Integrand size = 23, antiderivative size = 69 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\log \left (a \cos ^2(e+f x)+b \sin ^2(e+f x)\right )}{2 (a-b)^2 f}-\frac {a}{2 (a-b) b f \left (a+b \tan ^2(e+f x)\right )} \]
Time = 0.60 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {2 \log (\cos (e+f x))+\log \left (a+b \tan ^2(e+f x)\right )+\frac {a (-a+b)}{b \left (a+b \tan ^2(e+f x)\right )}}{2 (a-b)^2 f} \]
(2*Log[Cos[e + f*x]] + Log[a + b*Tan[e + f*x]^2] + (a*(-a + b))/(b*(a + b* Tan[e + f*x]^2)))/(2*(a - b)^2*f)
Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4153, 354, 86, 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.
| \(\displaystyle \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^3}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {\int \frac {\tan ^3(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (\frac {a}{(a-b) \left (b \tan ^2(e+f x)+a\right )^2}-\frac {1}{(a-b)^2 \left (\tan ^2(e+f x)+1\right )}+\frac {b}{(a-b)^2 \left (b \tan ^2(e+f x)+a\right )}\right )d\tan ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a}{b (a-b) \left (a+b \tan ^2(e+f x)\right )}-\frac {\log \left (\tan ^2(e+f x)+1\right )}{(a-b)^2}+\frac {\log \left (a+b \tan ^2(e+f x)\right )}{(a-b)^2}}{2 f}\) |
(-(Log[1 + Tan[e + f*x]^2]/(a - b)^2) + Log[a + b*Tan[e + f*x]^2]/(a - b)^ 2 - a/((a - b)*b*(a + b*Tan[e + f*x]^2)))/(2*f)
3.3.25.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(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] && IntegerQ [(m - 1)/2]
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 = 0.09 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}+\frac {-\frac {a \left (a -b \right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}+\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}}{f}\) | \(74\) |
| default | \(\frac {-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}+\frac {-\frac {a \left (a -b \right )}{b \left (a +b \tan \left (f x +e \right )^{2}\right )}+\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 \left (a -b \right )^{2}}}{f}\) | \(74\) |
| norman | \(\frac {\tan \left (f x +e \right )^{2}}{2 f \left (a -b \right ) \left (a +b \tan \left (f x +e \right )^{2}\right )}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left (a +b \tan \left (f x +e \right )^{2}\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}\) | \(96\) |
| parallelrisch | \(-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2} b^{2}-b^{2} \ln \left (a +b \tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right )^{2}+\ln \left (1+\tan \left (f x +e \right )^{2}\right ) a b -\ln \left (a +b \tan \left (f x +e \right )^{2}\right ) a b +a^{2}-a b}{2 \left (a^{2}-2 a b +b^{2}\right ) \left (a +b \tan \left (f x +e \right )^{2}\right ) b f}\) | \(124\) |
| risch | \(-\frac {i x}{a^{2}-2 a b +b^{2}}-\frac {2 i e}{f \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 a \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {\ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}\) | \(166\) |
1/f*(-1/2/(a-b)^2*ln(1+tan(f*x+e)^2)+1/2/(a-b)^2*(-a*(a-b)/b/(a+b*tan(f*x+ e)^2)+ln(a+b*tan(f*x+e)^2)))
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.42 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {a \tan \left (f x + e\right )^{2} + {\left (b \tan \left (f x + e\right )^{2} + a\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) + a}{2 \, {\left ({\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} f\right )}} \]
1/2*(a*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)) + a)/((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. 910 vs. \(2 (51) = 102\).
Time = 13.98 (sec) , antiderivative size = 910, normalized size of antiderivative = 13.19 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\begin {cases} \frac {\tilde {\infty } x}{\tan {\left (e \right )}} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {- \frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\tan ^{2}{\left (e + f x \right )}}{2 f}}{a^{2}} & \text {for}\: b = 0 \\- \frac {2 \tan ^{2}{\left (e + f x \right )}}{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} - \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 ^{3}{\left (e \right )}}{\left (a + b \tan ^{2}{\left (e \right )}\right )^{2}} & \text {for}\: f = 0 \\- \frac {a^{2}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} + \frac {a b \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} + \frac {a b \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} - \frac {a b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} + \frac {a b}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} + \frac {b^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} + \frac {b^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} - \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b f + 2 a^{2} b^{2} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{2} f - 4 a b^{3} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{3} f + 2 b^{4} f \tan ^{2}{\left (e + f x \right )}} & \text {otherwise} \end {cases} \]
Piecewise((zoo*x/tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((-log(tan(e + f *x)**2 + 1)/(2*f) + tan(e + f*x)**2/(2*f))/a**2, Eq(b, 0)), (-2*tan(e + f* x)**2/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) - 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)**3/(a + b*tan(e)**2)**2, Eq(f, 0)), (-a**2/(2*a**3*b*f + 2*a **2*b**2*f*tan(e + f*x)**2 - 4*a**2*b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*x)**2) + a*b*log(-sqrt(-a/b) + tan(e + f*x ))/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x)**2 - 4*a**2*b**2*f - 4*a*b**3* f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*x)**2) + a*b*log(sqrt( -a/b) + tan(e + f*x))/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x)**2 - 4*a**2 *b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*x)* *2) - a*b*log(tan(e + f*x)**2 + 1)/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x )**2 - 4*a**2*b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f* tan(e + f*x)**2) + a*b/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x)**2 - 4*a** 2*b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*x) **2) + b**2*log(-sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x)**2 - 4*a**2*b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*x)**2) + b**2*log(sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**3*b*f + 2*a**2*b**2*f*tan(e + f*x)**2 - 4*a**2 *b**2*f - 4*a*b**3*f*tan(e + f*x)**2 + 2*a*b**3*f + 2*b**4*f*tan(e + f*...
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.28 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {a}{a^{3} - 2 \, a^{2} b + a b^{2} - {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {\log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{2} - 2 \, a b + b^{2}}}{2 \, f} \]
1/2*(a/(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. 275 vs. \(2 (65) = 130\).
Time = 0.85 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.99 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {\log \left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {2 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {a + \frac {6 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (a + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}}{2 \, f} \]
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 + 6*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 8* 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)*(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 = 10.96 (sec) , antiderivative size = 270, normalized size of antiderivative = 3.91 \[ \int \frac {\tan ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {a^2\,{\cos \left (e+f\,x\right )}^2}{2}+b^2\,{\sin \left (e+f\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\sin \left (e+f\,x\right )}^2-b\,{\sin \left (e+f\,x\right )}^2}{a\,{\cos \left (e+f\,x\right )}^2\,2{}\mathrm {i}+a\,{\sin \left (e+f\,x\right )}^2\,1{}\mathrm {i}+b\,{\sin \left (e+f\,x\right )}^2\,1{}\mathrm {i}}\right )\,1{}\mathrm {i}-\frac {a\,b\,{\cos \left (e+f\,x\right )}^2}{2}+a\,b\,{\cos \left (e+f\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\sin \left (e+f\,x\right )}^2-b\,{\sin \left (e+f\,x\right )}^2}{a\,{\cos \left (e+f\,x\right )}^2\,2{}\mathrm {i}+a\,{\sin \left (e+f\,x\right )}^2\,1{}\mathrm {i}+b\,{\sin \left (e+f\,x\right )}^2\,1{}\mathrm {i}}\right )\,1{}\mathrm {i}}{f\,\left (a^3\,b\,{\cos \left (e+f\,x\right )}^2-2\,a^2\,b^2\,{\cos \left (e+f\,x\right )}^2+a^2\,b^2\,{\sin \left (e+f\,x\right )}^2+a\,b^3\,{\cos \left (e+f\,x\right )}^2-2\,a\,b^3\,{\sin \left (e+f\,x\right )}^2+b^4\,{\sin \left (e+f\,x\right )}^2\right )} \]
-((a^2*cos(e + f*x)^2)/2 + b^2*sin(e + f*x)^2*atan((a*sin(e + f*x)^2 - b*s in(e + f*x)^2)/(a*cos(e + f*x)^2*2i + a*sin(e + f*x)^2*1i + b*sin(e + f*x) ^2*1i))*1i - (a*b*cos(e + f*x)^2)/2 + a*b*cos(e + f*x)^2*atan((a*sin(e + f *x)^2 - b*sin(e + f*x)^2)/(a*cos(e + f*x)^2*2i + a*sin(e + f*x)^2*1i + b*s in(e + f*x)^2*1i))*1i)/(f*(b^4*sin(e + f*x)^2 + a*b^3*cos(e + f*x)^2 + a^3 *b*cos(e + f*x)^2 - 2*a*b^3*sin(e + f*x)^2 - 2*a^2*b^2*cos(e + f*x)^2 + a^ 2*b^2*sin(e + f*x)^2))