Integrand size = 21, antiderivative size = 65 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 (a-b)^2 f}+\frac {1}{2 (a-b) f \left (a+b \tan ^2(e+f x)\right )} \]
Time = 0.64 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {2 \log (\cos (e+f x))+\log \left (a+b \tan ^2(e+f x)\right )+\frac {-a+b}{a+b \tan ^2(e+f x)}}{2 (a-b)^2 f} \]
-1/2*(2*Log[Cos[e + f*x]] + Log[a + b*Tan[e + f*x]^2] + (-a + b)/(a + b*Ta n[e + f*x]^2))/((a - b)^2*f)
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4153, 353, 54, 2009}
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.
| \(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan (e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\int \left (-\frac {b}{(a-b)^2 \left (b \tan ^2(e+f x)+a\right )}-\frac {b}{(a-b) \left (b \tan ^2(e+f x)+a\right )^2}+\frac {1}{(a-b)^2 \left (\tan ^2(e+f x)+1\right )}\right )d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{(a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {\log \left (\tan ^2(e+f x)+1\right )}{(a-b)^2}-\frac {\log \left (a+b \tan ^2(e+f x)\right )}{(a-b)^2}}{2 f}\) |
(Log[1 + Tan[e + f*x]^2]/(a - b)^2 - Log[a + b*Tan[e + f*x]^2]/(a - b)^2 + 1/((a - b)*(a + b*Tan[e + f*x]^2)))/(2*f)
3.3.26.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 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 = 0.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a -b}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}}{f}\) | \(78\) |
| default | \(\frac {-\frac {b \left (\frac {\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{b}-\frac {a -b}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}\right )}{2 \left (a -b \right )^{2}}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}}{f}\) | \(78\) |
| norman | \(-\frac {b \tan \left (f x +e \right )^{2}}{2 a f \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}-\frac {\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}\) | \(100\) |
| parallelrisch | \(\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2} b^{2}-b^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2}+\ln \left (1+\tan \left (f x +e \right )^{2}\right ) a b -\ln \left (a +b \tan \left (f x +e \right )^{2}\right ) a b +a b -b^{2}}{2 \left (a^{2}-2 a b +b^{2}\right ) \left (a +b \tan \left (f x +e \right )^{2}\right ) b f}\) | \(125\) |
| risch | \(\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {2 i e}{f \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 b \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left (-a +b \right )^{2} \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )}-\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}\) | \(166\) |
1/f*(-1/2*b/(a-b)^2*(1/b*ln(a+b*tan(f*x+e)^2)-(a-b)/b/(a+b*tan(f*x+e)^2))+ 1/2/(a-b)^2*ln(1+tan(f*x+e)^2))
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.51 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {b \tan \left (f x + e\right )^{2} + {\left (b \tan \left (f x + e\right )^{2} + a\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) + b}{2 \, {\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f\right )}} \]
-1/2*(b*tan(f*x + e)^2 + (b*tan(f*x + e)^2 + a)*log((b*tan(f*x + e)^2 + a) /(tan(f*x + e)^2 + 1)) + b)/((a^2*b - 2*a*b^2 + b^3)*f*tan(f*x + e)^2 + (a ^3 - 2*a^2*b + a*b^2)*f)
Leaf count of result is larger than twice the leaf count of optimal. 796 vs. \(2 (49) = 98\).
Time = 14.18 (sec) , antiderivative size = 796, normalized size of antiderivative = 12.25 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\begin {cases} \frac {\tilde {\infty } x}{\tan ^{3}{\left (e \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{2} f} & \text {for}\: b = 0 \\- \frac {1}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} & \text {for}\: a = b \\\frac {x \tan {\left (e \right )}}{\left (a + b \tan ^{2}{\left (e \right )}\right )^{2}} & \text {for}\: f = 0 \\- \frac {a \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} - \frac {a \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} + \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} + \frac {a}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} - \frac {b \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} - \frac {b \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} + \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} - \frac {b}{2 a^{3} f + 2 a^{2} b f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b f - 4 a b^{2} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{2} f + 2 b^{3} f \tan ^{2}{\left (e + f x \right )}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x/tan(e)**3, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (log(tan(e + f*x)**2 + 1)/(2*a**2*f), Eq(b, 0)), (-1/(4*b**2*f*tan(e + f*x)**4 + 8*b**2 *f*tan(e + f*x)**2 + 4*b**2*f), Eq(a, b)), (x*tan(e)/(a + b*tan(e)**2)**2, Eq(f, 0)), (-a*log(-sqrt(-a/b) + tan(e + f*x))/(2*a**3*f + 2*a**2*b*f*tan (e + f*x)**2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2*a*b**2*f + 2*b* *3*f*tan(e + f*x)**2) - a*log(sqrt(-a/b) + tan(e + f*x))/(2*a**3*f + 2*a** 2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2*a*b**2 *f + 2*b**3*f*tan(e + f*x)**2) + a*log(tan(e + f*x)**2 + 1)/(2*a**3*f + 2* a**2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2*a*b **2*f + 2*b**3*f*tan(e + f*x)**2) + a/(2*a**3*f + 2*a**2*b*f*tan(e + f*x)* *2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2*a*b**2*f + 2*b**3*f*tan(e + f*x)**2) - b*log(-sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**3*f + 2*a**2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2 *a*b**2*f + 2*b**3*f*tan(e + f*x)**2) - b*log(sqrt(-a/b) + tan(e + f*x))*t an(e + f*x)**2/(2*a**3*f + 2*a**2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b **2*f*tan(e + f*x)**2 + 2*a*b**2*f + 2*b**3*f*tan(e + f*x)**2) + b*log(tan (e + f*x)**2 + 1)*tan(e + f*x)**2/(2*a**3*f + 2*a**2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b**2*f*tan(e + f*x)**2 + 2*a*b**2*f + 2*b**3*f*tan(e + f *x)**2) - b/(2*a**3*f + 2*a**2*b*f*tan(e + f*x)**2 - 4*a**2*b*f - 4*a*b**2 *f*tan(e + f*x)**2 + 2*a*b**2*f + 2*b**3*f*tan(e + f*x)**2), True))
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {b}{a^{3} - 2 \, a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {\log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{2} - 2 \, a b + b^{2}}}{2 \, f} \]
-1/2*(b/(a^3 - 2*a^2*b + a*b^2 - (a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sin(f*x + e)^2) + log(-(a - b)*sin(f*x + e)^2 + a)/(a^2 - 2*a*b + b^2))/f
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (61) = 122\).
Time = 0.59 (sec) , antiderivative size = 287, normalized size of antiderivative = 4.42 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {\log \left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {2 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {a^{2} + \frac {2 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}}{2 \, f} \]
-1/2*(log(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e ) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/( a^2 - 2*a*b + b^2) - 2*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1) )/(a^2 - 2*a*b + b^2) - (a^2 + 2*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a^2*(cos(f*x + e) - 1)^2/ (cos(f*x + e) + 1)^2)/((a^3 - 2*a^2*b + a*b^2)*(a + 2*a*(cos(f*x + e) - 1) /(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f *x + e) - 1)^2/(cos(f*x + e) + 1)^2)))/f
Time = 11.05 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.00 \[ \int \frac {\tan (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {b\,\left (1+{\mathrm {tan}\left (e+f\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,a+a\,{\mathrm {tan}\left (e+f\,x\right )}^2+b\,{\mathrm {tan}\left (e+f\,x\right )}^2}\right )\,2{}\mathrm {i}\right )+a\,\left (-1+\mathrm {atan}\left (\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-b\,{\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}}{2\,a+a\,{\mathrm {tan}\left (e+f\,x\right )}^2+b\,{\mathrm {tan}\left (e+f\,x\right )}^2}\right )\,2{}\mathrm {i}\right )}{f\,\left (2\,a^3+2\,a^2\,b\,{\mathrm {tan}\left (e+f\,x\right )}^2-4\,a^2\,b-4\,a\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+2\,a\,b^2+2\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\right )} \]
-(b*(tan(e + f*x)^2*atan((a*tan(e + f*x)^2*1i - b*tan(e + f*x)^2*1i)/(2*a + a*tan(e + f*x)^2 + b*tan(e + f*x)^2))*2i + 1) + a*(atan((a*tan(e + f*x)^ 2*1i - b*tan(e + f*x)^2*1i)/(2*a + a*tan(e + f*x)^2 + b*tan(e + f*x)^2))*2 i - 1))/(f*(2*a*b^2 - 4*a^2*b + 2*a^3 + 2*b^3*tan(e + f*x)^2 - 4*a*b^2*tan (e + f*x)^2 + 2*a^2*b*tan(e + f*x)^2))