Integrand size = 14, antiderivative size = 97 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {x}{(a-b)^2}-\frac {(3 a-b) \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^2 f}-\frac {b \tan (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )} \]
x/(a-b)^2-1/2*(3*a-b)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/a^(3/2)/( a-b)^2/f-1/2*b*tan(f*x+e)/a/(a-b)/f/(a+b*tan(f*x+e)^2)
Time = 1.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {2 \arctan (\tan (e+f x))+\frac {\sqrt {b} (-3 a+b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {b (-a+b) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )}}{2 (a-b)^2 f} \]
(2*ArcTan[Tan[e + f*x]] + (Sqrt[b]*(-3*a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x] )/Sqrt[a]])/a^(3/2) + (b*(-a + b)*Tan[e + f*x])/(a*(a + b*Tan[e + f*x]^2)) )/(2*(a - b)^2*f)
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.20, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4144, 316, 397, 216, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-b \tan ^2(e+f x)+2 a-b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 a (a-b)}-\frac {b \tan (e+f x)}{2 a (a-b) \left (a+b \tan ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {2 a \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a-b}-\frac {b (3 a-b) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 a (a-b)}-\frac {b \tan (e+f x)}{2 a (a-b) \left (a+b \tan ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {2 a \arctan (\tan (e+f x))}{a-b}-\frac {b (3 a-b) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 a (a-b)}-\frac {b \tan (e+f x)}{2 a (a-b) \left (a+b \tan ^2(e+f x)\right )}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {2 a \arctan (\tan (e+f x))}{a-b}-\frac {\sqrt {b} (3 a-b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b \tan (e+f x)}{2 a (a-b) \left (a+b \tan ^2(e+f x)\right )}}{f}\) |
(((2*a*ArcTan[Tan[e + f*x]])/(a - b) - ((3*a - b)*Sqrt[b]*ArcTan[(Sqrt[b]* Tan[e + f*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - (b*Tan[e + f*x] )/(2*a*(a - b)*(a + b*Tan[e + f*x]^2)))/f
3.1.76.3.1 Defintions of rubi rules used
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[((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_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((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[(a + b* (ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.96
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\left (a -b \right ) \tan \left (f x +e \right )}{2 a \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a -b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{\left (a -b \right )^{2}}}{f}\) | \(93\) |
| default | \(\frac {-\frac {b \left (\frac {\left (a -b \right ) \tan \left (f x +e \right )}{2 a \left (a +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a -b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a -b \right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{\left (a -b \right )^{2}}}{f}\) | \(93\) |
| risch | \(\frac {x}{a^{2}-2 a b +b^{2}}+\frac {i b \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}{f a \left (-a +b \right )^{2} \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{4 a \left (a -b \right )^{2} f}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{4 a^{2} \left (a -b \right )^{2} f}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{4 a \left (a -b \right )^{2} f}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{4 a^{2} \left (a -b \right )^{2} f}\) | \(335\) |
1/f*(-b/(a-b)^2*(1/2/a*(a-b)*tan(f*x+e)/(a+b*tan(f*x+e)^2)+1/2*(3*a-b)/a/( a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2)))+1/(a-b)^2*arctan(tan(f*x+e)))
Time = 0.30 (sec) , antiderivative size = 390, normalized size of antiderivative = 4.02 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [\frac {8 \, a b f x \tan \left (f x + e\right )^{2} + 8 \, a^{2} f x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) - 4 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{8 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, \frac {4 \, a b f x \tan \left (f x + e\right )^{2} + 4 \, a^{2} f x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (f x + e\right )}\right ) - 2 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \]
[1/8*(8*a*b*f*x*tan(f*x + e)^2 + 8*a^2*f*x - ((3*a*b - b^2)*tan(f*x + e)^2 + 3*a^2 - a*b)*sqrt(-b/a)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 + 4*(a*b*tan(f*x + e)^3 - a^2*tan(f*x + e))*sqrt(-b/a))/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)) - 4*(a*b - b^2)*tan(f*x + e))/((a^3 *b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f), 1 /4*(4*a*b*f*x*tan(f*x + e)^2 + 4*a^2*f*x - ((3*a*b - b^2)*tan(f*x + e)^2 + 3*a^2 - a*b)*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(b/a)/(b*tan (f*x + e))) - 2*(a*b - b^2)*tan(f*x + e))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*t an(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f)]
Leaf count of result is larger than twice the leaf count of optimal. 2125 vs. \(2 (78) = 156\).
Time = 14.42 (sec) , antiderivative size = 2125, normalized size of antiderivative = 21.91 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \]
Piecewise((zoo*x/tan(e)**4, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (x/a**2, Eq(b , 0)), ((x + 1/(f*tan(e + f*x)) - 1/(3*f*tan(e + f*x)**3))/b**2, Eq(a, 0)) , (3*f*x*tan(e + f*x)**4/(8*b**2*f*tan(e + f*x)**4 + 16*b**2*f*tan(e + f*x )**2 + 8*b**2*f) + 6*f*x*tan(e + f*x)**2/(8*b**2*f*tan(e + f*x)**4 + 16*b* *2*f*tan(e + f*x)**2 + 8*b**2*f) + 3*f*x/(8*b**2*f*tan(e + f*x)**4 + 16*b* *2*f*tan(e + f*x)**2 + 8*b**2*f) + 3*tan(e + f*x)**3/(8*b**2*f*tan(e + f*x )**4 + 16*b**2*f*tan(e + f*x)**2 + 8*b**2*f) + 5*tan(e + f*x)/(8*b**2*f*ta n(e + f*x)**4 + 16*b**2*f*tan(e + f*x)**2 + 8*b**2*f), Eq(a, b)), (x/(a + b*tan(e)**2)**2, Eq(f, 0)), (4*a**2*f*x*sqrt(-a/b)/(4*a**4*f*sqrt(-a/b) + 4*a**3*b*f*sqrt(-a/b)*tan(e + f*x)**2 - 8*a**3*b*f*sqrt(-a/b) - 8*a**2*b** 2*f*sqrt(-a/b)*tan(e + f*x)**2 + 4*a**2*b**2*f*sqrt(-a/b) + 4*a*b**3*f*sqr t(-a/b)*tan(e + f*x)**2) - 3*a**2*log(-sqrt(-a/b) + tan(e + f*x))/(4*a**4* f*sqrt(-a/b) + 4*a**3*b*f*sqrt(-a/b)*tan(e + f*x)**2 - 8*a**3*b*f*sqrt(-a/ b) - 8*a**2*b**2*f*sqrt(-a/b)*tan(e + f*x)**2 + 4*a**2*b**2*f*sqrt(-a/b) + 4*a*b**3*f*sqrt(-a/b)*tan(e + f*x)**2) + 3*a**2*log(sqrt(-a/b) + tan(e + f*x))/(4*a**4*f*sqrt(-a/b) + 4*a**3*b*f*sqrt(-a/b)*tan(e + f*x)**2 - 8*a** 3*b*f*sqrt(-a/b) - 8*a**2*b**2*f*sqrt(-a/b)*tan(e + f*x)**2 + 4*a**2*b**2* f*sqrt(-a/b) + 4*a*b**3*f*sqrt(-a/b)*tan(e + f*x)**2) + 4*a*b*f*x*sqrt(-a/ b)*tan(e + f*x)**2/(4*a**4*f*sqrt(-a/b) + 4*a**3*b*f*sqrt(-a/b)*tan(e + f* x)**2 - 8*a**3*b*f*sqrt(-a/b) - 8*a**2*b**2*f*sqrt(-a/b)*tan(e + f*x)**...
Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {b \tan \left (f x + e\right )}{a^{3} - a^{2} b + {\left (a^{2} b - a b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {{\left (3 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}}}{2 \, f} \]
-1/2*(b*tan(f*x + e)/(a^3 - a^2*b + (a^2*b - a*b^2)*tan(f*x + e)^2) + (3*a *b - b^2)*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^3 - 2*a^2*b + a*b^2)*sqrt(a *b)) - 2*(f*x + e)/(a^2 - 2*a*b + b^2))/f
Time = 0.43 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b - b^{2}\right )}}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac {b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} {\left (a^{2} - a b\right )}}}{2 \, f} \]
-1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a* b)))*(3*a*b - b^2)/((a^3 - 2*a^2*b + a*b^2)*sqrt(a*b)) - 2*(f*x + e)/(a^2 - 2*a*b + b^2) + b*tan(f*x + e)/((b*tan(f*x + e)^2 + a)*(a^2 - a*b)))/f
Time = 12.40 (sec) , antiderivative size = 2489, normalized size of antiderivative = 25.66 \[ \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \]
(2*atan((((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(16* a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2)) /(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))/(2*a^2 - 4*a*b + 2 *b^2) + (tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^ 2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - ((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3 *b^2) + (tan(e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2 *b^2)))/(2*a^2 - 4*a*b + 2*b^2) - (tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^ 3))/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^2 - 4*a*b + 2*b^2))/((((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^ 4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i)/(2*a^2 - 4*a*b + 2*b^2) + (tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3)*1i)/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^ 2 - 4*a*b + 2*b^2) + (((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) + (tan( e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i...