Integrand size = 25, antiderivative size = 141 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {2 \sqrt {2} a^3 \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{d^{7/2} f}-\frac {8 a^3}{5 d^2 f (d \tan (e+f x))^{3/2}}-\frac {4 a^3}{d^3 f \sqrt {d \tan (e+f x)}}-\frac {2 \left (a^3+a^3 \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \] Output:
2*2^(1/2)*a^3*arctan(1/2*(d^(1/2)-d^(1/2)*tan(f*x+e))*2^(1/2)/(d*tan(f*x+e ))^(1/2))/d^(7/2)/f-8/5*a^3/d^2/f/(d*tan(f*x+e))^(3/2)-4*a^3/d^3/f/(d*tan( f*x+e))^(1/2)-2/5*(a^3+a^3*tan(f*x+e))/d/f/(d*tan(f*x+e))^(5/2)
Leaf count is larger than twice the leaf count of optimal. \(415\) vs. \(2(141)=282\).
Time = 3.57 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.94 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {a^3 (1+\cot (e+f x))^3 \left (8 \cos ^2(e+f x) \sin (e+f x)+40 \cos (e+f x) \sin ^2(e+f x)+80 \sin ^3(e+f x)+10 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^3(e+f x) \tan ^{\frac {7}{2}}(e+f x)-10 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (e+f x)}\right ) \cos ^3(e+f x) \tan ^{\frac {7}{2}}(e+f x)+5 \sqrt {2} \cos ^3(e+f x) \log \left (1-\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {7}{2}}(e+f x)-5 \sqrt {2} \cos ^3(e+f x) \log \left (1+\sqrt {2} \sqrt {\tan (e+f x)}+\tan (e+f x)\right ) \tan ^{\frac {7}{2}}(e+f x)-20 \text {arctanh}\left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \left (2 \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {13}{4}}(e+f x)-3 \left (-\tan ^2(e+f x)\right )^{7/4}\right )+20 \arctan \left (\sqrt [4]{-\tan ^2(e+f x)}\right ) \cos ^3(e+f x) \left (2 \sqrt [4]{-\tan (e+f x)} \tan ^{\frac {13}{4}}(e+f x)+3 \left (-\tan ^2(e+f x)\right )^{7/4}\right )\right )}{20 d^3 f (\cos (e+f x)+\sin (e+f x))^3 \sqrt {d \tan (e+f x)}} \] Input:
Integrate[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]
Output:
-1/20*(a^3*(1 + Cot[e + f*x])^3*(8*Cos[e + f*x]^2*Sin[e + f*x] + 40*Cos[e + f*x]*Sin[e + f*x]^2 + 80*Sin[e + f*x]^3 + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]* Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^3*Tan[e + f*x]^(7/2) - 10*Sqrt[2]*ArcTan[ 1 + Sqrt[2]*Sqrt[Tan[e + f*x]]]*Cos[e + f*x]^3*Tan[e + f*x]^(7/2) + 5*Sqrt [2]*Cos[e + f*x]^3*Log[1 - Sqrt[2]*Sqrt[Tan[e + f*x]] + Tan[e + f*x]]*Tan[ e + f*x]^(7/2) - 5*Sqrt[2]*Cos[e + f*x]^3*Log[1 + Sqrt[2]*Sqrt[Tan[e + f*x ]] + Tan[e + f*x]]*Tan[e + f*x]^(7/2) - 20*ArcTanh[(-Tan[e + f*x]^2)^(1/4) ]*Cos[e + f*x]^3*(2*(-Tan[e + f*x])^(1/4)*Tan[e + f*x]^(13/4) - 3*(-Tan[e + f*x]^2)^(7/4)) + 20*ArcTan[(-Tan[e + f*x]^2)^(1/4)]*Cos[e + f*x]^3*(2*(- Tan[e + f*x])^(1/4)*Tan[e + f*x]^(13/4) + 3*(-Tan[e + f*x]^2)^(7/4))))/(d^ 3*f*(Cos[e + f*x] + Sin[e + f*x])^3*Sqrt[d*Tan[e + f*x]])
Time = 0.78 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.14, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4048, 3042, 4111, 27, 3042, 4012, 25, 3042, 4015, 218}
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.
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^3}{(d \tan (e+f x))^{7/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \tan (e+f x)+a)^3}{(d \tan (e+f x))^{7/2}}dx\) |
| \(\Big \downarrow \) 4048 |
| \(\displaystyle \frac {2 \int \frac {6 d^2 a^3+d^2 \tan ^2(e+f x) a^3+5 d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{5/2}}dx}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {6 d^2 a^3+d^2 \tan (e+f x)^2 a^3+5 d^2 \tan (e+f x) a^3}{(d \tan (e+f x))^{5/2}}dx}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4111 |
| \(\displaystyle \frac {2 \left (\frac {\int \frac {5 \left (a^3 d^3-a^3 d^3 \tan (e+f x)\right )}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {5 \int \frac {a^3 d^3-a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {5 \int \frac {a^3 d^3-a^3 d^3 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {\int -\frac {a^3 d^4+a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 a^3 d^2}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (-\frac {\int \frac {a^3 d^4+a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 a^3 d^2}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (-\frac {\int \frac {a^3 d^4+a^3 \tan (e+f x) d^4}{\sqrt {d \tan (e+f x)}}dx}{d^2}-\frac {2 a^3 d^2}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4015 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {2 a^6 d^6 \int \frac {1}{2 a^6 d^8+\cot (e+f x) \left (a^3 d^4-a^3 d^4 \tan (e+f x)\right )^2}d\frac {a^3 d^4-a^3 d^4 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}}{f}-\frac {2 a^3 d^2}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 \left (\frac {5 \left (\frac {\sqrt {2} a^3 d^{3/2} \arctan \left (\frac {a^3 d^4-a^3 d^4 \tan (e+f x)}{\sqrt {2} a^3 d^{7/2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {2 a^3 d^2}{f \sqrt {d \tan (e+f x)}}\right )}{d^2}-\frac {4 a^3 d}{f (d \tan (e+f x))^{3/2}}\right )}{5 d^3}-\frac {2 \left (a^3 \tan (e+f x)+a^3\right )}{5 d f (d \tan (e+f x))^{5/2}}\) |
Input:
Int[(a + a*Tan[e + f*x])^3/(d*Tan[e + f*x])^(7/2),x]
Output:
(-2*(a^3 + a^3*Tan[e + f*x]))/(5*d*f*(d*Tan[e + f*x])^(5/2)) + (2*((-4*a^3 *d)/(f*(d*Tan[e + f*x])^(3/2)) + (5*((Sqrt[2]*a^3*d^(3/2)*ArcTan[(a^3*d^4 - a^3*d^4*Tan[e + f*x])/(Sqrt[2]*a^3*d^(7/2)*Sqrt[d*Tan[e + f*x]])])/f - ( 2*a^3*d^2)/(f*Sqrt[d*Tan[e + f*x]])))/d^2))/(5*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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (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 + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
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. \(322\) vs. \(2(120)=240\).
Time = 1.79 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.29
| method | result | size |
| derivativedivides | \(\frac {2 a^{3} \left (-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {1}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{d \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d}\right )}{f \,d^{2}}\) | \(323\) |
| default | \(\frac {2 a^{3} \left (-\frac {d}{5 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}-\frac {1}{\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2}{d \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d}\right )}{f \,d^{2}}\) | \(323\) |
| parts | \(\frac {2 a^{3} d \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 )}{8 d^{4} \left (d^{2}\right )^{\frac {1}{4}}}-\frac {1}{5 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f}+\frac {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^{4}}+\frac {3 a^{3} \left (-\frac {2}{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 )}{4 d^{4}}\right )}{f}+\frac {6 a^{3} \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 d}\) | \(634\) |
Input:
int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
2/f*a^3/d^2*(-1/5*d/(d*tan(f*x+e))^(5/2)-1/(d*tan(f*x+e))^(3/2)-2/d/(d*tan (f*x+e))^(1/2)+1/d*(-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*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))- 1/4/(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*ar ctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.82 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\left [\frac {5 \, \sqrt {2} a^{3} d \sqrt {-\frac {1}{d}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-\frac {1}{d}} {\left (\tan \left (f x + e\right ) - 1\right )} - \tan \left (f x + e\right )^{2} + 4 \, \tan \left (f x + e\right ) - 1}{\tan \left (f x + e\right )^{2} + 1}\right ) \tan \left (f x + e\right )^{3} - 2 \, {\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}, -\frac {2 \, {\left (5 \, \sqrt {2} a^{3} \sqrt {d} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{3} + {\left (10 \, a^{3} \tan \left (f x + e\right )^{2} + 5 \, a^{3} \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \tan \left (f x + e\right )}\right )}}{5 \, d^{4} f \tan \left (f x + e\right )^{3}}\right ] \] Input:
integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="fricas")
Output:
[1/5*(5*sqrt(2)*a^3*d*sqrt(-1/d)*log(-(2*sqrt(2)*sqrt(d*tan(f*x + e))*sqrt (-1/d)*(tan(f*x + e) - 1) - tan(f*x + e)^2 + 4*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1))*tan(f*x + e)^3 - 2*(10*a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e ) + a^3)*sqrt(d*tan(f*x + e)))/(d^4*f*tan(f*x + e)^3), -2/5*(5*sqrt(2)*a^3 *sqrt(d)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt( d)*tan(f*x + e)))*tan(f*x + e)^3 + (10*a^3*tan(f*x + e)^2 + 5*a^3*tan(f*x + e) + a^3)*sqrt(d*tan(f*x + e)))/(d^4*f*tan(f*x + e)^3)]
\[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=a^{3} \left (\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {3 \tan ^{2}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \] Input:
integrate((a+a*tan(f*x+e))**3/(d*tan(f*x+e))**(7/2),x)
Output:
a**3*(Integral((d*tan(e + f*x))**(-7/2), x) + Integral(3*tan(e + f*x)/(d*t an(e + f*x))**(7/2), x) + Integral(3*tan(e + f*x)**2/(d*tan(e + f*x))**(7/ 2), x) + Integral(tan(e + f*x)**3/(d*tan(e + f*x))**(7/2), x))
Time = 0.13 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {2 \, {\left (\frac {5 \, a^{3} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )}}{d^{2}} + \frac {10 \, a^{3} d^{2} \tan \left (f x + e\right )^{2} + 5 \, a^{3} d^{2} \tan \left (f x + e\right ) + a^{3} d^{2}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}\right )}}{5 \, d f} \] Input:
integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="maxima")
Output:
-2/5*(5*a^3*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(f* x + e)))/sqrt(d))/sqrt(d) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d))/d^2 + (10*a^3*d^2*tan(f*x + e)^ 2 + 5*a^3*d^2*tan(f*x + e) + a^3*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)
Timed out. \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x, algorithm="giac")
Output:
Timed out
Time = 2.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91 \[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {4\,d\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,d\,a^3\,\mathrm {tan}\left (e+f\,x\right )+\frac {2\,d\,a^3}{5}}{d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}-\frac {\sqrt {2}\,a^3\,\left (2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}\right )+2\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {d}}+\frac {\sqrt {2}\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{2\,d^{3/2}}\right )\right )}{d^{7/2}\,f} \] Input:
int((a + a*tan(e + f*x))^3/(d*tan(e + f*x))^(7/2),x)
Output:
- ((2*a^3*d)/5 + 4*a^3*d*tan(e + f*x)^2 + 2*a^3*d*tan(e + f*x))/(d^2*f*(d* tan(e + f*x))^(5/2)) - (2^(1/2)*a^3*(2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2 ))/(2*d^(1/2))) + 2*atan((2^(1/2)*(d*tan(e + f*x))^(1/2))/(2*d^(1/2)) + (2 ^(1/2)*(d*tan(e + f*x))^(3/2))/(2*d^(3/2)))))/(d^(7/2)*f)
\[ \int \frac {(a+a \tan (e+f x))^3}{(d \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {d}\, a^{3} \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{4}}d x +3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3}}d x \right )+3 \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{2}}d x \right )+\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )}d x \right )}{d^{4}} \] Input:
int((a+a*tan(f*x+e))^3/(d*tan(f*x+e))^(7/2),x)
Output:
(sqrt(d)*a**3*(int(sqrt(tan(e + f*x))/tan(e + f*x)**4,x) + 3*int(sqrt(tan( e + f*x))/tan(e + f*x)**3,x) + 3*int(sqrt(tan(e + f*x))/tan(e + f*x)**2,x) + int(sqrt(tan(e + f*x))/tan(e + f*x),x)))/d**4