\(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx\) [375]

Optimal result

Integrand size = 25, antiderivative size = 165 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\frac {11 \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{8 a^3 \sqrt {d} f}+\frac {\arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{2 \sqrt {2} a^3 \sqrt {d} f}+\frac {7 \sqrt {d \tan (e+f x)}}{8 a^3 d f (1+\tan (e+f x))}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a+a \tan (e+f x))^2} \] Output:

11/8*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a^3/d^(1/2)/f+1/4*arctan(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/d^(1/2 
)/f+7/8*(d*tan(f*x+e))^(1/2)/a^3/d/f/(1+tan(f*x+e))+1/4*(d*tan(f*x+e))^(1/ 
2)/a/d/f/(a+a*tan(f*x+e))^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.75 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\frac {11 \sqrt {d} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+(2-2 i) \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+(2+2 i) \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )+\frac {2 \sqrt {d \tan (e+f x)}}{(1+\tan (e+f x))^2}+\frac {7 \sqrt {d \tan (e+f x)}}{1+\tan (e+f x)}}{8 a^3 d f} \] Input:

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

Output:

(11*Sqrt[d]*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]] + (2 - 2*I)*(-1)^(1/4)*Sq 
rt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + (2 + 2*I)*(-1)^( 
1/4)*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]] + (2*Sqrt[ 
d*Tan[e + f*x]])/(1 + Tan[e + f*x])^2 + (7*Sqrt[d*Tan[e + f*x]])/(1 + Tan[ 
e + f*x]))/(8*a^3*d*f)
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.13, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4052, 27, 3042, 4132, 3042, 4136, 27, 3042, 4015, 218, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

\(\displaystyle \frac {\int \frac {3 d \tan ^2(e+f x) a^2+7 d a^2-4 d \tan (e+f x) a^2}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)^2}dx}{4 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 d \tan ^2(e+f x) a^2+7 d a^2-4 d \tan (e+f x) a^2}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {3 d \tan (e+f x)^2 a^2+7 d a^2-4 d \tan (e+f x) a^2}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)^2}dx}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4132

\(\displaystyle \frac {\frac {\int \frac {7 d^2 a^4+7 d^2 \tan ^2(e+f x) a^4-8 d^2 \tan (e+f x) a^4}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {7 d^2 a^4+7 d^2 \tan (e+f x)^2 a^4-8 d^2 \tan (e+f x) a^4}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {\frac {11 a^4 d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {\int -\frac {8 \left (d^2 a^5+d^2 \tan (e+f x) a^5\right )}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {11 a^4 d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {d^2 a^5+d^2 \tan (e+f x) a^5}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {11 a^4 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {4 \int \frac {d^2 a^5+d^2 \tan (e+f x) a^5}{\sqrt {d \tan (e+f x)}}dx}{a^2}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4015

\(\displaystyle \frac {\frac {11 a^4 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {8 a^8 d^4 \int \frac {1}{2 d^4 a^{10}+\cot (e+f x) \left (a^5 d^2-a^5 d^2 \tan (e+f x)\right )^2}d\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {11 a^4 d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx+\frac {4 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {2} a^5 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {\frac {\frac {11 a^4 d^2 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {2} a^5 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {11 a^3 d^2 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{f}+\frac {4 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {2} a^5 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {22 a^3 d \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{f}+\frac {4 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {2} a^5 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {22 a^3 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}+\frac {4 \sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^5 d^2-a^5 d^2 \tan (e+f x)}{\sqrt {2} a^5 d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}}{2 a^3 d}+\frac {7 \sqrt {d \tan (e+f x)}}{f (\tan (e+f x)+1)}}{8 a^3 d}+\frac {\sqrt {d \tan (e+f x)}}{4 a d f (a \tan (e+f x)+a)^2}\)

Input:

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

Output:

Sqrt[d*Tan[e + f*x]]/(4*a*d*f*(a + a*Tan[e + f*x])^2) + (((22*a^3*d^(3/2)* 
ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/f + (4*Sqrt[2]*a^3*d^(3/2)*ArcTan[(a 
^5*d^2 - a^5*d^2*Tan[e + f*x])/(Sqrt[2]*a^5*d^(3/2)*Sqrt[d*Tan[e + f*x]])] 
)/f)/(2*a^3*d) + (7*Sqrt[d*Tan[e + f*x]])/(f*(1 + Tan[e + f*x])))/(8*a^3*d 
)
 

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 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 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 4132

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*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(348\) vs. \(2(136)=272\).

Time = 1.56 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.12

method	result	size
derivativedivides	\(\frac {2 d^{4} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{4}}+\frac {\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{4}}\right )}{f \,a^{3}}\)	\(349\)
default	\(\frac {2 d^{4} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{4 d^{4}}+\frac {\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{4}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{4}}{\left (d \tan \left (f x +e \right )+d \right )^{2}}+\frac {11 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{4 \sqrt {d}}}{4 d^{4}}\right )}{f \,a^{3}}\)	\(349\)

Input:

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

Output:

2/f/a^3*d^4*(1/4/d^4*(-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)))+1/4/d^4*((7/4*(d*tan 
(f*x+e))^(3/2)+9/4*d*(d*tan(f*x+e))^(1/2))/(d*tan(f*x+e)+d)^2+11/4/d^(1/2) 
*arctan((d*tan(f*x+e))^(1/2)/d^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 387, normalized size of antiderivative = 2.35 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\left [-\frac {2 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 11 \, {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} {\left (7 \, \tan \left (f x + e\right ) + 9\right )}}{16 \, {\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}, -\frac {2 \, \sqrt {2} {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) + 11 \, {\left (\tan \left (f x + e\right )^{2} + 2 \, \tan \left (f x + e\right ) + 1\right )} \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) - \sqrt {d \tan \left (f x + e\right )} {\left (7 \, \tan \left (f x + e\right ) + 9\right )}}{8 \, {\left (a^{3} d f \tan \left (f x + e\right )^{2} + 2 \, a^{3} d f \tan \left (f x + e\right ) + a^{3} d f\right )}}\right ] \] Input:

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

Output:

[-1/16*(2*sqrt(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(-d)*log((d*ta 
n(f*x + e)^2 + 2*sqrt(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)) + 11*(tan(f*x + e)^2 + 2*tan 
(f*x + e) + 1)*sqrt(-d)*log((d*tan(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt( 
-d) - d)/(tan(f*x + e) + 1)) - 2*sqrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9) 
)/(a^3*d*f*tan(f*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f), -1/8*(2*sqr 
t(2)*(tan(f*x + e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(1/2*sqrt(2)*sqrt 
(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e))) + 11*(tan(f*x 
+ e)^2 + 2*tan(f*x + e) + 1)*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqrt(d)* 
tan(f*x + e))) - sqrt(d*tan(f*x + e))*(7*tan(f*x + e) + 9))/(a^3*d*f*tan(f 
*x + e)^2 + 2*a^3*d*f*tan(f*x + e) + a^3*d*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\frac {\frac {7 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d + 9 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{a^{3} d^{2} \tan \left (f x + e\right )^{2} + 2 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}} - \frac {2 \, d {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{a^{3}} + \frac {11 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a^{3}}}{8 \, d f} \] Input:

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

Output:

1/8*((7*(d*tan(f*x + e))^(3/2)*d + 9*sqrt(d*tan(f*x + e))*d^2)/(a^3*d^2*ta 
n(f*x + e)^2 + 2*a^3*d^2*tan(f*x + e) + a^3*d^2) - 2*d*(sqrt(2)*arctan(1/2 
*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + sqr 
t(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d 
))/sqrt(d))/a^3 + 11*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/a^3)/(d* 
f)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 1.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^3} \, dx=\frac {\frac {9\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8}+\frac {7\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8}}{f\,a^3\,d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,f\,a^3\,d^2\,\mathrm {tan}\left (e+f\,x\right )+f\,a^3\,d^2}+\frac {11\,\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{8\,a^3\,\sqrt {d}\,f}-\frac {\sqrt {2}\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{8\,a^3\,\sqrt {d}\,f} \] Input:

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

Output:

((9*d*(d*tan(e + f*x))^(1/2))/8 + (7*(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)) + (11*atan((d*tan( 
e + f*x))^(1/2)/d^(1/2)))/(8*a^3*d^(1/2)*f) - (2^(1/2)*(2*atan((2^(1/2)*(d 
*tan(e + f*x))^(1/2))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2 
))/(2*d^(1/2)) + (2^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(8*a^3*d^ 
(1/2)*f)
 

Reduce [F]

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

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

Output:

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