Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\frac {\arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a d^{5/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a d^{5/2} f}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}+\frac {2}{a d^2 f \sqrt {d \tan (e+f x)}} \] Output:
arctan((d*tan(f*x+e))^(1/2)/d^(1/2))/a/d^(5/2)/f-1/2*arctanh(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/d^(5/2)/f-2/3/ a/d/f/(d*tan(f*x+e))^(3/2)+2/a/d^2/f/(d*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.01 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\frac {6 d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {9}{4},-\tan ^2(e+f x)\right ) \tan ^3(e+f x)+5 \left (12 d^2-4 d^2 \cot (e+f x)-6 d^2 \tan (e+f x)+6 d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) \sqrt {d \tan (e+f x)}-3 \left (-d^2\right )^{3/4} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) \sqrt {d \tan (e+f x)}+3 \left (-d^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt [4]{-d^2}}\right ) \sqrt {d \tan (e+f x)}\right )}{30 a d^4 f \sqrt {d \tan (e+f x)}} \] Input:
Integrate[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]
Output:
(6*d^2*Hypergeometric2F1[1, 5/4, 9/4, -Tan[e + f*x]^2]*Tan[e + f*x]^3 + 5* (12*d^2 - 4*d^2*Cot[e + f*x] - 6*d^2*Tan[e + f*x] + 6*d^(3/2)*ArcTan[Sqrt[ d*Tan[e + f*x]]/Sqrt[d]]*Sqrt[d*Tan[e + f*x]] - 3*(-d^2)^(3/4)*ArcTan[Sqrt [d*Tan[e + f*x]]/(-d^2)^(1/4)]*Sqrt[d*Tan[e + f*x]] + 3*(-d^2)^(3/4)*ArcTa nh[Sqrt[d*Tan[e + f*x]]/(-d^2)^(1/4)]*Sqrt[d*Tan[e + f*x]]))/(30*a*d^4*f*S qrt[d*Tan[e + f*x]])
Time = 1.07 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.13, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4052, 27, 3042, 4132, 27, 2030, 3042, 4056, 25, 3042, 4015, 221, 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.
| \(\displaystyle \int \frac {1}{(a \tan (e+f x)+a) (d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \tan (e+f x)+a) (d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4052 |
| \(\displaystyle -\frac {2 \int \frac {3 \left (a \tan ^2(e+f x) d^2+a d^2+a \tan (e+f x) d^2\right )}{2 (d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{3 a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {a \tan ^2(e+f x) d^2+a d^2+a \tan (e+f x) d^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {a \tan (e+f x)^2 d^2+a d^2+a \tan (e+f x) d^2}{(d \tan (e+f x))^{3/2} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle -\frac {-\frac {2 \int \frac {a^2 d^4 \tan ^2(e+f x)}{2 \sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx}{a d^3}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-a d \int \frac {\tan ^2(e+f x)}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle -\frac {-\frac {a \int \frac {(d \tan (e+f x))^{3/2}}{\tan (e+f x) a+a}dx}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {a \int \frac {(d \tan (e+f x))^{3/2}}{\tan (e+f x) a+a}dx}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4056 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {\int -\frac {a d^2-a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}+\frac {1}{2} d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {1}{2} d^2 \int \frac {\tan ^2(e+f x)+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {\int \frac {a d^2-a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {1}{2} d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {\int \frac {a d^2-a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4015 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {d^4 \int \frac {1}{\cot (e+f x) \left (a d^2+a \tan (e+f x) d^2\right )^2-2 a^2 d^4}d\frac {a d^2+a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}}{f}+\frac {1}{2} d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {1}{2} d^2 \int \frac {\tan (e+f x)^2+1}{\sqrt {d \tan (e+f x)} (\tan (e+f x) a+a)}dx-\frac {d^{3/2} \text {arctanh}\left (\frac {a d^2 \tan (e+f x)+a d^2}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {d^2 \int \frac {1}{a \sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 f}-\frac {d^{3/2} \text {arctanh}\left (\frac {a d^2 \tan (e+f x)+a d^2}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {d^2 \int \frac {1}{\sqrt {d \tan (e+f x)} (\tan (e+f x)+1)}d\tan (e+f x)}{2 a f}-\frac {d^{3/2} \text {arctanh}\left (\frac {a d^2 \tan (e+f x)+a d^2}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {d \int \frac {1}{\tan (e+f x)+1}d\sqrt {d \tan (e+f x)}}{a f}-\frac {d^{3/2} \text {arctanh}\left (\frac {a d^2 \tan (e+f x)+a d^2}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {-\frac {a \left (\frac {d^{3/2} \arctan \left (\frac {\sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{a f}-\frac {d^{3/2} \text {arctanh}\left (\frac {a d^2 \tan (e+f x)+a d^2}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{\sqrt {2} a f}\right )}{d}-\frac {2 d}{f \sqrt {d \tan (e+f x)}}}{a d^3}-\frac {2}{3 a d f (d \tan (e+f x))^{3/2}}\) |
Input:
Int[1/((d*Tan[e + f*x])^(5/2)*(a + a*Tan[e + f*x])),x]
Output:
-2/(3*a*d*f*(d*Tan[e + f*x])^(3/2)) - (-((a*((d^(3/2)*ArcTan[Sqrt[d*Tan[e + f*x]]/Sqrt[d]])/(a*f) - (d^(3/2)*ArcTanh[(a*d^2 + a*d^2*Tan[e + f*x])/(S qrt[2]*a*d^(3/2)*Sqrt[d*Tan[e + f*x]])])/(Sqrt[2]*a*f)))/d) - (2*d)/(f*Sqr t[d*Tan[e + f*x]]))/(a*d^3)
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_))^(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]
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])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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[((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]
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])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x ]], x], x] + Simp[(b*c - a*d)^2/(c^2 + d^2) Int[(1 + Tan[e + f*x]^2)/(Sqr t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e , f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
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]
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])))
Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(112)=224\).
Time = 1.36 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.47
| method | result | size |
| derivativedivides | \(\frac {2 d^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{4}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {9}{2}}}-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f a}\) | \(333\) |
| default | \(\frac {2 d^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}}{2 d^{4}}+\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {d}}\right )}{2 d^{\frac {9}{2}}}-\frac {1}{3 d^{3} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f a}\) | \(333\) |
Input:
int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x,method=_RETURNVERBOSE)
Output:
2/f/a*d^2*(1/2/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*ta n(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*ar ctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))+1/2/d^(9/2)*arctan((d* tan(f*x+e))^(1/2)/d^(1/2))-1/3/d^3/(d*tan(f*x+e))^(3/2)+1/d^4/(d*tan(f*x+e ))^(1/2))
Time = 0.10 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.31 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\left [\frac {3 \, \sqrt {2} \sqrt {-d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} - 3 \, \sqrt {-d} \log \left (\frac {d \tan \left (f x + e\right ) - 2 \, \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} - d}{\tan \left (f x + e\right ) + 1}\right ) \tan \left (f x + e\right )^{2} + 4 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{6 \, a d^{3} f \tan \left (f x + e\right )^{2}}, \frac {3 \, \sqrt {2} \sqrt {d} \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} - 12 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} + 8 \, \sqrt {d \tan \left (f x + e\right )} {\left (3 \, \tan \left (f x + e\right ) - 1\right )}}{12 \, a d^{3} f \tan \left (f x + e\right )^{2}}\right ] \] Input:
integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(2)*sqrt(-d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(-d)* (tan(f*x + e) + 1)/(d*tan(f*x + e)))*tan(f*x + e)^2 - 3*sqrt(-d)*log((d*ta n(f*x + e) - 2*sqrt(d*tan(f*x + e))*sqrt(-d) - d)/(tan(f*x + e) + 1))*tan( f*x + e)^2 + 4*sqrt(d*tan(f*x + e))*(3*tan(f*x + e) - 1))/(a*d^3*f*tan(f*x + e)^2), 1/12*(3*sqrt(2)*sqrt(d)*log((d*tan(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))*tan(f*x + e)^2 - 12*sqrt(d)*arctan(sqrt(d*tan(f*x + e))/(sqr t(d)*tan(f*x + e)))*tan(f*x + e)^2 + 8*sqrt(d*tan(f*x + e))*(3*tan(f*x + e ) - 1))/(a*d^3*f*tan(f*x + e)^2)]
\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )} + \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \] Input:
integrate(1/(d*tan(f*x+e))**(5/2)/(a+a*tan(f*x+e)),x)
Output:
Integral(1/((d*tan(e + f*x))**(5/2)*tan(e + f*x) + (d*tan(e + f*x))**(5/2) ), x)/a
Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=-\frac {\frac {3 \, {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{a d} - \frac {12 \, \arctan \left (\frac {\sqrt {d \tan \left (f x + e\right )}}{\sqrt {d}}\right )}{a d^{\frac {3}{2}}} - \frac {8 \, {\left (3 \, d \tan \left (f x + e\right ) - d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d}}{12 \, d f} \] Input:
integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="maxima")
Output:
-1/12*(3*(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))/(a*d) - 12*arctan(sqrt(d*tan(f*x + e))/sqrt(d))/(a* d^(3/2)) - 8*(3*d*tan(f*x + e) - d)/((d*tan(f*x + e))^(3/2)*a*d))/(d*f)
Timed out. \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\text {Timed out} \] Input:
integrate(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x, algorithm="giac")
Output:
Timed out
Time = 1.72 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\frac {\frac {2\,\mathrm {tan}\left (e+f\,x\right )}{d}-\frac {2}{3\,d}}{a\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {\mathrm {atan}\left (\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{a\,d^{5/2}\,f}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,a^3\,d^{21/2}\,f^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{12\,a^3\,d^{11}\,f^3+12\,a^3\,d^{11}\,f^3\,\mathrm {tan}\left (e+f\,x\right )}\right )}{2\,a\,d^{5/2}\,f} \] Input:
int(1/((d*tan(e + f*x))^(5/2)*(a + a*tan(e + f*x))),x)
Output:
((2*tan(e + f*x))/d - 2/(3*d))/(a*f*(d*tan(e + f*x))^(3/2)) + atan((d*tan( e + f*x))^(1/2)/d^(1/2))/(a*d^(5/2)*f) - (2^(1/2)*atanh((12*2^(1/2)*a^3*d^ (21/2)*f^3*(d*tan(e + f*x))^(1/2))/(12*a^3*d^11*f^3 + 12*a^3*d^11*f^3*tan( e + f*x))))/(2*a*d^(5/2)*f)
\[ \int \frac {1}{(d \tan (e+f x))^{5/2} (a+a \tan (e+f x))} \, dx=\frac {\sqrt {d}\, \left (6 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )-2 \sqrt {\tan \left (f x +e \right )}+3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )+1}d x \right ) \tan \left (f x +e \right )^{2} f \right )}{3 \tan \left (f x +e \right )^{2} a \,d^{3} f} \] Input:
int(1/(d*tan(f*x+e))^(5/2)/(a+a*tan(f*x+e)),x)
Output:
(sqrt(d)*(6*sqrt(tan(e + f*x))*tan(e + f*x) - 2*sqrt(tan(e + f*x)) + 3*int ((sqrt(tan(e + f*x))*tan(e + f*x))/(tan(e + f*x) + 1),x)*tan(e + f*x)**2*f ))/(3*tan(e + f*x)**2*a*d**3*f)