\(\int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx\) [346]

Optimal result

Integrand size = 25, antiderivative size = 169 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}+\frac {\sqrt {2} a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}-\frac {\sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {d} f}+\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 5.50 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.40 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2 a^2 \left ((\cos (e+f x)+\sin (e+f x))^2 \tan (e+f x) \left (-\tan ^2(e+f x)\right )^{3/4}+\text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \left (\tan ^2(e+f x)+2 \cos ^2(e+f x) (-\tan (e+f x))^{5/4} \sqrt [4]{\tan (e+f x)} \left (-\tan ^2(e+f x)\right )^{3/4}\right )+\arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \left (-\sin ^2(e+f x)+\cos ^2(e+f x) \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {5}{4}}(e+f x) \left (-\tan ^2(e+f x)\right )^{3/4} (2+\tan (e+f x))\right )\right )}{f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)} \left (-\tan ^2(e+f x)\right )^{3/4}} \] Input:

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

Output:

(2*a^2*((Cos[e + f*x] + Sin[e + f*x])^2*Tan[e + f*x]*(-Tan[e + f*x]^2)^(3/ 
4) + ArcTanh[(-Tan[e + f*x]^2)^(1/4)]*(Tan[e + f*x]^2 + 2*Cos[e + f*x]^2*( 
-Tan[e + f*x])^(5/4)*Tan[e + f*x]^(1/4)*(-Tan[e + f*x]^2)^(3/4)) + ArcTan[ 
(-Tan[e + f*x]^2)^(1/4)]*(-Sin[e + f*x]^2 + Cos[e + f*x]^2*(-Tan[e + f*x]) 
^(1/4)*Tan[e + f*x]^(5/4)*(-Tan[e + f*x]^2)^(3/4)*(2 + Tan[e + f*x]))))/(f 
*(Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f*x]]*(-Tan[e + f*x]^2)^(3 
/4))
 

Rubi [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.18, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4026, 27, 2030, 3042, 3957, 266, 826, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 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 = 1.68 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\sqrt {d \tan \left (f x +e \right )}+\frac {d \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(155\)
default	\(\frac {2 a^{2} \left (\sqrt {d \tan \left (f x +e \right )}+\frac {d \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f d}\)	\(155\)
parts	\(\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f d}+\frac {2 a^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f d}+\frac {a^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}\)	\(432\)

Input:

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

Output:

2/f*a^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 = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + 1\right ) + 2 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} - 1\right ) - \sqrt {2} a^{2} \sqrt {d} \log \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right ) + \sqrt {2} a^{2} \sqrt {d} \log \left (-\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right ) + 4 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{2 \, d f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {a^{2} d {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 4 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{2 \, d f} \] Input:

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

Output:

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

Giac [F(-2)]

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

integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(1/2),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,[3,9]%%%}+%%%{4,[3,7]%%%}+%%%{6,[3,5]%%%}+%%%{4,[3,3 
]%%%}+%%%
 

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.51 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2\,a^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{d\,f}+\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{\sqrt {d}\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{\sqrt {d}\,f} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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