x+e))^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 793, normalized size of antiderivative = 2.82 \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/250*(250*(4*b^2*tan(f*x + e) + 3*b^2)*sqrt(-b/(4*a - 3*b))*log(((4*a - 
 3*b)*sqrt(-b/(4*a - 3*b))*sqrt(4*tan(f*x + e) + 3) + 2*b*tan(f*x + e) - 2 
*a + 3*b)/(b*tan(f*x + e) + a)) - 2*(24*a^2 - 150*a*b + 99*b^2 + 4*(8*a^2 
- 50*a*b + 33*b^2)*tan(f*x + e))*arctan(sqrt(4*tan(f*x + e) + 3) + 2) - 2* 
(24*a^2 - 150*a*b + 99*b^2 + 4*(8*a^2 - 50*a*b + 33*b^2)*tan(f*x + e))*arc 
tan(sqrt(4*tan(f*x + e) + 3) - 2) - (132*a^2 - 75*a*b - 18*b^2 + 4*(44*a^2 
 - 25*a*b - 6*b^2)*tan(f*x + e))*log(sqrt(4*tan(f*x + e) + 3) + tan(f*x + 
e) + 2) + (132*a^2 - 75*a*b - 18*b^2 + 4*(44*a^2 - 25*a*b - 6*b^2)*tan(f*x 
 + e))*log(-sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + 320*(a^2 + b^2) 
*sqrt(4*tan(f*x + e) + 3))/(4*(4*a^3 - 3*a^2*b + 4*a*b^2 - 3*b^3)*f*tan(f* 
x + e) + 3*(4*a^3 - 3*a^2*b + 4*a*b^2 - 3*b^3)*f), -1/250*(500*(4*b^2*tan( 
f*x + e) + 3*b^2)*sqrt(b/(4*a - 3*b))*arctan(sqrt(b/(4*a - 3*b))*sqrt(4*ta 
n(f*x + e) + 3)) - 2*(24*a^2 - 150*a*b + 99*b^2 + 4*(8*a^2 - 50*a*b + 33*b 
^2)*tan(f*x + e))*arctan(sqrt(4*tan(f*x + e) + 3) + 2) - 2*(24*a^2 - 150*a 
*b + 99*b^2 + 4*(8*a^2 - 50*a*b + 33*b^2)*tan(f*x + e))*arctan(sqrt(4*tan( 
f*x + e) + 3) - 2) - (132*a^2 - 75*a*b - 18*b^2 + 4*(44*a^2 - 25*a*b - 6*b 
^2)*tan(f*x + e))*log(sqrt(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + (132* 
a^2 - 75*a*b - 18*b^2 + 4*(44*a^2 - 25*a*b - 6*b^2)*tan(f*x + e))*log(-sqr 
t(4*tan(f*x + e) + 3) + tan(f*x + e) + 2) + 320*(a^2 + b^2)*sqrt(4*tan(f*x 
 + e) + 3))/(4*(4*a^3 - 3*a^2*b + 4*a*b^2 - 3*b^3)*f*tan(f*x + e) + 3*(...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(3+4*tan(f*x+e))^(3/2)/(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. \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(3+4*tan(f*x+e))^(3/2)/(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 = 5.95 (sec) , antiderivative size = 6361, normalized size of antiderivative = 22.64 \[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((4*tan(e + f*x) + 3)^(1/2)*(a*f*1i - b*f)*(((((74323008000000000000 
000000*b^18*f^9 - 876759552000000000000000000*a*b^17*f^9 + 467227180800000 
0000000000000*a^2*b^16*f^9 - 14949702144000000000000000000*a^3*b^15*f^9 + 
32221893312000000000000000000*a^4*b^14*f^9 - 49460958720000000000000000000 
*a^5*b^13*f^9 + 54198621504000000000000000000*a^6*b^12*f^9 - 3704689612800 
0000000000000000*a^7*b^11*f^9 + 251126784000000000000000000*a^8*b^10*f^9 + 
 40306966528000000000000000000*a^9*b^9*f^9 - 65042743296000000000000000000 
*a^10*b^8*f^9 + 65445560320000000000000000000*a^11*b^7*f^9 - 4784180428800 
0000000000000000*a^12*b^6*f^9 + 25526534144000000000000000000*a^13*b^5*f^9 
 - 9323937792000000000000000000*a^14*b^4*f^9 + 204682035200000000000000000 
0*a^15*b^3*f^9 - 201326592000000000000000000*a^16*b^2*f^9)*(117/62500 + 11 
i/15625))/(a*f*1i - b*f)^2 + 173504332800000000000000*b^16*f^7 + 348912230 
400000000000000*a*b^15*f^7 - 13047070464000000000000000*a^2*b^14*f^7 + 696 
54148300800000000000000*a^3*b^13*f^7 - 184983199718400000000000000*a^4*b^1 
2*f^7 + 287826623078400000000000000*a^5*b^11*f^7 - 26796790702080000000000 
0000*a^6*b^10*f^7 + 135149636812800000000000000*a^7*b^9*f^7 - 193903591424 
00000000000000*a^8*b^8*f^7 - 9696077414400000000000000*a^9*b^7*f^7 - 33485 
22598400000000000000*a^10*b^6*f^7 + 9190558924800000000000000*a^11*b^5*f^7 
 - 5132150374400000000000000*a^12*b^4*f^7 + 1409286144000000000000000*a^13 
*b^3*f^7 - 187904819200000000000000*a^14*b^2*f^7)*(117/62500 + 11i/1562...
 

Reduce [F]

\[ \int \frac {1}{(3+4 \tan (e+f x))^{3/2} (a+b \tan (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(4*tan(e + f*x) + 3) + 16*int(sqrt(4*tan(e + f*x) + 3)/(16*tan(e + f* 
x)**3*b + 16*tan(e + f*x)**2*a + 24*tan(e + f*x)**2*b + 24*tan(e + f*x)*a 
+ 9*tan(e + f*x)*b + 9*a),x)*tan(e + f*x)*a*f + 12*int(sqrt(4*tan(e + f*x) 
 + 3)/(16*tan(e + f*x)**3*b + 16*tan(e + f*x)**2*a + 24*tan(e + f*x)**2*b 
+ 24*tan(e + f*x)*a + 9*tan(e + f*x)*b + 9*a),x)*a*f + 8*int((sqrt(4*tan(e 
 + f*x) + 3)*tan(e + f*x)**3)/(16*tan(e + f*x)**3*b + 16*tan(e + f*x)**2*a 
 + 24*tan(e + f*x)**2*b + 24*tan(e + f*x)*a + 9*tan(e + f*x)*b + 9*a),x)*t 
an(e + f*x)*b*f + 6*int((sqrt(4*tan(e + f*x) + 3)*tan(e + f*x)**3)/(16*tan 
(e + f*x)**3*b + 16*tan(e + f*x)**2*a + 24*tan(e + f*x)**2*b + 24*tan(e + 
f*x)*a + 9*tan(e + f*x)*b + 9*a),x)*b*f + 8*int((sqrt(4*tan(e + f*x) + 3)* 
tan(e + f*x)**2)/(16*tan(e + f*x)**3*b + 16*tan(e + f*x)**2*a + 24*tan(e + 
 f*x)**2*b + 24*tan(e + f*x)*a + 9*tan(e + f*x)*b + 9*a),x)*tan(e + f*x)*a 
*f + 6*int((sqrt(4*tan(e + f*x) + 3)*tan(e + f*x)**2)/(16*tan(e + f*x)**3* 
b + 16*tan(e + f*x)**2*a + 24*tan(e + f*x)**2*b + 24*tan(e + f*x)*a + 9*ta 
n(e + f*x)*b + 9*a),x)*a*f + 8*int((sqrt(4*tan(e + f*x) + 3)*tan(e + f*x)) 
/(16*tan(e + f*x)**3*b + 16*tan(e + f*x)**2*a + 24*tan(e + f*x)**2*b + 24* 
tan(e + f*x)*a + 9*tan(e + f*x)*b + 9*a),x)*tan(e + f*x)*b*f + 6*int((sqrt 
(4*tan(e + f*x) + 3)*tan(e + f*x))/(16*tan(e + f*x)**3*b + 16*tan(e + f*x) 
**2*a + 24*tan(e + f*x)**2*b + 24*tan(e + f*x)*a + 9*tan(e + f*x)*b + 9*a) 
,x)*b*f)/(2*a*f*(4*tan(e + f*x) + 3))