\(\int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx\) [1275]

Optimal result

Integrand size = 27, antiderivative size = 162 \[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=-\frac {2 \left (a^2+11 a b-b^2\right ) \arctan \left (\frac {a^2+11 a b-b^2-2 \left (a^2+11 a b-b^2\right ) \tan (e+f x)}{\left (a^2+11 a b-b^2\right ) \sqrt {3+4 \tan (e+f x)}}\right )}{125 f}+\frac {\left (11 a^2-4 a b-11 b^2\right ) \text {arctanh}\left (\frac {2+\tan (e+f x)}{\sqrt {3+4 \tan (e+f x)}}\right )}{125 f}-\frac {(4 a-3 b)^2}{50 f \sqrt {3+4 \tan (e+f x)}} \] Output:

-2/125*(a^2+11*a*b-b^2)*arctan((a^2+11*a*b-b^2-2*(a^2+11*a*b-b^2)*tan(f*x+ 
e))/(a^2+11*a*b-b^2)/(3+4*tan(f*x+e))^(1/2))/f+1/125*(11*a^2-4*a*b-11*b^2) 
*arctanh((2+tan(f*x+e))/(3+4*tan(f*x+e))^(1/2))/f-1/50*(4*a-3*b)^2/f/(3+4* 
tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.69 \[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=\frac {(-8-6 i) (a+i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\left (\frac {3}{25}-\frac {4 i}{25}\right ) (3+4 \tan (e+f x))\right )+(3+4 i) \left ((-3+4 i) b^2+2 i (a-i b)^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\left (\frac {3}{25}+\frac {4 i}{25}\right ) (3+4 \tan (e+f x))\right )\right )}{50 f \sqrt {3+4 \tan (e+f x)}} \] Input:

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

Output:

((-8 - 6*I)*(a + I*b)^2*Hypergeometric2F1[-1/2, 1, 1/2, (3/25 - (4*I)/25)* 
(3 + 4*Tan[e + f*x])] + (3 + 4*I)*((-3 + 4*I)*b^2 + (2*I)*(a - I*b)^2*Hype 
rgeometric2F1[-1/2, 1, 1/2, (3/25 + (4*I)/25)*(3 + 4*Tan[e + f*x])]))/(50* 
f*Sqrt[3 + 4*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4025, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

\(\displaystyle \frac {1}{25} \int \frac {(3 a-b) (a+3 b)-2 (a-2 b) (2 a+b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx-\frac {(4 a-3 b)^2}{50 f \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{25} \int \frac {(3 a-b) (a+3 b)-2 (a-2 b) (2 a+b) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx-\frac {(4 a-3 b)^2}{50 f \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4019

\(\displaystyle \frac {1}{25} \left (\frac {1}{10} \int \frac {4 \left (\tan (e+f x) \left (a^2+11 b a-b^2\right )+2 \left (a^2+11 b a-b^2\right )\right )}{\sqrt {4 \tan (e+f x)+3}}dx-\frac {1}{10} \int -\frac {2 \left (11 a^2-4 b a-11 b^2-2 \left (11 a^2-4 b a-11 b^2\right ) \tan (e+f x)\right )}{\sqrt {4 \tan (e+f x)+3}}dx\right )-\frac {(4 a-3 b)^2}{50 f \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{25} \left (\frac {1}{5} \int \frac {11 a^2-4 b a-11 b^2-2 \left (11 a^2-4 b a-11 b^2\right ) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {2}{5} \int \frac {\tan (e+f x) \left (a^2+11 b a-b^2\right )+2 \left (a^2+11 b a-b^2\right )}{\sqrt {4 \tan (e+f x)+3}}dx\right )-\frac {(4 a-3 b)^2}{50 f \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{25} \left (\frac {1}{5} \int \frac {11 a^2-4 b a-11 b^2-2 \left (11 a^2-4 b a-11 b^2\right ) \tan (e+f x)}{\sqrt {4 \tan (e+f x)+3}}dx+\frac {2}{5} \int \frac {\tan (e+f x) \left (a^2+11 b a-b^2\right )+2 \left (a^2+11 b a-b^2\right )}{\sqrt {4 \tan (e+f x)+3}}dx\right )-\frac {(4 a-3 b)^2}{50 f \sqrt {4 \tan (e+f x)+3}}\)

\(\Big \downarrow \) 4018

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

((-2*(a^2 + 11*a*b - b^2)*ArcTan[(a^2 + 11*a*b - b^2 - 2*(a^2 + 11*a*b - b 
^2)*Tan[e + f*x])/((a^2 + 11*a*b - b^2)*Sqrt[3 + 4*Tan[e + f*x]])])/(5*f) 
+ ((11*a^2 - 4*a*b - 11*b^2)*ArcTanh[(2*(11*a^2 - 4*a*b - 11*b^2) + (11*a^ 
2 - 4*a*b - 11*b^2)*Tan[e + f*x])/((11*a^2 - 4*a*b - 11*b^2)*Sqrt[3 + 4*Ta 
n[e + f*x]])])/(5*f))/25 - (4*a - 3*b)^2/(50*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 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 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 4025

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\left (22 a^{2}-8 a b -22 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{500}+\frac {\left (4 a^{2}+44 a b -4 b^{2}\right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}+\frac {\left (-22 a^{2}+8 a b +22 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{500}+\frac {\left (4 a^{2}+44 a b -4 b^{2}\right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}-\frac {\frac {16}{25} a^{2}-\frac {24}{25} a b +\frac {9}{25} b^{2}}{2 \sqrt {3+4 \tan \left (f x +e \right )}}}{f}\)	\(183\)
default	\(\frac {\frac {\left (22 a^{2}-8 a b -22 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{500}+\frac {\left (4 a^{2}+44 a b -4 b^{2}\right ) \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}+\frac {\left (-22 a^{2}+8 a b +22 b^{2}\right ) \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{500}+\frac {\left (4 a^{2}+44 a b -4 b^{2}\right ) \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}-\frac {\frac {16}{25} a^{2}-\frac {24}{25} a b +\frac {9}{25} b^{2}}{2 \sqrt {3+4 \tan \left (f x +e \right )}}}{f}\)	\(183\)
parts	\(\frac {a^{2} \left (-\frac {11 \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}+\frac {2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}-\frac {8}{25 \sqrt {3+4 \tan \left (f x +e \right )}}+\frac {11 \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}+\frac {2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}\right )}{f}+\frac {b^{2} \left (\frac {11 \ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}-\frac {2 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}-\frac {9}{50 \sqrt {3+4 \tan \left (f x +e \right )}}-\frac {11 \ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{250}-\frac {2 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}\right )}{f}+\frac {2 a b \left (\frac {\ln \left (8+4 \tan \left (f x +e \right )-4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}+\frac {11 \arctan \left (-2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}+\frac {6}{25 \sqrt {3+4 \tan \left (f x +e \right )}}-\frac {\ln \left (8+4 \tan \left (f x +e \right )+4 \sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}+\frac {11 \arctan \left (2+\sqrt {3+4 \tan \left (f x +e \right )}\right )}{125}\right )}{f}\)	\(332\)

Input:

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

Output:

1/f*(1/500*(22*a^2-8*a*b-22*b^2)*ln(8+4*tan(f*x+e)+4*(3+4*tan(f*x+e))^(1/2 
))+1/250*(4*a^2+44*a*b-4*b^2)*arctan(2+(3+4*tan(f*x+e))^(1/2))+1/500*(-22* 
a^2+8*a*b+22*b^2)*ln(8+4*tan(f*x+e)-4*(3+4*tan(f*x+e))^(1/2))+1/250*(4*a^2 
+44*a*b-4*b^2)*arctan(-2+(3+4*tan(f*x+e))^(1/2))-1/2*(16/25*a^2-24/25*a*b+ 
9/25*b^2)/(3+4*tan(f*x+e))^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.70 \[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=\frac {4 \, {\left (3 \, a^{2} + 33 \, a b - 3 \, b^{2} + 4 \, {\left (a^{2} + 11 \, a b - b^{2}\right )} \tan \left (f x + e\right )\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 4 \, {\left (3 \, a^{2} + 33 \, a b - 3 \, b^{2} + 4 \, {\left (a^{2} + 11 \, a b - b^{2}\right )} \tan \left (f x + e\right )\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) + {\left (33 \, a^{2} - 12 \, a b - 33 \, b^{2} + 4 \, {\left (11 \, a^{2} - 4 \, a b - 11 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \log \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) - {\left (33 \, a^{2} - 12 \, a b - 33 \, b^{2} + 4 \, {\left (11 \, a^{2} - 4 \, a b - 11 \, b^{2}\right )} \tan \left (f x + e\right )\right )} \log \left (-\sqrt {4 \, \tan \left (f x + e\right ) + 3} + \tan \left (f x + e\right ) + 2\right ) - 5 \, {\left (16 \, a^{2} - 24 \, a b + 9 \, b^{2}\right )} \sqrt {4 \, \tan \left (f x + e\right ) + 3}}{250 \, {\left (4 \, f \tan \left (f x + e\right ) + 3 \, f\right )}} \] Input:

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

Output:

1/250*(4*(3*a^2 + 33*a*b - 3*b^2 + 4*(a^2 + 11*a*b - b^2)*tan(f*x + e))*ar 
ctan(sqrt(4*tan(f*x + e) + 3) + 2) + 4*(3*a^2 + 33*a*b - 3*b^2 + 4*(a^2 + 
11*a*b - b^2)*tan(f*x + e))*arctan(sqrt(4*tan(f*x + e) + 3) - 2) + (33*a^2 
 - 12*a*b - 33*b^2 + 4*(11*a^2 - 4*a*b - 11*b^2)*tan(f*x + e))*log(sqrt(4* 
tan(f*x + e) + 3) + tan(f*x + e) + 2) - (33*a^2 - 12*a*b - 33*b^2 + 4*(11* 
a^2 - 4*a*b - 11*b^2)*tan(f*x + e))*log(-sqrt(4*tan(f*x + e) + 3) + tan(f* 
x + e) + 2) - 5*(16*a^2 - 24*a*b + 9*b^2)*sqrt(4*tan(f*x + e) + 3))/(4*f*t 
an(f*x + e) + 3*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.10 \[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=\frac {4 \, {\left (a^{2} + 11 \, a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} + 2\right ) + 4 \, {\left (a^{2} + 11 \, a b - b^{2}\right )} \arctan \left (\sqrt {4 \, \tan \left (f x + e\right ) + 3} - 2\right ) + {\left (11 \, a^{2} - 4 \, a b - 11 \, b^{2}\right )} \log \left (4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right ) - {\left (11 \, a^{2} - 4 \, a b - 11 \, b^{2}\right )} \log \left (-4 \, \sqrt {4 \, \tan \left (f x + e\right ) + 3} + 4 \, \tan \left (f x + e\right ) + 8\right ) - \frac {5 \, {\left (16 \, a^{2} - 24 \, a b + 9 \, b^{2}\right )}}{\sqrt {4 \, \tan \left (f x + e\right ) + 3}}}{250 \, f} \] Input:

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

Output:

1/250*(4*(a^2 + 11*a*b - b^2)*arctan(sqrt(4*tan(f*x + e) + 3) + 2) + 4*(a^ 
2 + 11*a*b - b^2)*arctan(sqrt(4*tan(f*x + e) + 3) - 2) + (11*a^2 - 4*a*b - 
 11*b^2)*log(4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8) - (11*a^2 - 
4*a*b - 11*b^2)*log(-4*sqrt(4*tan(f*x + e) + 3) + 4*tan(f*x + e) + 8) - 5* 
(16*a^2 - 24*a*b + 9*b^2)/sqrt(4*tan(f*x + e) + 3))/f
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [B] (verification not implemented)

Time = 4.13 (sec) , antiderivative size = 1629, normalized size of antiderivative = 10.06 \[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((4*tan(e + f*x) + 3)^(1/2)*(28000000*a^4*f^3 + 28000000*b^4*f^3 + 
 384000000*a*b^3*f^3 - 384000000*a^3*b*f^3 - 168000000*a^2*b^2*f^3) + ((a* 
b*(4 - 22i) - a^2*(11 + 2i) + b^2*(11 + 2i))*(4800000000*a^2*f^4 - 4800000 
000*b^2*f^4 - 120000000*f^4*(4*tan(e + f*x) + 3)^(1/2)*(a*b*(4 - 22i) - a^ 
2*(11 + 2i) + b^2*(11 + 2i)) + 2800000000*a*b*f^4))/(250*f))*(a*b*(4 - 22i 
) - a^2*(11 + 2i) + b^2*(11 + 2i))*1i)/(250*f) + (((4*tan(e + f*x) + 3)^(1 
/2)*(28000000*a^4*f^3 + 28000000*b^4*f^3 + 384000000*a*b^3*f^3 - 384000000 
*a^3*b*f^3 - 168000000*a^2*b^2*f^3) - ((a*b*(4 - 22i) - a^2*(11 + 2i) + b^ 
2*(11 + 2i))*(4800000000*a^2*f^4 - 4800000000*b^2*f^4 + 120000000*f^4*(4*t 
an(e + f*x) + 3)^(1/2)*(a*b*(4 - 22i) - a^2*(11 + 2i) + b^2*(11 + 2i)) + 2 
800000000*a*b*f^4))/(250*f))*(a*b*(4 - 22i) - a^2*(11 + 2i) + b^2*(11 + 2i 
))*1i)/(250*f))/(16000000*b^6*f^2 - 16000000*a^6*f^2 + (((4*tan(e + f*x) + 
 3)^(1/2)*(28000000*a^4*f^3 + 28000000*b^4*f^3 + 384000000*a*b^3*f^3 - 384 
000000*a^3*b*f^3 - 168000000*a^2*b^2*f^3) + ((a*b*(4 - 22i) - a^2*(11 + 2i 
) + b^2*(11 + 2i))*(4800000000*a^2*f^4 - 4800000000*b^2*f^4 - 120000000*f^ 
4*(4*tan(e + f*x) + 3)^(1/2)*(a*b*(4 - 22i) - a^2*(11 + 2i) + b^2*(11 + 2i 
)) + 2800000000*a*b*f^4))/(250*f))*(a*b*(4 - 22i) - a^2*(11 + 2i) + b^2*(1 
1 + 2i)))/(250*f) - (((4*tan(e + f*x) + 3)^(1/2)*(28000000*a^4*f^3 + 28000 
000*b^4*f^3 + 384000000*a*b^3*f^3 - 384000000*a^3*b*f^3 - 168000000*a^2*b^ 
2*f^3) - ((a*b*(4 - 22i) - a^2*(11 + 2i) + b^2*(11 + 2i))*(4800000000*a...
 

Reduce [F]

\[ \int \frac {(a+b \tan (e+f x))^2}{(3+4 \tan (e+f x))^{3/2}} \, dx=\frac {-\sqrt {4 \tan \left (f x +e \right )+3}\, a^{2}-8 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) \tan \left (f x +e \right ) a^{2} f +8 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) \tan \left (f x +e \right ) b^{2} f -6 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) a^{2} f +6 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )^{2}}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) b^{2} f +16 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) \tan \left (f x +e \right ) a b f +12 \left (\int \frac {\sqrt {4 \tan \left (f x +e \right )+3}\, \tan \left (f x +e \right )}{16 \tan \left (f x +e \right )^{2}+24 \tan \left (f x +e \right )+9}d x \right ) a b f}{2 f \left (4 \tan \left (f x +e \right )+3\right )} \] Input:

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

Output:

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