Integrand size = 25, antiderivative size = 222 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=-\frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}+\frac {\sqrt {2} a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {d} f}+\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} \sqrt {d} f}-\frac {a^2 \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} \sqrt {d} f}+\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f} \]
1/2*a^2*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/f*2^(1 /2)/d^(1/2)-1/2*a^2*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f* x+e))/f*2^(1/2)/d^(1/2)-a^2*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2)) *2^(1/2)/f/d^(1/2)+a^2*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1 /2)/f/d^(1/2)+2*a^2*(d*tan(f*x+e))^(1/2)/d/f
Time = 6.09 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.14 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2 \cos (e+f x) \sin (e+f x) (a+a \tan (e+f x))^2}{f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)}}+\frac {2 \arctan \left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^2(e+f x) \sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)} (a+a \tan (e+f x))^2}{f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)}}-\frac {2 \text {arctanh}\left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^2(e+f x) \sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)} (a+a \tan (e+f x))^2}{f (\cos (e+f x)+\sin (e+f x))^2 \sqrt {d \tan (e+f x)}} \]
(2*Cos[e + f*x]*Sin[e + f*x]*(a + a*Tan[e + f*x])^2)/(f*(Cos[e + f*x] + Si n[e + f*x])^2*Sqrt[d*Tan[e + f*x]]) + (2*ArcTan[(-Tan[e + f*x])^(1/4)*Tan[ e + f*x]^(1/4)]*Cos[e + f*x]^2*(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)*(a + a*Tan[e + f*x])^2)/(f*(Cos[e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f* x]]) - (2*ArcTanh[(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)]*Cos[e + f*x]^2 *(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)*(a + a*Tan[e + f*x])^2)/(f*(Cos[ e + f*x] + Sin[e + f*x])^2*Sqrt[d*Tan[e + f*x]])
Time = 0.55 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.90, 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 |
\(\displaystyle \frac {4 a^2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}+\frac {2 a^2 \sqrt {d \tan (e+f x)}}{d f}\) |
(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)
3.4.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]]))
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]
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]
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]
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]
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]
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]
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])
Time = 0.84 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.70
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 | \(\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f d}+\frac {2 a^{2} \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f d}+\frac {a^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 f \left (d^{2}\right )^{\frac {1}{4}}}\) | \(432\) |
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)))
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.07 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {\left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {1}{4}} d f \log \left (\sqrt {d \tan \left (f x + e\right )} a^{6} + \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {3}{4}} d^{2} f^{3}\right ) - i \, \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {1}{4}} d f \log \left (\sqrt {d \tan \left (f x + e\right )} a^{6} + i \, \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {3}{4}} d^{2} f^{3}\right ) + i \, \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {1}{4}} d f \log \left (\sqrt {d \tan \left (f x + e\right )} a^{6} - i \, \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {3}{4}} d^{2} f^{3}\right ) - \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {1}{4}} d f \log \left (\sqrt {d \tan \left (f x + e\right )} a^{6} - \left (-\frac {a^{8}}{d^{2} f^{4}}\right )^{\frac {3}{4}} d^{2} f^{3}\right ) + 2 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{d f} \]
((-a^8/(d^2*f^4))^(1/4)*d*f*log(sqrt(d*tan(f*x + e))*a^6 + (-a^8/(d^2*f^4) )^(3/4)*d^2*f^3) - I*(-a^8/(d^2*f^4))^(1/4)*d*f*log(sqrt(d*tan(f*x + e))*a ^6 + I*(-a^8/(d^2*f^4))^(3/4)*d^2*f^3) + I*(-a^8/(d^2*f^4))^(1/4)*d*f*log( sqrt(d*tan(f*x + e))*a^6 - I*(-a^8/(d^2*f^4))^(3/4)*d^2*f^3) - (-a^8/(d^2* f^4))^(1/4)*d*f*log(sqrt(d*tan(f*x + e))*a^6 - (-a^8/(d^2*f^4))^(3/4)*d^2* f^3) + 2*sqrt(d*tan(f*x + e))*a^2)/(d*f)
\[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \frac {2 \tan {\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt {d \tan {\left (e + f x \right )}}}\, dx\right ) \]
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))
Time = 0.38 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.79 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {a^{2} d {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 4 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{2 \, d f} \]
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)
Timed out. \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\text {Timed out} \]
Time = 5.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.39 \[ \int \frac {(a+a \tan (e+f x))^2}{\sqrt {d \tan (e+f x)}} \, dx=\frac {2\,a^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{d\,f}+\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{\sqrt {d}\,f}-\frac {2\,{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{\sqrt {d}\,f} \]