\(\int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx\) [370]

Optimal result

Integrand size = 25, antiderivative size = 189 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=-\frac {31 d^{9/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 f}+\frac {d^{9/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 f}+\frac {27 d^4 \sqrt {d \tan (e+f x)}}{8 a^3 f}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{8 a^3 f (1+\tan (e+f x))}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a+a \tan (e+f x))^2} \] Output:

-31/8*d^(9/2)*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/f+1/4*d^(9/2)*arcta 
nh(1/2*(d^(1/2)+d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e))^(1/2))*2^(1/2)/ 
a^3/f+27/8*d^4*(d*tan(f*x+e))^(1/2)/a^3/f-9/8*d^3*(d*tan(f*x+e))^(3/2)/a^3 
/f/(1+tan(f*x+e))-1/4*d^2*(d*tan(f*x+e))^(5/2)/a/f/(a+a*tan(f*x+e))^2
 

Mathematica [A] (warning: unable to verify)

Time = 7.87 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.83 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\cot (e+f x) \csc ^3(e+f x) (\cos (e+f x)+\sin (e+f x))^3 \left (\frac {7}{2}-\frac {1}{8 (\cos (e+f x)+\sin (e+f x))^2}-\frac {11 \sin (e+f x)}{8 (\cos (e+f x)+\sin (e+f x))}\right ) (d \tan (e+f x))^{9/2}}{f (a+a \tan (e+f x))^3}+\frac {\sec ^3(e+f x) (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{9/2} \left (-\frac {62 \arctan \left (\sqrt {\tan (e+f x)}\right ) \csc (e+f x) \sec ^3(e+f x) (1+\tan (e+f x))}{(1+\cot (e+f x)) \left (1+\tan ^2(e+f x)\right )^2}+\frac {2 \sqrt {2} \cos (2 (e+f x)) \csc (e+f x) \left (-\log \left (-1+\sqrt {2} \sqrt {\tan (e+f x)}-\tan (e+f x)\right )+\log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right ) \sec ^3(e+f x)}{(1+\cot (e+f x)) (1-\tan (e+f x)) \left (1+\tan ^2(e+f x)\right )}\right )}{16 f \tan ^{\frac {9}{2}}(e+f x) (a+a \tan (e+f x))^3} \] Input:

Integrate[(d*Tan[e + f*x])^(9/2)/(a + a*Tan[e + f*x])^3,x]
 

Output:

(Cot[e + f*x]*Csc[e + f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*(7/2 - 1/(8*( 
Cos[e + f*x] + Sin[e + f*x])^2) - (11*Sin[e + f*x])/(8*(Cos[e + f*x] + Sin 
[e + f*x])))*(d*Tan[e + f*x])^(9/2))/(f*(a + a*Tan[e + f*x])^3) + (Sec[e + 
 f*x]^3*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(9/2)*((-62*ArcTa 
n[Sqrt[Tan[e + f*x]]]*Csc[e + f*x]*Sec[e + f*x]^3*(1 + Tan[e + f*x]))/((1 
+ Cot[e + f*x])*(1 + Tan[e + f*x]^2)^2) + (2*Sqrt[2]*Cos[2*(e + f*x)]*Csc[ 
e + f*x]*(-Log[-1 + Sqrt[2]*Sqrt[Tan[e + f*x]] - Tan[e + f*x]] + Log[1 + S 
qrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Sec[e + f*x]^3)/((1 + Cot[e + f 
*x])*(1 - Tan[e + f*x])*(1 + Tan[e + f*x]^2))))/(16*f*Tan[e + f*x]^(9/2)*( 
a + a*Tan[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.58 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.12, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.720, Rules used = {3042, 4048, 27, 3042, 4128, 3042, 4130, 27, 3042, 4137, 27, 3042, 4015, 221, 4117, 27, 73, 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.

\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a \tan (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \tan (e+f x))^{9/2}}{(a \tan (e+f x)+a)^3}dx\)

\(\Big \downarrow \) 4048

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a^2 d^3+9 a^2 \tan ^2(e+f x) d^3-4 a^2 \tan (e+f x) d^3\right )}{2 (\tan (e+f x) a+a)^2}dx}{4 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a^2 d^3+9 a^2 \tan ^2(e+f x) d^3-4 a^2 \tan (e+f x) d^3\right )}{(\tan (e+f x) a+a)^2}dx}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a^2 d^3+9 a^2 \tan (e+f x)^2 d^3-4 a^2 \tan (e+f x) d^3\right )}{(\tan (e+f x) a+a)^2}dx}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (27 a^4 d^4+27 a^4 \tan ^2(e+f x) d^4-8 a^4 \tan (e+f x) d^4\right )}{\tan (e+f x) a+a}dx}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {d \tan (e+f x)} \left (27 a^4 d^4+27 a^4 \tan (e+f x)^2 d^4-8 a^4 \tan (e+f x) d^4\right )}{\tan (e+f x) a+a}dx}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4130

\(\displaystyle \frac {\frac {\frac {2 \int -\frac {27 a^5 d^5+35 a^5 \tan ^2(e+f x) d^5}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}+\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\int \frac {27 a^5 d^5+35 a^5 \tan ^2(e+f x) d^5}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\int \frac {27 a^5 d^5+35 a^5 \tan (e+f x)^2 d^5}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4137

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {31 a^5 d^5 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int -\frac {8 \left (a^6 d^5-a^6 d^5 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {31 a^5 d^5 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {a^6 d^5-a^6 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {31 a^5 d^5 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {a^6 d^5-a^6 d^5 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {31 a^5 d^5 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {8 a^{10} d^{10} \int \frac {1}{\cot (e+f x) \left (d^5 a^6+d^5 \tan (e+f x) a^6\right )^2-2 a^{12} d^{10}}d\frac {d^5 a^6+d^5 \tan (e+f x) a^6}{\sqrt {d \tan (e+f x)}}}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {31 a^5 d^5 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \sqrt {2} a^4 d^{9/2} \text {arctanh}\left (\frac {a^6 d^5 \tan (e+f x)+a^6 d^5}{\sqrt {2} a^6 d^{9/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\frac {31 a^5 d^5 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}-\frac {4 \sqrt {2} a^4 d^{9/2} \text {arctanh}\left (\frac {a^6 d^5 \tan (e+f x)+a^6 d^5}{\sqrt {2} a^6 d^{9/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\frac {31 a^4 d^5 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}-\frac {4 \sqrt {2} a^4 d^{9/2} \text {arctanh}\left (\frac {a^6 d^5 \tan (e+f x)+a^6 d^5}{\sqrt {2} a^6 d^{9/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\frac {62 a^4 d^4 \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}-\frac {4 \sqrt {2} a^4 d^{9/2} \text {arctanh}\left (\frac {a^6 d^5 \tan (e+f x)+a^6 d^5}{\sqrt {2} a^6 d^{9/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {54 a^3 d^4 \sqrt {d \tan (e+f x)}}{f}-\frac {\frac {62 a^4 d^{9/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {4 \sqrt {2} a^4 d^{9/2} \text {arctanh}\left (\frac {a^6 d^5 \tan (e+f x)+a^6 d^5}{\sqrt {2} a^6 d^{9/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{a}}{2 a^3}-\frac {9 d^3 (d \tan (e+f x))^{3/2}}{f (\tan (e+f x)+1)}}{8 a^3}-\frac {d^2 (d \tan (e+f x))^{5/2}}{4 a f (a \tan (e+f x)+a)^2}\)

Input:

Int[(d*Tan[e + f*x])^(9/2)/(a + a*Tan[e + f*x])^3,x]
 

Output:

-1/4*(d^2*(d*Tan[e + f*x])^(5/2))/(a*f*(a + a*Tan[e + f*x])^2) + ((-9*d^3* 
(d*Tan[e + f*x])^(3/2))/(f*(1 + Tan[e + f*x])) + (-(((62*a^4*d^(9/2)*ArcTa 
n[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f - (4*Sqrt[2]*a^4*d^(9/2)*ArcTanh[(a^6*d 
^5 + a^6*d^5*Tan[e + f*x])/(Sqrt[2]*a^6*d^(9/2)*Sqrt[d*Tan[e + f*x]])])/f) 
/a) + (54*a^3*d^4*Sqrt[d*Tan[e + f*x]])/f)/(2*a^3))/(8*a^3)
 

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 221

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

rule 3042

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

rule 4015

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*c*d + b*x^2), x], x, (c 
- d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && 
 EqQ[c^2 - d^2, 0]
 

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], 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] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*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]
 

rule 4128

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

rule 4130

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_. 
) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, 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[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4137

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Sim 
p[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*C)*T 
an[e + f*x], x], x], x] + Simp[(A*b^2 + 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, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(354\) vs. \(2(156)=312\).

Time = 1.82 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.88

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-\frac {13 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {11 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}\right )}{4}+\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{4}\right )}{f \,a^{3}}\)	\(355\)
default	\(\frac {2 d^{4} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {-\frac {13 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}-\frac {11 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {31 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}\right )}{4}+\frac {d \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{4}\right )}{f \,a^{3}}\)	\(355\)

Input:

int((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

2/f/a^3*d^4*((d*tan(f*x+e))^(1/2)-1/4*d*((-13/4*(d*tan(f*x+e))^(3/2)-11/4* 
d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+31/4/d^(1/2)*arctan((d*tan(f*x+ 
e))^(1/2)/d^(1/2)))+1/4*d*(1/8/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^ 
2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/ 
4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4) 
*(d*tan(f*x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2 
)+1))-1/8/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e)) 
^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2) 
*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1 
)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.42 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\left [-\frac {8 \, \sqrt {\frac {1}{2}} {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{d \tan \left (f x + e\right )}\right ) - 31 \, {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, {\left (16 \, d^{4} \tan \left (f x + e\right )^{2} + 45 \, d^{4} \tan \left (f x + e\right ) + 27 \, d^{4}\right )} \sqrt {d \tan \left (f x + e\right )}}{16 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}, \frac {2 \, \sqrt {\frac {1}{2}} {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 31 \, {\left (d^{4} \tan \left (f x + e\right )^{2} + 2 \, d^{4} \tan \left (f x + e\right ) + d^{4}\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) + {\left (16 \, d^{4} \tan \left (f x + e\right )^{2} + 45 \, d^{4} \tan \left (f x + e\right ) + 27 \, d^{4}\right )} \sqrt {d \tan \left (f x + e\right )}}{8 \, {\left (a^{3} f \tan \left (f x + e\right )^{2} + 2 \, a^{3} f \tan \left (f x + e\right ) + a^{3} f\right )}}\right ] \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x, algorithm="fricas")
 

Output:

[-1/16*(8*sqrt(1/2)*(d^4*tan(f*x + e)^2 + 2*d^4*tan(f*x + e) + d^4)*sqrt(- 
d)*arctan(sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*ta 
n(f*x + e))) - 31*(d^4*tan(f*x + e)^2 + 2*d^4*tan(f*x + e) + d^4)*sqrt(-d) 
*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) 
+ 1)) - 2*(16*d^4*tan(f*x + e)^2 + 45*d^4*tan(f*x + e) + 27*d^4)*sqrt(d*ta 
n(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + e) + a^3*f), 1/8*(2 
*sqrt(1/2)*(d^4*tan(f*x + e)^2 + 2*d^4*tan(f*x + e) + d^4)*sqrt(d)*log((d* 
tan(f*x + e)^2 + 4*sqrt(1/2)*sqrt(d*tan(f*x + e))*sqrt(d)*(tan(f*x + e) + 
1) + 4*d*tan(f*x + e) + d)/(tan(f*x + e)^2 + 1)) + 31*(d^4*tan(f*x + e)^2 
+ 2*d^4*tan(f*x + e) + d^4)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)*t 
an(f*x + e))) + (16*d^4*tan(f*x + e)^2 + 45*d^4*tan(f*x + e) + 27*d^4)*sqr 
t(d*tan(f*x + e)))/(a^3*f*tan(f*x + e)^2 + 2*a^3*f*tan(f*x + e) + a^3*f)]
 

Sympy [F]

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

integrate((d*tan(f*x+e))**(9/2)/(a+a*tan(f*x+e))**3,x)
 

Output:

Integral((d*tan(e + f*x))**(9/2)/(tan(e + f*x)**3 + 3*tan(e + f*x)**2 + 3* 
tan(e + f*x) + 1), x)/a**3
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.06 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\frac {d^{6} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a^{3}} - \frac {31 \, d^{\frac {11}{2}} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}} + \frac {16 \, \sqrt {d \tan \left (f x + e\right )} d^{5}}{a^{3}} + \frac {13 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d^{6} + 11 \, \sqrt {d \tan \left (f x + e\right )} d^{7}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}}{8 \, d f} \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x, algorithm="maxima")
 

Output:

1/8*(d^6*(sqrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d 
) + d)/sqrt(d) - sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e)) 
*sqrt(d) + d)/sqrt(d))/a^3 - 31*d^(11/2)*arctan(sqrt(d*tan(f*x + e))/sqrt( 
d))/a^3 + 16*sqrt(d*tan(f*x + e))*d^5/a^3 + (13*(d*tan(f*x + e))^(3/2)*d^6 
 + 11*sqrt(d*tan(f*x + e))*d^7)/(a^3*d^2*tan(f*x + e)^2 + 2*a^3*d^2*tan(f* 
x + e) + a^3*d^2))/(d*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,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%%%{%%%{1,[6,21]%%%}+%%%{10,[6,19]%%%}+%%%{45,[6,17]%%%}+%%%{1 
20,[6,15]
 

Mupad [B] (verification not implemented)

Time = 1.96 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.93 \[ \int \frac {(d \tan (e+f x))^{9/2}}{(a+a \tan (e+f x))^3} \, dx=\frac {\frac {11\,d^6\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {13\,d^5\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {2\,d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a^3\,f}-\frac {31\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,f}-\frac {\sqrt {2}\,d^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,d^{49/2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,969{}\mathrm {i}}{32\,\left (\frac {969\,d^{25}\,\mathrm {tan}\left (e+f\,x\right )}{32}+\frac {969\,d^{25}}{32}\right )}\right )\,1{}\mathrm {i}}{4\,a^3\,f} \] Input:

int((d*tan(e + f*x))^(9/2)/(a + a*tan(e + f*x))^3,x)
 

Output:

((11*d^6*(d*tan(e + f*x))^(1/2))/8 + (13*d^5*(d*tan(e + f*x))^(3/2))/8)/(a 
^3*d^2*f + a^3*d^2*f*tan(e + f*x)^2 + 2*a^3*d^2*f*tan(e + f*x)) + (2*d^4*( 
d*tan(e + f*x))^(1/2))/(a^3*f) - (31*d^(9/2)*atan((d*tan(e + f*x))^(1/2)/d 
^(1/2)))/(8*a^3*f) - (2^(1/2)*d^(9/2)*atan((2^(1/2)*d^(49/2)*(d*tan(e + f* 
x))^(1/2)*969i)/(32*((969*d^25*tan(e + f*x))/32 + (969*d^25)/32)))*1i)/(4* 
a^3*f)
 

Reduce [F]

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

int((d*tan(f*x+e))^(9/2)/(a+a*tan(f*x+e))^3,x)
 

Output:

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