Integrand size = 23, antiderivative size = 115 \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\frac {\sqrt {2} a d^{5/2} \text {arctanh}\left (\frac {\sqrt {d}+\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {2 a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f} \]
a*d^(5/2)*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)/f-2*a*d^2*(d*tan(f*x+e))^(1/2)/f+2/3*a*d*(d*tan(f*x+e))^(3/ 2)/f+2/5*a*(d*tan(f*x+e))^(5/2)/f
Result contains complex when optimal does not.
Time = 0.91 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\frac {\left (\frac {1}{15}+\frac {i}{15}\right ) a (d \tan (e+f x))^{5/2} \left (-15 \sqrt [4]{-1} \arctan \left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+15 (-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+(1-i) \sqrt {\tan (e+f x)} \left (-15+5 \tan (e+f x)+3 \tan ^2(e+f x)\right )\right )}{f \tan ^{\frac {5}{2}}(e+f x)} \]
((1/15 + I/15)*a*(d*Tan[e + f*x])^(5/2)*(-15*(-1)^(1/4)*ArcTan[(-1)^(3/4)* Sqrt[Tan[e + f*x]]] + 15*(-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + (1 - I)*Sqrt[Tan[e + f*x]]*(-15 + 5*Tan[e + f*x] + 3*Tan[e + f*x]^2)))/ (f*Tan[e + f*x]^(5/2))
Time = 0.60 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 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.
| \(\displaystyle \int (a \tan (e+f x)+a) (d \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \tan (e+f x)+a) (d \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (a d \tan (e+f x)-a d)dx+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \tan (e+f x))^{3/2} (a d \tan (e+f x)-a d)dx+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (-a d^2-a \tan (e+f x) d^2\right )dx+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {d \tan (e+f x)} \left (-a d^2-a \tan (e+f x) d^2\right )dx+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \int \frac {a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a d^3-a d^3 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}dx-\frac {2 a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}\) |
| \(\Big \downarrow \) 4015 |
| \(\displaystyle -\frac {2 a^2 d^6 \int \frac {1}{\cot (e+f x) \left (a d^3+a \tan (e+f x) d^3\right )^2-2 a^2 d^6}d\frac {a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}}{f}-\frac {2 a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt {2} a d^{5/2} \text {arctanh}\left (\frac {a d^3 \tan (e+f x)+a d^3}{\sqrt {2} a d^{5/2} \sqrt {d \tan (e+f x)}}\right )}{f}-\frac {2 a d^2 \sqrt {d \tan (e+f x)}}{f}+\frac {2 a d (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a (d \tan (e+f x))^{5/2}}{5 f}\) |
(Sqrt[2]*a*d^(5/2)*ArcTanh[(a*d^3 + a*d^3*Tan[e + f*x])/(Sqrt[2]*a*d^(5/2) *Sqrt[d*Tan[e + f*x]])])/f - (2*a*d^2*Sqrt[d*Tan[e + f*x]])/f + (2*a*d*(d* Tan[e + f*x])^(3/2))/(3*f) + (2*a*(d*Tan[e + f*x])^(5/2))/(5*f)
3.4.36.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*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] && GtQ[m, 0]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(317\) vs. \(2(94)=188\).
Time = 1.14 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.77
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(318\) |
| default | \(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 d \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+2 d^{3} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(318\) |
| parts | \(\frac {2 a d \left (\frac {\left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}+\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-2 d^{2} \sqrt {d \tan \left (f x +e \right )}+\frac {d^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4}\right )}{f}\) | \(323\) |
1/f*a*(2/5*(d*tan(f*x+e))^(5/2)+2/3*d*(d*tan(f*x+e))^(3/2)-2*d^2*(d*tan(f* x+e))^(1/2)+2*d^3*(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*ta n(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/ 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*arct an(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))))
Time = 0.25 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.97 \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\left [\frac {15 \, \sqrt {2} a d^{\frac {5}{2}} \log \left (\frac {d \tan \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} {\left (\tan \left (f x + e\right ) + 1\right )} + 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (3 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 \, a d^{2} \tan \left (f x + e\right ) - 15 \, a d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{30 \, f}, -\frac {15 \, \sqrt {2} a \sqrt {-d} d^{2} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) + 1\right )}}{2 \, d \tan \left (f x + e\right )}\right ) - 2 \, {\left (3 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 \, a d^{2} \tan \left (f x + e\right ) - 15 \, a d^{2}\right )} \sqrt {d \tan \left (f x + e\right )}}{15 \, f}\right ] \]
[1/30*(15*sqrt(2)*a*d^(5/2)*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)) + 4*(3*a*d^2*tan(f*x + e)^2 + 5*a*d^2*tan(f*x + e) - 15*a*d^2)*sqr t(d*tan(f*x + e)))/f, -1/15*(15*sqrt(2)*a*sqrt(-d)*d^2*arctan(1/2*sqrt(2)* sqrt(d*tan(f*x + e))*sqrt(-d)*(tan(f*x + e) + 1)/(d*tan(f*x + e))) - 2*(3* a*d^2*tan(f*x + e)^2 + 5*a*d^2*tan(f*x + e) - 15*a*d^2)*sqrt(d*tan(f*x + e )))/f]
\[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=a \left (\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}} \tan {\left (e + f x \right )}\, dx\right ) \]
a*(Integral((d*tan(e + f*x))**(5/2), x) + Integral((d*tan(e + f*x))**(5/2) *tan(e + f*x), x))
Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\frac {15 \, a d^{4} {\left (\frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )} + 12 \, \left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} a d + 20 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d^{2} - 60 \, \sqrt {d \tan \left (f x + e\right )} a d^{3}}{30 \, d f} \]
1/30*(15*a*d^4*(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)) + 12*(d*tan(f*x + e))^(5/2)*a*d + 20*(d*tan(f *x + e))^(3/2)*a*d^2 - 60*sqrt(d*tan(f*x + e))*a*d^3)/(d*f)
Timed out. \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\text {Timed out} \]
Time = 6.53 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.25 \[ \int (d \tan (e+f x))^{5/2} (a+a \tan (e+f x)) \, dx=\frac {2\,a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{5\,f}+\frac {2\,a\,d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}-\frac {2\,a\,d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}-\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{5/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1-\mathrm {i}\right )}{f} \]
(2*a*(d*tan(e + f*x))^(5/2))/(5*f) + (2*a*d*(d*tan(e + f*x))^(3/2))/(3*f) - (2*a*d^2*(d*tan(e + f*x))^(1/2))/f - ((-1)^(1/4)*a*d^(5/2)*atan(((-1)^(1 /4)*(d*tan(e + f*x))^(1/2))/d^(1/2))*(1 + 1i))/f - ((-1)^(1/4)*a*d^(5/2)*a tan(((-1)^(1/4)*(d*tan(e + f*x))^(1/2)*1i)/d^(1/2)))/f + ((-1)^(1/4)*a*d^( 5/2)*atanh(((-1)^(1/4)*(d*tan(e + f*x))^(1/2))/d^(1/2)))/f