Integrand size = 15, antiderivative size = 66 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=-32 \arctan \left (\frac {1}{2} \sqrt {1+5 \tan ^2(x)}\right )+16 \sqrt {1+5 \tan ^2(x)}-\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2} \]
-32*arctan(1/2*(1+5*tan(x)^2)^(1/2))+16*(1+5*tan(x)^2)^(1/2)-4/3*(1+5*tan( x)^2)^(3/2)+1/5*(1+5*tan(x)^2)^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {5 \sqrt {5} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{2},-\frac {3}{2},\frac {4 \cos ^2(x)}{5}\right ) \left (1+5 \tan ^2(x)\right )^{5/2}}{(3-2 \cos (2 x))^{5/2}} \]
(5*Sqrt[5]*Hypergeometric2F1[-5/2, -5/2, -3/2, (4*Cos[x]^2)/5]*(1 + 5*Tan[ x]^2)^(5/2))/(3 - 2*Cos[2*x])^(5/2)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 4153, 353, 60, 60, 60, 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.
| \(\displaystyle \int \tan (x) \left (5 \tan ^2(x)+1\right )^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (x) \left (5 \tan (x)^2+1\right )^{5/2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \int \frac {\tan (x) \left (5 \tan ^2(x)+1\right )^{5/2}}{\tan ^2(x)+1}d\tan (x)\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (5 \tan ^2(x)+1\right )^{5/2}}{\tan ^2(x)+1}d\tan ^2(x)\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-4 \int \frac {\left (5 \tan ^2(x)+1\right )^{3/2}}{\tan ^2(x)+1}d\tan ^2(x)\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-4 \left (\frac {2}{3} \left (5 \tan ^2(x)+1\right )^{3/2}-4 \int \frac {\sqrt {5 \tan ^2(x)+1}}{\tan ^2(x)+1}d\tan ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-4 \left (\frac {2}{3} \left (5 \tan ^2(x)+1\right )^{3/2}-4 \left (2 \sqrt {5 \tan ^2(x)+1}-4 \int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan ^2(x)\right )\right )\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-4 \left (\frac {2}{3} \left (5 \tan ^2(x)+1\right )^{3/2}-4 \left (2 \sqrt {5 \tan ^2(x)+1}-\frac {8}{5} \int \frac {1}{\frac {\tan ^4(x)}{5}+\frac {4}{5}}d\sqrt {5 \tan ^2(x)+1}\right )\right )\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{2} \left (\frac {2}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-4 \left (\frac {2}{3} \left (5 \tan ^2(x)+1\right )^{3/2}-4 \left (2 \sqrt {5 \tan ^2(x)+1}-4 \arctan \left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right )\right )\right )\right )\) |
((2*(1 + 5*Tan[x]^2)^(5/2))/5 - 4*((2*(1 + 5*Tan[x]^2)^(3/2))/3 - 4*(-4*Ar cTan[Sqrt[1 + 5*Tan[x]^2]/2] + 2*Sqrt[1 + 5*Tan[x]^2])))/2
3.5.40.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {223 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{15}+5 \left (\tan ^{4}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}-\frac {14 \left (\tan ^{2}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{3}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )\) | \(61\) |
| default | \(\frac {223 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{15}+5 \left (\tan ^{4}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}-\frac {14 \left (\tan ^{2}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{3}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )\) | \(61\) |
223/15*(1+5*tan(x)^2)^(1/2)+5*tan(x)^4*(1+5*tan(x)^2)^(1/2)-14/3*tan(x)^2* (1+5*tan(x)^2)^(1/2)-32*arctan(1/2*(1+5*tan(x)^2)^(1/2))
Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {1}{15} \, {\left (75 \, \tan \left (x\right )^{4} - 70 \, \tan \left (x\right )^{2} + 223\right )} \sqrt {5 \, \tan \left (x\right )^{2} + 1} - 16 \, \arctan \left (\frac {5 \, \tan \left (x\right )^{2} - 3}{4 \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}}\right ) \]
1/15*(75*tan(x)^4 - 70*tan(x)^2 + 223)*sqrt(5*tan(x)^2 + 1) - 16*arctan(1/ 4*(5*tan(x)^2 - 3)/sqrt(5*tan(x)^2 + 1))
\[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\int \left (5 \tan ^{2}{\left (x \right )} + 1\right )^{\frac {5}{2}} \tan {\left (x \right )}\, dx \]
\[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\int { {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} \tan \left (x\right ) \,d x } \]
Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.79 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {1}{5} \, {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} - \frac {4}{3} \, {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} + 16 \, \sqrt {5 \, \tan \left (x\right )^{2} + 1} - 32 \, \arctan \left (\frac {1}{2} \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}\right ) \]
1/5*(5*tan(x)^2 + 1)^(5/2) - 4/3*(5*tan(x)^2 + 1)^(3/2) + 16*sqrt(5*tan(x) ^2 + 1) - 32*arctan(1/2*sqrt(5*tan(x)^2 + 1))
Time = 1.62 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.36 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}\,\left (25\,{\mathrm {tan}\left (x\right )}^4-\frac {70\,{\mathrm {tan}\left (x\right )}^2}{3}+\frac {223}{3}\right )}{5}-\ln \left (\mathrm {tan}\left (x\right )-\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}+\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}-\ln \left (\mathrm {tan}\left (x\right )+\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}-\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i} \]
log(tan(x) - 1i)*16i - log(tan(x) + (2*5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/5 - 1i/5)*16i - log(tan(x) - (2*5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/5 + 1i/5)*16i + log(tan(x) + 1i)*16i + (5^(1/2)*(tan(x)^2 + 1/5)^(1/2)*(25*tan(x)^4 - ( 70*tan(x)^2)/3 + 223/3))/5
\[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=5 \sqrt {5 \tan \left (x \right )^{2}+1}\, \tan \left (x \right )^{4}-\frac {14 \sqrt {5 \tan \left (x \right )^{2}+1}\, \tan \left (x \right )^{2}}{3}+\frac {31 \sqrt {5 \tan \left (x \right )^{2}+1}}{15}+64 \left (\int \frac {\sqrt {5 \tan \left (x \right )^{2}+1}\, \tan \left (x \right )^{3}}{5 \tan \left (x \right )^{2}+1}d x \right ) \]