\(\int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx\) [207]

Optimal result

Integrand size = 23, antiderivative size = 177 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{9/2} f}+\frac {91 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{32 \sqrt {2} a^{9/2} f}+\frac {\tan (e+f x)}{3 a f (a+a \sec (e+f x))^{7/2}}+\frac {11 \tan (e+f x)}{24 a^2 f (a+a \sec (e+f x))^{5/2}}+\frac {27 \tan (e+f x)}{32 a^3 f (a+a \sec (e+f x))^{3/2}} \] Output:

-2*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(9/2)/f+91/64*arcta 
n(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/(a+a*sec(f*x+e))^(1/2))*2^(1/2)/a^(9/2)/f 
+1/3*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(7/2)+11/24*tan(f*x+e)/a^2/f/(a+a*sec 
(f*x+e))^(5/2)+27/32*tan(f*x+e)/a^3/f/(a+a*sec(f*x+e))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 6.46 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.56 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\frac {\left (91 \arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right )-64 \sqrt {2} \arctan \left (\frac {\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}}}\right )\right ) \cos ^8\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sec ^{\frac {9}{2}}(e+f x) \sqrt {1+\sec (e+f x)}}{2 f \sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )} (a (1+\sec (e+f x)))^{9/2}}+\frac {\cos ^9\left (\frac {1}{2} (e+f x)\right ) \sec ^5(e+f x) \left (\frac {157}{3} \sin \left (\frac {1}{2} (e+f x)\right )-\frac {265}{6} \sec \left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )+\frac {35}{3} \sec ^3\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )-\frac {4}{3} \sec ^5\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f (a (1+\sec (e+f x)))^{9/2}} \] Input:

Integrate[Tan[e + f*x]^2/(a + a*Sec[e + f*x])^(9/2),x]
 

Output:

((91*ArcSin[Tan[(e + f*x)/2]] - 64*Sqrt[2]*ArcTan[Tan[(e + f*x)/2]/Sqrt[Co 
s[e + f*x]/(1 + Cos[e + f*x])]])*Cos[(e + f*x)/2]^8*Sqrt[Cos[e + f*x]/(1 + 
 Cos[e + f*x])]*Sec[e + f*x]^(9/2)*Sqrt[1 + Sec[e + f*x]])/(2*f*Sqrt[Sec[( 
e + f*x)/2]^2]*(a*(1 + Sec[e + f*x]))^(9/2)) + (Cos[(e + f*x)/2]^9*Sec[e + 
 f*x]^5*((157*Sin[(e + f*x)/2])/3 - (265*Sec[(e + f*x)/2]*Tan[(e + f*x)/2] 
)/6 + (35*Sec[(e + f*x)/2]^3*Tan[(e + f*x)/2])/3 - (4*Sec[(e + f*x)/2]^5*T 
an[(e + f*x)/2])/3))/(f*(a*(1 + Sec[e + f*x]))^(9/2))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.46, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4375, 373, 402, 27, 402, 27, 397, 216}

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.

\(\displaystyle \int \frac {\tan ^2(e+f x)}{(a \sec (e+f x)+a)^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cot \left (e+f x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{9/2}}dx\)

\(\Big \downarrow \) 4375

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

\(\Big \downarrow \) 373

\(\displaystyle -\frac {2 \left (-\frac {\int \frac {1-\frac {5 a \tan ^2(e+f x)}{\sec (e+f x) a+a}}{\left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1\right ) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^3}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 402

\(\displaystyle -\frac {2 \left (-\frac {\frac {\int \frac {3 a \left (5-\frac {11 a \tan ^2(e+f x)}{\sec (e+f x) a+a}\right )}{\left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1\right ) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^2}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{8 a}+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (-\frac {\frac {3}{8} \int \frac {5-\frac {11 a \tan ^2(e+f x)}{\sec (e+f x) a+a}}{\left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1\right ) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^2}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 402

\(\displaystyle -\frac {2 \left (-\frac {\frac {3}{8} \left (\frac {\int \frac {a \left (37-\frac {27 a \tan ^2(e+f x)}{\sec (e+f x) a+a}\right )}{\left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1\right ) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{4 a}+\frac {27 \tan (e+f x)}{4 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )}\right )+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (-\frac {\frac {3}{8} \left (\frac {1}{4} \int \frac {37-\frac {27 a \tan ^2(e+f x)}{\sec (e+f x) a+a}}{\left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1\right ) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )+\frac {27 \tan (e+f x)}{4 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )}\right )+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 397

\(\displaystyle -\frac {2 \left (-\frac {\frac {3}{8} \left (\frac {1}{4} \left (64 \int \frac {1}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )-91 \int \frac {1}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )\right )+\frac {27 \tan (e+f x)}{4 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )}\right )+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 \left (-\frac {\frac {3}{8} \left (\frac {1}{4} \left (\frac {91 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {2} \sqrt {a}}-\frac {64 \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{\sqrt {a}}\right )+\frac {27 \tan (e+f x)}{4 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )}\right )+\frac {11 \tan (e+f x)}{8 \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^2}}{6 a}-\frac {\tan (e+f x)}{6 a \sqrt {a \sec (e+f x)+a} \left (\frac {a \tan ^2(e+f x)}{a \sec (e+f x)+a}+2\right )^3}\right )}{a^3 f}\)

Input:

Int[Tan[e + f*x]^2/(a + a*Sec[e + f*x])^(9/2),x]
 

Output:

(-2*(-1/6*Tan[e + f*x]/(a*Sqrt[a + a*Sec[e + f*x]]*(2 + (a*Tan[e + f*x]^2) 
/(a + a*Sec[e + f*x]))^3) - ((11*Tan[e + f*x])/(8*Sqrt[a + a*Sec[e + f*x]] 
*(2 + (a*Tan[e + f*x]^2)/(a + a*Sec[e + f*x]))^2) + (3*(((-64*ArcTan[(Sqrt 
[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/Sqrt[a] + (91*ArcTan[(Sqrt[a] 
*Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(Sqrt[2]*Sqrt[a]))/4 + 
 (27*Tan[e + f*x])/(4*Sqrt[a + a*Sec[e + f*x]]*(2 + (a*Tan[e + f*x]^2)/(a 
+ a*Sec[e + f*x])))))/8)/(6*a)))/(a^3*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 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4375

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d)   Subst[Int[x^m*((2 + a*x^2 
)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] 
]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I 
ntegerQ[n - 1/2]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(330\) vs. \(2(148)=296\).

Time = 11.74 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.87

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \left (8 \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {5}{2}} \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-10 \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-12 \left (1-\cos \left (f x +e \right )\right )^{3} \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \csc \left (f x +e \right )^{3}+192 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\right )+93 \sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\, \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-273 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )+\sqrt {\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )\right )}{192 f \,a^{5}}\)	\(331\)

Input:

int(tan(f*x+e)^2/(a+a*sec(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/192/f/a^5*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)*((1-cos(f*x+e) 
)^2*csc(f*x+e)^2-1)^(1/2)*(8*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(5/2)*(-cot 
(f*x+e)+csc(f*x+e))-10*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(3/2)*(-cot(f*x+e 
)+csc(f*x+e))-12*(1-cos(f*x+e))^3*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2)* 
csc(f*x+e)^3+192*2^(1/2)*arctanh(2^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1) 
^(1/2)*(-cot(f*x+e)+csc(f*x+e)))+93*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)^(1/2 
)*(-cot(f*x+e)+csc(f*x+e))-273*ln(csc(f*x+e)-cot(f*x+e)+((1-cos(f*x+e))^2* 
csc(f*x+e)^2-1)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 674, normalized size of antiderivative = 3.81 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^2/(a+a*sec(f*x+e))^(9/2),x, algorithm="fricas")
 

Output:

[-1/384*(273*sqrt(2)*(cos(f*x + e)^4 + 4*cos(f*x + e)^3 + 6*cos(f*x + e)^2 
 + 4*cos(f*x + e) + 1)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + 
e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + 3*a*cos(f*x + e)^2 + 2*a 
*cos(f*x + e) - a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 384*(cos(f*x + 
 e)^4 + 4*cos(f*x + e)^3 + 6*cos(f*x + e)^2 + 4*cos(f*x + e) + 1)*sqrt(-a) 
*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + 
e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 
4*(157*cos(f*x + e)^3 + 206*cos(f*x + e)^2 + 81*cos(f*x + e))*sqrt((a*cos( 
f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^4 + 4*a^5*f* 
cos(f*x + e)^3 + 6*a^5*f*cos(f*x + e)^2 + 4*a^5*f*cos(f*x + e) + a^5*f), - 
1/192*(273*sqrt(2)*(cos(f*x + e)^4 + 4*cos(f*x + e)^3 + 6*cos(f*x + e)^2 + 
 4*cos(f*x + e) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos( 
f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 384*(cos(f*x + e)^4 + 4*c 
os(f*x + e)^3 + 6*cos(f*x + e)^2 + 4*cos(f*x + e) + 1)*sqrt(a)*arctan(sqrt 
((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 
 2*(157*cos(f*x + e)^3 + 206*cos(f*x + e)^2 + 81*cos(f*x + e))*sqrt((a*cos 
(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^4 + 4*a^5*f 
*cos(f*x + e)^3 + 6*a^5*f*cos(f*x + e)^2 + 4*a^5*f*cos(f*x + e) + a^5*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(f*x+e)**2/(a+a*sec(f*x+e))**(9/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate(tan(f*x+e)^2/(a+a*sec(f*x+e))^(9/2),x, algorithm="maxima")
 

Output:

integrate(tan(f*x + e)^2/(a*sec(f*x + e) + a)^(9/2), x)
 

Giac [A] (verification not implemented)

Time = 0.87 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.62 \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\frac {\sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} {\left (2 \, {\left (\frac {4 \, \sqrt {2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} - \frac {19 \, \sqrt {2}}{a^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + \frac {111 \, \sqrt {2}}{a^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{192 \, f} \] Input:

integrate(tan(f*x+e)^2/(a+a*sec(f*x+e))^(9/2),x, algorithm="giac")
 

Output:

1/192*sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a)*(2*(4*sqrt(2)*tan(1/2*f*x + 1/2* 
e)^2/(a^5*sgn(cos(f*x + e))) - 19*sqrt(2)/(a^5*sgn(cos(f*x + e))))*tan(1/2 
*f*x + 1/2*e)^2 + 111*sqrt(2)/(a^5*sgn(cos(f*x + e))))*tan(1/2*f*x + 1/2*e 
)/f
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \] Input:

int(tan(e + f*x)^2/(a + a/cos(e + f*x))^(9/2),x)
 

Output:

int(tan(e + f*x)^2/(a + a/cos(e + f*x))^(9/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{9/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )+1}\, \tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{5}+5 \sec \left (f x +e \right )^{4}+10 \sec \left (f x +e \right )^{3}+10 \sec \left (f x +e \right )^{2}+5 \sec \left (f x +e \right )+1}d x \right )}{a^{5}} \] Input:

int(tan(f*x+e)^2/(a+a*sec(f*x+e))^(9/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(e + f*x) + 1)*tan(e + f*x)**2)/(sec(e + f*x)**5 + 5 
*sec(e + f*x)**4 + 10*sec(e + f*x)**3 + 10*sec(e + f*x)**2 + 5*sec(e + f*x 
) + 1),x))/a**5