Integrand size = 25, antiderivative size = 117 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2 \sqrt {2} a^3 \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{5/2} f}-\frac {16 a^3}{3 d^2 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{3 d f (d \tan (e+f x))^{3/2}} \]
2*a^3*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)/d^(5/2)/f-16/3*a^3/d^2/f/(d*tan(f*x+e))^(1/2)-2/3*(a^3+a^3*tan( f*x+e))/d/f/(d*tan(f*x+e))^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(691\) vs. \(2(117)=234\).
Time = 6.20 (sec) , antiderivative size = 691, normalized size of antiderivative = 5.91 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {2 \cos ^2(e+f x) \sin (e+f x) (a+a \tan (e+f x))^3}{3 f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}-\frac {6 \cos (e+f x) \sin ^2(e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}+\frac {\arctan \left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^3(e+f x) (-\tan (e+f x))^{3/4} \tan ^{\frac {7}{4}}(e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}+\frac {\text {arctanh}\left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^3(e+f x) (-\tan (e+f x))^{3/4} \tan ^{\frac {7}{4}}(e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}-\frac {2 \arctan \left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {9}{4}}(e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}+\frac {2 \text {arctanh}\left (\sqrt [4]{-\tan (e+f x)} \sqrt [4]{\tan (e+f x)}\right ) \cos ^3(e+f x) \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {9}{4}}(e+f x) (a+a \tan (e+f x))^3}{f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}}-\frac {3 \cos ^3(e+f x) \left (2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right )-2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right )+\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right )\right ) \tan ^{\frac {5}{2}}(e+f x) (a+a \tan (e+f x))^3}{4 f (\cos (e+f x)+\sin (e+f x))^3 (d \tan (e+f x))^{5/2}} \]
(-2*Cos[e + f*x]^2*Sin[e + f*x]*(a + a*Tan[e + f*x])^3)/(3*f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(5/2)) - (6*Cos[e + f*x]*Sin[e + f*x]^ 2*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f* x])^(5/2)) + (ArcTan[(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)]*Cos[e + f*x ]^3*(-Tan[e + f*x])^(3/4)*Tan[e + f*x]^(7/4)*(a + a*Tan[e + f*x])^3)/(f*(C os[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(5/2)) + (ArcTanh[(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^(3/4)*Tan[ e + f*x]^(7/4)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3* (d*Tan[e + f*x])^(5/2)) - (2*ArcTan[(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/ 4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(9/4)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(5/2)) + (2* ArcTanh[(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(1/4)]*Cos[e + f*x]^3*(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(9/4)*(a + a*Tan[e + f*x])^3)/(f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Tan[e + f*x])^(5/2)) - (3*Cos[e + f*x]^3*(2*Sqrt[2]*Ar cTan[1 - Sqrt[2]*Sqrt[Tan[e + f*x]]] - 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[T an[e + f*x]]] + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]])*Tan[e + f*x ]^(5/2)*(a + a*Tan[e + f*x])^3)/(4*f*(Cos[e + f*x] + Sin[e + f*x])^3*(d*Ta n[e + f*x])^(5/2))
Time = 0.58 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4048, 3042, 4111, 27, 3042, 4015, 221}
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.
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^3}{(d \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^3}{(d \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2 \int \frac {4 d^2 a^3+d^2 \tan ^2(e+f x) a^3+3 d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{3/2}}dx}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {4 d^2 a^3+d^2 \tan (e+f x)^2 a^3+3 d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{3/2}}dx}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {2 \left (\frac {\int \frac {3 \left (a^3 d^3-a^3 d^3 \tan (e+f x)\right )}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {8 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \frac {a^3 d^3-a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {8 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {3 \int \frac {a^3 d^3-a^3 d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {8 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4015 |
| \(\displaystyle \frac {2 \left (-\frac {6 a^6 d^4 \int \frac {1}{\cot (e+f x) \left (a^3 d^3+a^3 \tan (e+f x) d^3\right )^2-2 a^6 d^6}d\frac {a^3 d^3+a^3 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}}{f}-\frac {8 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \left (\frac {3 \sqrt {2} a^3 \sqrt {d} \text {arctanh}\left (\frac {a^3 d^3 \tan (e+f x)+a^3 d^3}{\sqrt {2} a^3 d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {8 a^3 d}{f \sqrt {d \tan (e+f x)}}\right )}{3 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{3 d f (d \tan (e+f x))^{3/2}}\) |
(-2*(a^3 + a^3*Tan[e + f*x]))/(3*d*f*(d*Tan[e + f*x])^(3/2)) + (2*((3*Sqrt [2]*a^3*Sqrt[d]*ArcTanh[(a^3*d^3 + a^3*d^3*Tan[e + f*x])/(Sqrt[2]*a^3*d^(5 /2)*Sqrt[d*Tan[e + f*x]])])/f - (8*a^3*d)/(f*Sqrt[d*Tan[e + f*x]])))/(3*d^ 3)
3.4.55.3.1 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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 /(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c *(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) *Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( n + 1)))*Tan[e + f*x]^2, 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] && GtQ[m, 2] && LtQ [n, -1] && IntegerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*((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[b*B + a*(A - C) - (A*b - a*B - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B , C}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0 ]
Leaf count of result is larger than twice the leaf count of optimal. \(302\) vs. \(2(98)=196\).
Time = 0.96 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.59
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {d}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {3}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,d^{2}}\) | \(303\) |
| default | \(\frac {2 a^{3} \left (-\frac {d}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {3}{\sqrt {d \tan \left (f x +e \right )}}+\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f \,d^{2}}\) | \(303\) |
| parts | \(\frac {2 a^{3} d \left (-\frac {1}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\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^{4}}\right )}{f}+\frac {a^{3} \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^{2} \left (d^{2}\right )^{\frac {1}{4}}}+\frac {3 a^{3} \left (-\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 )}{4 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}+\frac {3 a^{3} \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^{3}}\) | \(599\) |
2/f*a^3/d^2*(-1/3*d/(d*tan(f*x+e))^(3/2)-3/(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))-1/4/(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*ta n(f*x+e))^(1/2)+1)))
Time = 0.27 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.95 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {2} a^{3} \sqrt {d} \log \left (\frac {\tan \left (f x + e\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) + 1\right )}}{\sqrt {d}} + 4 \, \tan \left (f x + e\right ) + 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{2} - 2 \, {\left (9 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}}, -\frac {2 \, {\left (3 \, \sqrt {2} a^{3} d \sqrt {-\frac {1}{d}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{2} + {\left (9 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{3 \, d^{3} f \tan \left (f x + e\right )^{2}}\right ] \]
[1/3*(3*sqrt(2)*a^3*sqrt(d)*log((tan(f*x + e)^2 + 2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) + 1)/sqrt(d) + 4*tan(f*x + e) + 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^2 - 2*(9*a^3*tan(f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/(d ^3*f*tan(f*x + e)^2), -2/3*(3*sqrt(2)*a^3*d*sqrt(-1/d)*arctan(1/2*sqrt(2)* sqrt(d*tan(f*x + e))*sqrt(-1/d)*(tan(f*x + e) + 1)/tan(f*x + e))*tan(f*x + e)^2 + (9*a^3*tan(f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/(d^3*f*tan(f*x + e)^2)]
\[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=a^{3} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]
a**3*(Integral((d*tan(e + f*x))**(-5/2), x) + Integral(3*tan(e + f*x)/(d*t an(e + f*x))**(5/2), x) + Integral(3*tan(e + f*x)**2/(d*tan(e + f*x))**(5/ 2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(5/2), x))
Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.05 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {3 \, a^{3} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {2 \, {\left (9 \, a^{3} d \tan \left (f x + e\right ) + a^{3} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{3 \, d f} \]
1/3*(3*a^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))/d - 2*(9*a^3*d*tan(f*x + e) + a^3*d)/((d*tan(f*x + e))^(3/2)*d))/(d*f)
Timed out. \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 5.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{5/2}} \, dx=\frac {2\,\sqrt {2}\,a^3\,\mathrm {atanh}\left (\frac {32\,\sqrt {2}\,a^6\,d^{5/2}\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{32\,a^6\,d^3\,f+32\,a^6\,d^3\,f\,\mathrm {tan}\left (e+f\,x\right )}\right )}{d^{5/2}\,f}-\frac {\frac {2\,a^3\,d}{3}+6\,a^3\,d\,\mathrm {tan}\left (e+f\,x\right )}{d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \]