\(\int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx\) [345]

Optimal result

Integrand size = 25, antiderivative size = 191 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=\frac {\sqrt {2} a^2 \sqrt {d} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 \sqrt {d} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {\sqrt {2} a^2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{f}+\frac {4 a^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a^2 (d \tan (e+f x))^{3/2}}{3 d f} \] Output:

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.92 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=\frac {a^2 \sqrt {d \tan (e+f x)} \left (6 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-6 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+3 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-3 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )+24 \sqrt {\tan (e+f x)}+4 \tan ^{\frac {3}{2}}(e+f x)\right )}{6 f \sqrt {\tan (e+f x)}} \] Input:

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

Output:

(a^2*Sqrt[d*Tan[e + f*x]]*(6*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] 
] - 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]] + 3*Sqrt[2]*Log[1 - S 
qrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - 3*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt 
[Tan[e + f*x]] + Tan[e + f*x]] + 24*Sqrt[Tan[e + f*x]] + 4*Tan[e + f*x]^(3 
/2)))/(6*f*Sqrt[Tan[e + f*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.68 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.23, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4026, 27, 2030, 3042, 3954, 3042, 3957, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \tan (e+f x)+a)^2 \sqrt {d \tan (e+f x)}dx\)

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*a^2*(d*Tan[e + f*x])^(3/2))/(3*d*f) + (2*a^2*((-2*d^3*((-(ArcTan[1 - Sq 
rt[2]*Sqrt[d]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + Sqrt[2]*Sqrt[d 
]*Tan[e + f*x]]/(Sqrt[2]*Sqrt[d]))/(2*d) + (-1/2*Log[d - Sqrt[2]*d^(3/2)*T 
an[e + f*x] + d^2*Tan[e + f*x]^2]/(Sqrt[2]*Sqrt[d]) + Log[d + Sqrt[2]*d^(3 
/2)*Tan[e + f*x] + d^2*Tan[e + f*x]^2]/(2*Sqrt[2]*Sqrt[d]))/(2*d)))/f + (2 
*d*Sqrt[d*Tan[e + f*x]])/f))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4026

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

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{f d}\)	\(170\)
default	\(\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d \sqrt {d \tan \left (f x +e \right )}-\frac {d \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{f d}\)	\(170\)
parts	\(\frac {a^{2} d \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 )}{4 f \left (d^{2}\right )^{\frac {1}{4}}}+\frac {2 a^{2} \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \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 )}{f d}+\frac {2 a^{2} \left (2 \sqrt {d \tan \left (f x +e \right )}-\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 )}{4}\right )}{f}\)	\(450\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.97 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=-\frac {6 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d}{d}\right ) + 6 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} - d}{d}\right ) + 3 \, \sqrt {2} a^{2} \sqrt {d} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - 3 \, \sqrt {2} a^{2} \sqrt {d} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - 4 \, {\left (a^{2} \tan \left (f x + e\right ) + 6 \, a^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{6 \, f} \] Input:

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

Output:

-1/6*(6*sqrt(2)*a^2*sqrt(d)*arctan((sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + 
 d)/d) + 6*sqrt(2)*a^2*sqrt(d)*arctan((sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d 
) - d)/d) + 3*sqrt(2)*a^2*sqrt(d)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan( 
f*x + e))*sqrt(d) + d) - 3*sqrt(2)*a^2*sqrt(d)*log(d*tan(f*x + e) - sqrt(2 
)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - 4*(a^2*tan(f*x + e) + 6*a^2)*sqrt(d* 
tan(f*x + e)))/f
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=\frac {4 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a^{2} + 24 \, \sqrt {d \tan \left (f x + e\right )} a^{2} d - 3 \, {\left (2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {3}{2}} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )\right )} a^{2}}{6 \, d f} \] Input:

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

Output:

1/6*(4*(d*tan(f*x + e))^(3/2)*a^2 + 24*sqrt(d*tan(f*x + e))*a^2*d - 3*(2*s 
qrt(2)*d^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f*x + e) 
))/sqrt(d)) + 2*sqrt(2)*d^(3/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*s 
qrt(d*tan(f*x + e)))/sqrt(d)) + sqrt(2)*d^(3/2)*log(d*tan(f*x + e) + sqrt( 
2)*sqrt(d*tan(f*x + e))*sqrt(d) + d) - sqrt(2)*d^(3/2)*log(d*tan(f*x + e) 
- sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d))*a^2)/(d*f)
 

Giac [F(-2)]

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

integrate((d*tan(f*x+e))^(1/2)*(a+a*tan(f*x+e))^2,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,[4,14]%%%}+%%%{6,[4,12]%%%}+%%%{15,[4,10]%%%}+%%%{20 
,[4,8]%%%
 

Mupad [B] (verification not implemented)

Time = 1.54 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.54 \[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=\frac {4\,a^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {2\,a^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,d\,f}+\frac {{\left (-1\right )}^{1/4}\,a^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{f}+\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\sqrt {d}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )}{f} \] Input:

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

Output:

(4*a^2*(d*tan(e + f*x))^(1/2))/f + (2*a^2*(d*tan(e + f*x))^(3/2))/(3*d*f) 
+ ((-1)^(1/4)*a^2*d^(1/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2) 
)*2i)/f + (2*(-1)^(1/4)*a^2*d^(1/2)*atan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2 
)*1i)/d^(1/2)))/f
 

Reduce [F]

\[ \int \sqrt {d \tan (e+f x)} (a+a \tan (e+f x))^2 \, dx=\frac {2 \sqrt {d}\, a^{2} \left (\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )+6 \sqrt {\tan \left (f x +e \right )}-3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right ) f \right )}{3 f} \] Input:

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

Output:

(2*sqrt(d)*a**2*(sqrt(tan(e + f*x))*tan(e + f*x) + 6*sqrt(tan(e + f*x)) - 
3*int(sqrt(tan(e + f*x))/tan(e + f*x),x)*f))/(3*f)