Integrand size = 23, antiderivative size = 93 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=\frac {\sqrt {2} a d^{3/2} \arctan \left (\frac {\sqrt {d}-\sqrt {d} \tan (e+f x)}{\sqrt {2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f} \]
a*d^(3/2)*arctan(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*(d*tan(f*x+e))^(1/2)/f+2/3*a*(d*tan(f*x+e))^(3/2)/f
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.13 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=\frac {\left (\frac {1}{3}+\frac {i}{3}\right ) a (d \tan (e+f x))^{3/2} \left (-3 (-1)^{3/4} \arctan \left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+3 \sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (e+f x)}\right )+(1-i) \sqrt {\tan (e+f x)} (3+\tan (e+f x))\right )}{f \tan ^{\frac {3}{2}}(e+f x)} \]
((1/3 + I/3)*a*(d*Tan[e + f*x])^(3/2)*(-3*(-1)^(3/4)*ArcTan[(-1)^(3/4)*Sqr t[Tan[e + f*x]]] + 3*(-1)^(1/4)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[e + f*x]]] + ( 1 - I)*Sqrt[Tan[e + f*x]]*(3 + Tan[e + f*x])))/(f*Tan[e + f*x]^(3/2))
Time = 0.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4011, 3042, 4011, 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.
\(\displaystyle \int (a \tan (e+f x)+a) (d \tan (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \tan (e+f x)+a) (d \tan (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \sqrt {d \tan (e+f x)} (a d \tan (e+f x)-a d)dx+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {d \tan (e+f x)} (a d \tan (e+f x)-a d)dx+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \frac {-a d^2-a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {-a d^2-a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}\) |
\(\Big \downarrow \) 4015 |
\(\displaystyle -\frac {2 a^2 d^4 \int \frac {1}{2 a^2 d^4+\cot (e+f x) \left (a d^2-a d^2 \tan (e+f x)\right )^2}d\left (-\frac {a d^2-a d^2 \tan (e+f x)}{\sqrt {d \tan (e+f x)}}\right )}{f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {2} a d^{3/2} \arctan \left (\frac {a d^2-a d^2 \tan (e+f x)}{\sqrt {2} a d^{3/2} \sqrt {d \tan (e+f x)}}\right )}{f}+\frac {2 a d \sqrt {d \tan (e+f x)}}{f}+\frac {2 a (d \tan (e+f x))^{3/2}}{3 f}\) |
(Sqrt[2]*a*d^(3/2)*ArcTan[(a*d^2 - a*d^2*Tan[e + f*x])/(Sqrt[2]*a*d^(3/2)* Sqrt[d*Tan[e + f*x]])])/f + (2*a*d*Sqrt[d*Tan[e + f*x]])/f + (2*a*(d*Tan[e + f*x])^(3/2))/(3*f)
3.4.37.3.1 Defintions of rubi rules used
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[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. \(302\) vs. \(2(76)=152\).
Time = 1.00 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.26
method | result | size |
derivativedivides | \(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(303\) |
default | \(\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 d \sqrt {d \tan \left (f x +e \right )}-2 d^{2} \left (\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (d^{2}\right )^{\frac {1}{4}}}\right )\right )}{f}\) | \(303\) |
parts | \(\frac {2 a d \left (\sqrt {d \tan \left (f x +e \right )}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{f}+\frac {a \left (\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {d^{2} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}\) | \(303\) |
1/f*a*(2/3*(d*tan(f*x+e))^(3/2)+2*d*(d*tan(f*x+e))^(1/2)-2*d^2*(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*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*arctan(-2^(1/2)/(d^2)^(1/4)*(d*ta n(f*x+e))^(1/2)+1))))
Time = 0.25 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.98 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=\left [\frac {3 \, \sqrt {2} a \sqrt {-d} d \log \left (\frac {d \tan \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {-d} {\left (\tan \left (f x + e\right ) - 1\right )} - 4 \, d \tan \left (f x + e\right ) + d}{\tan \left (f x + e\right )^{2} + 1}\right ) + 4 \, {\left (a d \tan \left (f x + e\right ) + 3 \, a d\right )} \sqrt {d \tan \left (f x + e\right )}}{6 \, f}, -\frac {3 \, \sqrt {2} a d^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} \sqrt {d \tan \left (f x + e\right )} {\left (\tan \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {d} \tan \left (f x + e\right )}\right ) - 2 \, {\left (a d \tan \left (f x + e\right ) + 3 \, a d\right )} \sqrt {d \tan \left (f x + e\right )}}{3 \, f}\right ] \]
[1/6*(3*sqrt(2)*a*sqrt(-d)*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)) + 4*(a*d*tan(f*x + e) + 3*a*d)*sqrt(d*tan(f*x + e)))/f, -1/3*(3* sqrt(2)*a*d^(3/2)*arctan(1/2*sqrt(2)*sqrt(d*tan(f*x + e))*(tan(f*x + e) - 1)/(sqrt(d)*tan(f*x + e))) - 2*(a*d*tan(f*x + e) + 3*a*d)*sqrt(d*tan(f*x + e)))/f]
\[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=a \left (\int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx + \int \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}\, dx\right ) \]
a*(Integral((d*tan(e + f*x))**(3/2), x) + Integral((d*tan(e + f*x))**(3/2) *tan(e + f*x), x))
Time = 0.30 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.27 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=-\frac {3 \, a d^{3} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}}\right )} - 2 \, \left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} a d - 6 \, \sqrt {d \tan \left (f x + e\right )} a d^{2}}{3 \, d f} \]
-1/3*(3*a*d^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)) - 2*(d*tan(f*x + e))^(3/2)*a* d - 6*sqrt(d*tan(f*x + e))*a*d^2)/(d*f)
Timed out. \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=\text {Timed out} \]
Time = 5.65 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int (d \tan (e+f x))^{3/2} (a+a \tan (e+f x)) \, dx=\frac {2\,a\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{3\,f}+\frac {2\,a\,d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (-1+1{}\mathrm {i}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,a\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )\,\left (1+1{}\mathrm {i}\right )}{f} \]