Integrand size = 15, antiderivative size = 54 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {1}{32} \arctan \left (\frac {1}{2} \sqrt {1+5 \tan ^2(x)}\right )-\frac {1}{12 \left (1+5 \tan ^2(x)\right )^{3/2}}+\frac {1}{16 \sqrt {1+5 \tan ^2(x)}} \]
Time = 0.66 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.31 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {(-3+2 \cos (2 x)) \left (-6 \cos (x)+8 \cos (3 x)-3 (-3+2 \cos (2 x))^{3/2} \log \left (2 \cos (x)+\sqrt {-3+2 \cos (2 x)}\right )\right ) \sec ^5(x)}{96 \left (1+5 \tan ^2(x)\right )^{5/2}} \]
((-3 + 2*Cos[2*x])*(-6*Cos[x] + 8*Cos[3*x] - 3*(-3 + 2*Cos[2*x])^(3/2)*Log [2*Cos[x] + Sqrt[-3 + 2*Cos[2*x]]])*Sec[x]^5)/(96*(1 + 5*Tan[x]^2)^(5/2))
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4153, 353, 61, 61, 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.
\(\displaystyle \int \frac {\tan (x)}{\left (5 \tan ^2(x)+1\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (x)}{\left (5 \tan (x)^2+1\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4153 |
\(\displaystyle \int \frac {\tan (x)}{\left (\tan ^2(x)+1\right ) \left (5 \tan ^2(x)+1\right )^{5/2}}d\tan (x)\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (\tan ^2(x)+1\right ) \left (5 \tan ^2(x)+1\right )^{5/2}}d\tan ^2(x)\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{4} \int \frac {1}{\left (\tan ^2(x)+1\right ) \left (5 \tan ^2(x)+1\right )^{3/2}}d\tan ^2(x)-\frac {1}{6 \left (5 \tan ^2(x)+1\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{4} \int \frac {1}{\left (\tan ^2(x)+1\right ) \sqrt {5 \tan ^2(x)+1}}d\tan ^2(x)+\frac {1}{2 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {1}{6 \left (5 \tan ^2(x)+1\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{10} \int \frac {1}{\frac {\tan ^4(x)}{5}+\frac {4}{5}}d\sqrt {5 \tan ^2(x)+1}+\frac {1}{2 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {1}{6 \left (5 \tan ^2(x)+1\right )^{3/2}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{4} \arctan \left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right )+\frac {1}{2 \sqrt {5 \tan ^2(x)+1}}\right )-\frac {1}{6 \left (5 \tan ^2(x)+1\right )^{3/2}}\right )\) |
(-1/6*1/(1 + 5*Tan[x]^2)^(3/2) + (ArcTan[Sqrt[1 + 5*Tan[x]^2]/2]/4 + 1/(2* Sqrt[1 + 5*Tan[x]^2]))/4)/2
3.5.41.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[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 = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {\arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )}{32}+\frac {1}{16 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}-\frac {1}{12 {\left (1+5 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}\) | \(41\) |
default | \(\frac {\arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )}{32}+\frac {1}{16 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}-\frac {1}{12 {\left (1+5 \left (\tan ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}\) | \(41\) |
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.41 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {3 \, {\left (25 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 1\right )} \arctan \left (\frac {5 \, \tan \left (x\right )^{2} - 3}{4 \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}}\right ) + 4 \, {\left (15 \, \tan \left (x\right )^{2} - 1\right )} \sqrt {5 \, \tan \left (x\right )^{2} + 1}}{192 \, {\left (25 \, \tan \left (x\right )^{4} + 10 \, \tan \left (x\right )^{2} + 1\right )}} \]
1/192*(3*(25*tan(x)^4 + 10*tan(x)^2 + 1)*arctan(1/4*(5*tan(x)^2 - 3)/sqrt( 5*tan(x)^2 + 1)) + 4*(15*tan(x)^2 - 1)*sqrt(5*tan(x)^2 + 1))/(25*tan(x)^4 + 10*tan(x)^2 + 1)
Time = 2.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {\operatorname {atan}{\left (\frac {\sqrt {5 \tan ^{2}{\left (x \right )} + 1}}{2} \right )}}{32} + \frac {1}{16 \sqrt {5 \tan ^{2}{\left (x \right )} + 1}} - \frac {1}{12 \left (5 \tan ^{2}{\left (x \right )} + 1\right )^{\frac {3}{2}}} \]
atan(sqrt(5*tan(x)**2 + 1)/2)/32 + 1/(16*sqrt(5*tan(x)**2 + 1)) - 1/(12*(5 *tan(x)**2 + 1)**(3/2))
\[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (x\right )}{{\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.67 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {15 \, \tan \left (x\right )^{2} - 1}{48 \, {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {3}{2}}} + \frac {1}{32} \, \arctan \left (\frac {1}{2} \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}\right ) \]
Time = 0.53 (sec) , antiderivative size = 172, normalized size of antiderivative = 3.19 \[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\frac {\ln \left (\mathrm {tan}\left (x\right )-\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}+\frac {1}{5}{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}+\frac {\ln \left (\mathrm {tan}\left (x\right )+\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}-\frac {1}{5}{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{64}-\frac {\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}\,1{}\mathrm {i}}{96\,\left (\mathrm {tan}\left (x\right )-\frac {\sqrt {5}\,1{}\mathrm {i}}{5}\right )}+\frac {\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}\,1{}\mathrm {i}}{96\,\left (\mathrm {tan}\left (x\right )+\frac {\sqrt {5}\,1{}\mathrm {i}}{5}\right )}+\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{240\,\left ({\mathrm {tan}\left (x\right )}^2+\frac {2{}\mathrm {i}\,\sqrt {5}\,\mathrm {tan}\left (x\right )}{5}-\frac {1}{5}\right )}-\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{240\,\left (-{\mathrm {tan}\left (x\right )}^2+\frac {2{}\mathrm {i}\,\sqrt {5}\,\mathrm {tan}\left (x\right )}{5}+\frac {1}{5}\right )} \]
(log(tan(x) - (2*5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/5 + 1i/5)*1i)/64 + (log(t an(x) + (2*5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/5 - 1i/5)*1i)/64 - (log(tan(x) - 1i)*1i)/64 - (log(tan(x) + 1i)*1i)/64 - ((tan(x)^2 + 1/5)^(1/2)*1i)/(96* (tan(x) - (5^(1/2)*1i)/5)) + ((tan(x)^2 + 1/5)^(1/2)*1i)/(96*(tan(x) + (5^ (1/2)*1i)/5)) + (5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/(240*(tan(x)^2 + (5^(1/2) *tan(x)*2i)/5 - 1/5)) - (5^(1/2)*(tan(x)^2 + 1/5)^(1/2))/(240*((5^(1/2)*ta n(x)*2i)/5 - tan(x)^2 + 1/5))
\[ \int \frac {\tan (x)}{\left (1+5 \tan ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {5 \tan \left (x \right )^{2}+1}\, \tan \left (x \right )}{125 \tan \left (x \right )^{6}+75 \tan \left (x \right )^{4}+15 \tan \left (x \right )^{2}+1}d x \]