Integrand size = 25, antiderivative size = 168 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}+\frac {\sqrt {2} a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{3/2} f}+\frac {\sqrt {2} a^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{d^{3/2} f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}} \] Output:
-2^(1/2)*a^2*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(3/2)/f+2^(1 /2)*a^2*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(3/2)/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^(3 /2)/f-2*a^2/d/f/(d*tan(f*x+e))^(1/2)
Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.10 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {a^2 \left (4+2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \sqrt {\tan (e+f x)}+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \sqrt {\tan (e+f x)}-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \sqrt {\tan (e+f x)}\right )}{2 d f \sqrt {d \tan (e+f x)}} \] Input:
Integrate[(a + a*Tan[e + f*x])^2/(d*Tan[e + f*x])^(3/2),x]
Output:
-1/2*(a^2*(4 + 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[Tan[e + f*x]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Sqrt[Tan[e + f *x]] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sqrt[Tan [e + f*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Sq rt[Tan[e + f*x]]))/(d*f*Sqrt[d*Tan[e + f*x]])
Time = 0.53 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.23, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4025, 27, 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.
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^2}{(d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^2}{(d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4025 |
| \(\displaystyle \frac {\int \frac {2 a^2 d}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a^2 \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{d}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a^2 \int \frac {1}{\sqrt {d \tan (e+f x)}}dx}{d}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3957 |
| \(\displaystyle \frac {2 a^2 \int \frac {1}{\sqrt {d \tan (e+f x)} \left (\tan ^2(e+f x) d^2+d^2\right )}d(d \tan (e+f x))}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {4 a^2 \int \frac {1}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d^2 \tan ^2(e+f x)+d}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d^2 \tan ^2(e+f x)-1}d\left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\int \frac {d-d^2 \tan ^2(e+f x)}{d^4 \tan ^4(e+f x)+d^2}d\sqrt {d \tan (e+f x)}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {4 a^2 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)-\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{d^2 \tan ^2(e+f x)+\sqrt {2} d^{3/2} \tan (e+f x)+d}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}}{2 d}+\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {4 a^2 \left (\frac {\frac {\arctan \left (\sqrt {2} \sqrt {d} \tan (e+f x)+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {d} \tan (e+f x)\right )}{\sqrt {2} \sqrt {d}}}{2 d}+\frac {\frac {\log \left (\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (-\sqrt {2} d^{3/2} \tan (e+f x)+d^2 \tan ^2(e+f x)+d\right )}{2 \sqrt {2} \sqrt {d}}}{2 d}\right )}{f}-\frac {2 a^2}{d f \sqrt {d \tan (e+f x)}}\) |
Input:
Int[(a + a*Tan[e + f*x])^2/(d*Tan[e + f*x])^(3/2),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*d) + (-1/2*L og[d - Sqrt[2]*d^(3/2)*Tan[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*a^2)/(d*f*Sqrt[d*Tan[e + f*x]])
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[((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]]))
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[((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[(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*c^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]
Time = 1.58 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {2 a^{2} \left (-\frac {1}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}\right )}{f d}\) | \(159\) |
| default | \(\frac {2 a^{2} \left (-\frac {1}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}\right )}{f d}\) | \(159\) |
| parts | \(\frac {2 a^{2} d \left (-\frac {1}{d^{2} \sqrt {d \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}+\frac {a^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f d \left (d^{2}\right )^{\frac {1}{4}}}+\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{2 f \,d^{2}}\) | \(441\) |
Input:
int((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/f*a^2/d*(-1/(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)))
Time = 0.09 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=\frac {2 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + 1\right ) \tan \left (f x + e\right ) + 2 \, \sqrt {2} a^{2} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} - 1\right ) \tan \left (f x + e\right ) + \sqrt {2} a^{2} \sqrt {d} \log \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right ) \tan \left (f x + e\right ) - \sqrt {2} a^{2} \sqrt {d} \log \left (-\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}} + \tan \left (f x + e\right ) + 1\right ) \tan \left (f x + e\right ) - 4 \, \sqrt {d \tan \left (f x + e\right )} a^{2}}{2 \, d^{2} f \tan \left (f x + e\right )} \] Input:
integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(3/2),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 )*tan(f*x + e) + 2*sqrt(2)*a^2*sqrt(d)*arctan(sqrt(2)*sqrt(d*tan(f*x + e)) /sqrt(d) - 1)*tan(f*x + e) + sqrt(2)*a^2*sqrt(d)*log(sqrt(2)*sqrt(d*tan(f* x + e))/sqrt(d) + tan(f*x + e) + 1)*tan(f*x + e) - sqrt(2)*a^2*sqrt(d)*log (-sqrt(2)*sqrt(d*tan(f*x + e))/sqrt(d) + tan(f*x + e) + 1)*tan(f*x + e) - 4*sqrt(d*tan(f*x + e))*a^2)/(d^2*f*tan(f*x + e))
\[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=a^{2} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {2 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \] Input:
integrate((a+a*tan(f*x+e))**2/(d*tan(f*x+e))**(3/2),x)
Output:
a**2*(Integral((d*tan(e + f*x))**(-3/2), x) + Integral(2*tan(e + f*x)/(d*t an(e + f*x))**(3/2), x) + Integral(tan(e + f*x)**2/(d*tan(e + f*x))**(3/2) , x))
Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=\frac {a^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} - \frac {4 \, a^{2}}{\sqrt {d \tan \left (f x + e\right )}}}{2 \, d f} \] Input:
integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
1/2*(a^2*(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*a^2/sqrt(d*t an(f*x + e)))/(d*f)
Timed out. \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(3/2),x, algorithm="giac")
Output:
Timed out
Time = 1.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.51 \[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=-\frac {2\,a^2}{d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}-\frac {{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{d^{3/2}\,f}-\frac {{\left (-1\right )}^{1/4}\,a^2\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,2{}\mathrm {i}}{d^{3/2}\,f} \] Input:
int((a + a*tan(e + f*x))^2/(d*tan(e + f*x))^(3/2),x)
Output:
- (2*a^2)/(d*f*(d*tan(e + f*x))^(1/2)) - ((-1)^(1/4)*a^2*atan(((-1)^(1/4)* (d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/(d^(3/2)*f) - ((-1)^(1/4)*a^2*atanh(( (-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*2i)/(d^(3/2)*f)
\[ \int \frac {(a+a \tan (e+f x))^2}{(d \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {d}\, a^{2} \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right )+\int \sqrt {\tan \left (f x +e \right )}d x \right )}{d^{2}} \] Input:
int((a+a*tan(f*x+e))^2/(d*tan(f*x+e))^(3/2),x)
Output:
(sqrt(d)*a**2*(int(sqrt(tan(e + f*x))/tan(e + f*x)**2,x) + 2*int(sqrt(tan( e + f*x))/tan(e + f*x),x) + int(sqrt(tan(e + f*x)),x)))/d**2