Integrand size = 14, antiderivative size = 150 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\frac {x}{(a-b)^3}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{5/2} (a-b)^3 d}-\frac {b \tan (c+d x)}{4 a (a-b) d \left (a+b \tan ^2(c+d x)\right )^2}-\frac {(7 a-3 b) b \tan (c+d x)}{8 a^2 (a-b)^2 d \left (a+b \tan ^2(c+d x)\right )} \]
x/(a-b)^3-1/8*(15*a^2-10*a*b+3*b^2)*arctan(b^(1/2)*tan(d*x+c)/a^(1/2))*b^( 1/2)/a^(5/2)/(a-b)^3/d-1/4*b*tan(d*x+c)/a/(a-b)/d/(a+b*tan(d*x+c)^2)^2-1/8 *(7*a-3*b)*b*tan(d*x+c)/a^2/(a-b)^2/d/(a+b*tan(d*x+c)^2)
Time = 2.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {-8 \arctan (\tan (c+d x))+\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 (a-b)^2 b \tan (c+d x)}{a \left (a+b \tan ^2(c+d x)\right )^2}+\frac {(7 a-3 b) (a-b) b \tan (c+d x)}{a^2 \left (a+b \tan ^2(c+d x)\right )}}{8 (a-b)^3 d} \]
-1/8*(-8*ArcTan[Tan[c + d*x]] + (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[ (Sqrt[b]*Tan[c + d*x])/Sqrt[a]])/a^(5/2) + (2*(a - b)^2*b*Tan[c + d*x])/(a *(a + b*Tan[c + d*x]^2)^2) + ((7*a - 3*b)*(a - b)*b*Tan[c + d*x])/(a^2*(a + b*Tan[c + d*x]^2)))/((a - b)^3*d)
Time = 0.33 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4144, 316, 402, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \tan (c+d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^3}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \tan ^2(c+d x)+4 a-3 b}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )^2}d\tan (c+d x)}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2-7 b a+3 b^2-(7 a-3 b) b \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left (b \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\tan (c+d x))}{a-b}-\frac {b \left (15 a^2-10 a b+3 b^2\right ) \int \frac {1}{b \tan ^2(c+d x)+a}d\tan (c+d x)}{a-b}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^2 \arctan (\tan (c+d x))}{a-b}-\frac {\sqrt {b} \left (15 a^2-10 a b+3 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{2 a (a-b)}-\frac {b (7 a-3 b) \tan (c+d x)}{2 a (a-b) \left (a+b \tan ^2(c+d x)\right )}}{4 a (a-b)}-\frac {b \tan (c+d x)}{4 a (a-b) \left (a+b \tan ^2(c+d x)\right )^2}}{d}\) |
(-1/4*(b*Tan[c + d*x])/(a*(a - b)*(a + b*Tan[c + d*x]^2)^2) + (((8*a^2*Arc Tan[Tan[c + d*x]])/(a - b) - (Sqrt[b]*(15*a^2 - 10*a*b + 3*b^2)*ArcTan[(Sq rt[b]*Tan[c + d*x])/Sqrt[a]])/(Sqrt[a]*(a - b)))/(2*a*(a - b)) - ((7*a - 3 *b)*b*Tan[c + d*x])/(2*a*(a - b)*(a + b*Tan[c + d*x]^2)))/(4*a*(a - b)))/d
3.3.56.3.1 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_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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 = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \tan \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tan \left (d x +c \right )}{8 a}}{\left (a +b \tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{3}}}{d}\) | \(142\) |
| default | \(\frac {-\frac {b \left (\frac {\frac {b \left (7 a^{2}-10 a b +3 b^{2}\right ) \tan \left (d x +c \right )^{3}}{8 a^{2}}+\frac {\left (9 a^{2}-14 a b +5 b^{2}\right ) \tan \left (d x +c \right )}{8 a}}{\left (a +b \tan \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}-10 a b +3 b^{2}\right ) \arctan \left (\frac {b \tan \left (d x +c \right )}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}\right )}{\left (a -b \right )^{3}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{\left (a -b \right )^{3}}}{d}\) | \(142\) |
| risch | \(\frac {x}{a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}}+\frac {i \left (9 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-13 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+9 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+21 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+27 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-13 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-23 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+9 a^{3}-21 a^{2} b +15 a \,b^{2}-3 b^{3}\right ) b}{4 \left (-a \,{\mathrm e}^{4 i \left (d x +c \right )}+b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-a +b \right )^{2} d \left (-a +b \right )^{3} a^{2}}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{16 a \left (a -b \right )^{3} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{8 a^{2} \left (a -b \right )^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b^{2}}{16 a^{3} \left (a -b \right )^{3} d}\) | \(629\) |
1/d*(-b/(a-b)^3*((1/8*b*(7*a^2-10*a*b+3*b^2)/a^2*tan(d*x+c)^3+1/8*(9*a^2-1 4*a*b+5*b^2)/a*tan(d*x+c))/(a+b*tan(d*x+c)^2)^2+1/8*(15*a^2-10*a*b+3*b^2)/ a^2/(a*b)^(1/2)*arctan(b*tan(d*x+c)/(a*b)^(1/2)))+1/(a-b)^3*arctan(tan(d*x +c)))
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (136) = 272\).
Time = 0.33 (sec) , antiderivative size = 742, normalized size of antiderivative = 4.95 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\left [\frac {32 \, a^{2} b^{2} d x \tan \left (d x + c\right )^{4} + 64 \, a^{3} b d x \tan \left (d x + c\right )^{2} + 32 \, a^{4} d x - 4 \, {\left (7 \, a^{2} b^{2} - 10 \, a b^{3} + 3 \, b^{4}\right )} \tan \left (d x + c\right )^{3} - {\left ({\left (15 \, a^{2} b^{2} - 10 \, a b^{3} + 3 \, b^{4}\right )} \tan \left (d x + c\right )^{4} + 15 \, a^{4} - 10 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, {\left (15 \, a^{3} b - 10 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \tan \left (d x + c\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{4} - 6 \, a b \tan \left (d x + c\right )^{2} + a^{2} + 4 \, {\left (a b \tan \left (d x + c\right )^{3} - a^{2} \tan \left (d x + c\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (d x + c\right )^{4} + 2 \, a b \tan \left (d x + c\right )^{2} + a^{2}}\right ) - 4 \, {\left (9 \, a^{3} b - 14 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \tan \left (d x + c\right )}{32 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} d \tan \left (d x + c\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} d\right )}}, \frac {16 \, a^{2} b^{2} d x \tan \left (d x + c\right )^{4} + 32 \, a^{3} b d x \tan \left (d x + c\right )^{2} + 16 \, a^{4} d x - 2 \, {\left (7 \, a^{2} b^{2} - 10 \, a b^{3} + 3 \, b^{4}\right )} \tan \left (d x + c\right )^{3} - {\left ({\left (15 \, a^{2} b^{2} - 10 \, a b^{3} + 3 \, b^{4}\right )} \tan \left (d x + c\right )^{4} + 15 \, a^{4} - 10 \, a^{3} b + 3 \, a^{2} b^{2} + 2 \, {\left (15 \, a^{3} b - 10 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \tan \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (d x + c\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (d x + c\right )}\right ) - 2 \, {\left (9 \, a^{3} b - 14 \, a^{2} b^{2} + 5 \, a b^{3}\right )} \tan \left (d x + c\right )}{16 \, {\left ({\left (a^{5} b^{2} - 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} - a^{2} b^{5}\right )} d \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{6} b - 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} - a^{3} b^{4}\right )} d \tan \left (d x + c\right )^{2} + {\left (a^{7} - 3 \, a^{6} b + 3 \, a^{5} b^{2} - a^{4} b^{3}\right )} d\right )}}\right ] \]
[1/32*(32*a^2*b^2*d*x*tan(d*x + c)^4 + 64*a^3*b*d*x*tan(d*x + c)^2 + 32*a^ 4*d*x - 4*(7*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^3 - ((15*a^2*b^2 - 1 0*a*b^3 + 3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*a^ 3*b - 10*a^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*sqrt(-b/a)*log((b^2*tan(d*x + c)^4 - 6*a*b*tan(d*x + c)^2 + a^2 + 4*(a*b*tan(d*x + c)^3 - a^2*tan(d*x + c))*sqrt(-b/a))/(b^2*tan(d*x + c)^4 + 2*a*b*tan(d*x + c)^2 + a^2)) - 4*(9* a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d*x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3* b^4 - a^2*b^5)*d*tan(d*x + c)^4 + 2*(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b ^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*d), 1/16*(16* a^2*b^2*d*x*tan(d*x + c)^4 + 32*a^3*b*d*x*tan(d*x + c)^2 + 16*a^4*d*x - 2* (7*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^3 - ((15*a^2*b^2 - 10*a*b^3 + 3*b^4)*tan(d*x + c)^4 + 15*a^4 - 10*a^3*b + 3*a^2*b^2 + 2*(15*a^3*b - 10*a ^2*b^2 + 3*a*b^3)*tan(d*x + c)^2)*sqrt(b/a)*arctan(1/2*(b*tan(d*x + c)^2 - a)*sqrt(b/a)/(b*tan(d*x + c))) - 2*(9*a^3*b - 14*a^2*b^2 + 5*a*b^3)*tan(d *x + c))/((a^5*b^2 - 3*a^4*b^3 + 3*a^3*b^4 - a^2*b^5)*d*tan(d*x + c)^4 + 2 *(a^6*b - 3*a^5*b^2 + 3*a^4*b^3 - a^3*b^4)*d*tan(d*x + c)^2 + (a^7 - 3*a^6 *b + 3*a^5*b^2 - a^4*b^3)*d)]
Leaf count of result is larger than twice the leaf count of optimal. 8964 vs. \(2 (133) = 266\).
Time = 72.26 (sec) , antiderivative size = 8964, normalized size of antiderivative = 59.76 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Piecewise((zoo*x/tan(c)**6, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (x/a**3, Eq(b , 0)), ((-x - 1/(d*tan(c + d*x)) + 1/(3*d*tan(c + d*x)**3) - 1/(5*d*tan(c + d*x)**5))/b**3, Eq(a, 0)), (15*d*x*tan(c + d*x)**6/(48*b**3*d*tan(c + d* x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3* d) + 45*d*x*tan(c + d*x)**4/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d) + 45*d*x*tan(c + d*x)* *2/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*ta n(c + d*x)**2 + 48*b**3*d) + 15*d*x/(48*b**3*d*tan(c + d*x)**6 + 144*b**3* d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d) + 15*tan(c + d *x)**5/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3* d*tan(c + d*x)**2 + 48*b**3*d) + 40*tan(c + d*x)**3/(48*b**3*d*tan(c + d*x )**6 + 144*b**3*d*tan(c + d*x)**4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d ) + 33*tan(c + d*x)/(48*b**3*d*tan(c + d*x)**6 + 144*b**3*d*tan(c + d*x)** 4 + 144*b**3*d*tan(c + d*x)**2 + 48*b**3*d), Eq(a, b)), (x/(a + b*tan(c)** 2)**3, Eq(d, 0)), (16*a**4*d*x*sqrt(-a/b)/(16*a**7*d*sqrt(-a/b) + 32*a**6* b*d*sqrt(-a/b)*tan(c + d*x)**2 - 48*a**6*b*d*sqrt(-a/b) + 16*a**5*b**2*d*s qrt(-a/b)*tan(c + d*x)**4 - 96*a**5*b**2*d*sqrt(-a/b)*tan(c + d*x)**2 + 48 *a**5*b**2*d*sqrt(-a/b) - 48*a**4*b**3*d*sqrt(-a/b)*tan(c + d*x)**4 + 96*a **4*b**3*d*sqrt(-a/b)*tan(c + d*x)**2 - 16*a**4*b**3*d*sqrt(-a/b) + 48*a** 3*b**4*d*sqrt(-a/b)*tan(c + d*x)**4 - 32*a**3*b**4*d*sqrt(-a/b)*tan(c +...
Time = 0.44 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt {a b}} + \frac {{\left (7 \, a b^{2} - 3 \, b^{3}\right )} \tan \left (d x + c\right )^{3} + {\left (9 \, a^{2} b - 5 \, a b^{2}\right )} \tan \left (d x + c\right )}{a^{6} - 2 \, a^{5} b + a^{4} b^{2} + {\left (a^{4} b^{2} - 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{5} b - 2 \, a^{4} b^{2} + a^{3} b^{3}\right )} \tan \left (d x + c\right )^{2}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}}}{8 \, d} \]
-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*arctan(b*tan(d*x + c)/sqrt(a*b))/((a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)*sqrt(a*b)) + ((7*a*b^2 - 3*b^3)*tan(d*x + c)^3 + (9*a^2*b - 5*a*b^2)*tan(d*x + c))/(a^6 - 2*a^5*b + a^4*b^2 + (a^4 *b^2 - 2*a^3*b^3 + a^2*b^4)*tan(d*x + c)^4 + 2*(a^5*b - 2*a^4*b^2 + a^3*b^ 3)*tan(d*x + c)^2) - 8*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3))/d
Time = 0.50 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b - 10 \, a b^{2} + 3 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (d x + c\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sqrt {a b}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {7 \, a b^{2} \tan \left (d x + c\right )^{3} - 3 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a^{2} b \tan \left (d x + c\right ) - 5 \, a b^{2} \tan \left (d x + c\right )}{{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \]
-1/8*((15*a^2*b - 10*a*b^2 + 3*b^3)*(pi*floor((d*x + c)/pi + 1/2)*sgn(b) + arctan(b*tan(d*x + c)/sqrt(a*b)))/((a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b^3)* sqrt(a*b)) - 8*(d*x + c)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (7*a*b^2*tan(d* x + c)^3 - 3*b^3*tan(d*x + c)^3 + 9*a^2*b*tan(d*x + c) - 5*a*b^2*tan(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*(b*tan(d*x + c)^2 + a)^2))/d
Time = 13.28 (sec) , antiderivative size = 3901, normalized size of antiderivative = 26.01 \[ \int \frac {1}{\left (a+b \tan ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
(atan((((-a^5*b)^(1/2)*((tan(c + d*x)*(9*b^7 - 60*a*b^6 + 190*a^2*b^5 - 30 0*a^3*b^4 + 289*a^4*b^3))/(32*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6 *b^2)) - (((96*a^2*b^10 - 800*a^3*b^9 + 3040*a^4*b^8 - 6816*a^5*b^7 + 9760 *a^6*b^6 - 9056*a^7*b^5 + 5280*a^8*b^4 - 1760*a^9*b^3 + 256*a^10*b^2)/(64* (a^10 - 6*a^9*b + a^4*b^6 - 6*a^5*b^5 + 15*a^6*b^4 - 20*a^7*b^3 + 15*a^8*b ^2)) - (tan(c + d*x)*(-a^5*b)^(1/2)*(15*a^2 - 10*a*b + 3*b^2)*(256*a^4*b^9 - 1280*a^5*b^8 + 2304*a^6*b^7 - 1280*a^7*b^6 - 1280*a^8*b^5 + 2304*a^9*b^ 4 - 1280*a^10*b^3 + 256*a^11*b^2))/(512*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b ^2)*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)))*(-a^5*b)^(1/2)*(15 *a^2 - 10*a*b + 3*b^2))/(16*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)))*(15*a^ 2 - 10*a*b + 3*b^2)*1i)/(16*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)) + ((-a^ 5*b)^(1/2)*((tan(c + d*x)*(9*b^7 - 60*a*b^6 + 190*a^2*b^5 - 300*a^3*b^4 + 289*a^4*b^3))/(32*(a^8 - 4*a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)) + ((( 96*a^2*b^10 - 800*a^3*b^9 + 3040*a^4*b^8 - 6816*a^5*b^7 + 9760*a^6*b^6 - 9 056*a^7*b^5 + 5280*a^8*b^4 - 1760*a^9*b^3 + 256*a^10*b^2)/(64*(a^10 - 6*a^ 9*b + a^4*b^6 - 6*a^5*b^5 + 15*a^6*b^4 - 20*a^7*b^3 + 15*a^8*b^2)) + (tan( c + d*x)*(-a^5*b)^(1/2)*(15*a^2 - 10*a*b + 3*b^2)*(256*a^4*b^9 - 1280*a^5* b^8 + 2304*a^6*b^7 - 1280*a^7*b^6 - 1280*a^8*b^5 + 2304*a^9*b^4 - 1280*a^1 0*b^3 + 256*a^11*b^2))/(512*(3*a^7*b - a^8 + a^5*b^3 - 3*a^6*b^2)*(a^8 - 4 *a^7*b + a^4*b^4 - 4*a^5*b^3 + 6*a^6*b^2)))*(-a^5*b)^(1/2)*(15*a^2 - 10...