Integrand size = 19, antiderivative size = 155 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b-a \tan (c+d x)}{\sqrt {a^2+b^2} \sqrt {\sec ^2(c+d x)}}\right ) \sec (c+d x)}{2 \left (a^2+b^2\right )^{5/2} d \sqrt {\sec ^2(c+d x)}}-\frac {b \sec (c+d x)}{2 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {3 a b \sec (c+d x)}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \]
-1/2*(2*a^2-b^2)*arctanh((b-a*tan(d*x+c))/(a^2+b^2)^(1/2)/(sec(d*x+c)^2)^( 1/2))*sec(d*x+c)/(a^2+b^2)^(5/2)/d/(sec(d*x+c)^2)^(1/2)-1/2*b*sec(d*x+c)/( a^2+b^2)/d/(a+b*tan(d*x+c))^2-3/2*a*b*sec(d*x+c)/(a^2+b^2)^2/d/(a+b*tan(d* x+c))
Time = 1.66 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {2 \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {-b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \sec (c+d x) \left (4 a^2+b^2+3 a b \tan (c+d x)\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}}{2 d} \]
((2*(2*a^2 - b^2)*ArcTanh[(-b + a*Tan[(c + d*x)/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (b*Sec[c + d*x]*(4*a^2 + b^2 + 3*a*b*Tan[c + d*x]))/((a^2 + b^2)^2*(a + b*Tan[c + d*x])^2))/(2*d)
Time = 0.37 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3992, 498, 25, 679, 488, 219}
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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3992 |
| \(\displaystyle \frac {\sec (c+d x) \int \frac {1}{(a+b \tan (c+d x))^3 \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\) |
| \(\Big \downarrow \) 498 |
| \(\displaystyle \frac {\sec (c+d x) \left (-\frac {\int -\frac {2 a-b \tan (c+d x)}{(a+b \tan (c+d x))^2 \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt {\tan ^2(c+d x)+1}}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sec (c+d x) \left (\frac {\int \frac {2 a-b \tan (c+d x)}{(a+b \tan (c+d x))^2 \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt {\tan ^2(c+d x)+1}}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
| \(\Big \downarrow \) 679 |
| \(\displaystyle \frac {\sec (c+d x) \left (\frac {\frac {\left (2 a^2-b^2\right ) \int \frac {1}{(a+b \tan (c+d x)) \sqrt {\tan ^2(c+d x)+1}}d(b \tan (c+d x))}{a^2+b^2}-\frac {3 a b^2 \sqrt {\tan ^2(c+d x)+1}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt {\tan ^2(c+d x)+1}}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
| \(\Big \downarrow \) 488 |
| \(\displaystyle \frac {\sec (c+d x) \left (\frac {-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{\frac {a^2}{b^2}-b^2 \tan ^2(c+d x)+1}d\frac {1-\frac {a \tan (c+d x)}{b}}{\sqrt {\tan ^2(c+d x)+1}}}{a^2+b^2}-\frac {3 a b^2 \sqrt {\tan ^2(c+d x)+1}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt {\tan ^2(c+d x)+1}}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sec (c+d x) \left (\frac {-\frac {b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {b^2 \tan (c+d x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}-\frac {3 a b^2 \sqrt {\tan ^2(c+d x)+1}}{\left (a^2+b^2\right ) (a+b \tan (c+d x))}}{2 \left (a^2+b^2\right )}-\frac {b^2 \sqrt {\tan ^2(c+d x)+1}}{2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\right )}{b d \sqrt {\sec ^2(c+d x)}}\) |
(Sec[c + d*x]*(-1/2*(b^2*Sqrt[1 + Tan[c + d*x]^2])/((a^2 + b^2)*(a + b*Tan [c + d*x])^2) + (-((b*(2*a^2 - b^2)*ArcTanh[(b^2*Tan[c + d*x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2)) - (3*a*b^2*Sqrt[1 + Tan[c + d*x]^2])/((a^2 + b^2 )*(a + b*Tan[c + d*x])))/(2*(a^2 + b^2))))/(b*d*Sqrt[Sec[c + d*x]^2])
3.6.75.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))), x] + S imp[b/((n + 1)*(b*c^2 + a*d^2)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p*(c*(n + 1) - d*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[n , -1] && ((LtQ[n, -1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]) || (SumSimp lerQ[n, 1] && IntegerQ[p]) || ILtQ[Simplify[n + 2*p + 3], 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[Sec[e + f*x]/(b*f*Sqrt[Sec[e + f*x]^2]) Subst[Int[( a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b , e, f, n}, x] && NeQ[a^2 + b^2, 0] && IntegerQ[(m - 1)/2]
Time = 3.62 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.81
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(280\) |
| default | \(\frac {-\frac {2 \left (-\frac {b^{2} \left (5 a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (4 a^{4}-7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}+2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{2}+b^{2}\right )}{2 a^{4}+4 a^{2} b^{2}+2 b^{4}}\right )}{{\left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-a \right )}^{2}}+\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}}}{d}\) | \(280\) |
| risch | \(-\frac {b \,{\mathrm e}^{i \left (d x +c \right )} \left (-3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 i a b +4 a^{2}+b^{2}\right )}{\left (i b +a \right )^{2} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+a \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +a \right )^{2} d \left (-i b +a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) a^{2}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right ) b^{2}}{2 \left (a^{2}+b^{2}\right )^{\frac {5}{2}} d}\) | \(451\) |
1/d*(-2*(-1/2*b^2*(5*a^2+2*b^2)/a/(a^4+2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^3 -1/2*b*(4*a^4-7*a^2*b^2-2*b^4)/(a^4+2*a^2*b^2+b^4)/a^2*tan(1/2*d*x+1/2*c)^ 2+1/2*b^2*(11*a^2+2*b^2)/(a^4+2*a^2*b^2+b^4)/a*tan(1/2*d*x+1/2*c)+1/2*b*(4 *a^2+b^2)/(a^4+2*a^2*b^2+b^4))/(tan(1/2*d*x+1/2*c)^2*a-2*b*tan(1/2*d*x+1/2 *c)-a)^2+(2*a^2-b^2)/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a* tan(1/2*d*x+1/2*c)-2*b)/(a^2+b^2)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (143) = 286\).
Time = 0.29 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.27 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {{\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{4} - 3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) - a \sin \left (d x + c\right )\right )}}{2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b + 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} d\right )}} \]
-1/4*((2*a^2*b^2 - b^4 + (2*a^4 - 3*a^2*b^2 + b^4)*cos(d*x + c)^2 + 2*(2*a ^3*b - a*b^3)*cos(d*x + c)*sin(d*x + c))*sqrt(a^2 + b^2)*log((2*a*b*cos(d* x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 - 2*a^2 - b^2 - 2*sqrt(a^ 2 + b^2)*(b*cos(d*x + c) - a*sin(d*x + c)))/(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)*cos(d*x + c)^2 + b^2)) + 2*(4*a^4*b + 5*a^2*b^3 + b^5)*co s(d*x + c) + 6*(a^3*b^2 + a*b^4)*sin(d*x + c))/((a^8 + 2*a^6*b^2 - 2*a^2*b ^6 - b^8)*d*cos(d*x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*d*c os(d*x + c)*sin(d*x + c) + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*d)
\[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (143) = 286\).
Time = 0.30 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.66 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (4 \, a^{4} b + a^{2} b^{3} + \frac {{\left (11 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {{\left (4 \, a^{4} b - 7 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {{\left (5 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4} + \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, {\left (a^{8} - 3 \, a^{4} b^{4} - 2 \, a^{2} b^{6}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {4 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {{\left (a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}}}{2 \, d} \]
-1/2*((2*a^2 - b^2)*log((b - a*sin(d*x + c)/(cos(d*x + c) + 1) + sqrt(a^2 + b^2))/(b - a*sin(d*x + c)/(cos(d*x + c) + 1) - sqrt(a^2 + b^2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + 2*(4*a^4*b + a^2*b^3 + (11*a^3*b^2 + 2*a*b^4)*sin(d*x + c)/(cos(d*x + c) + 1) - (4*a^4*b - 7*a^2*b^3 - 2*b^5)*s in(d*x + c)^2/(cos(d*x + c) + 1)^2 - (5*a^3*b^2 + 2*a*b^4)*sin(d*x + c)^3/ (cos(d*x + c) + 1)^3)/(a^8 + 2*a^6*b^2 + a^4*b^4 + 4*(a^7*b + 2*a^5*b^3 + a^3*b^5)*sin(d*x + c)/(cos(d*x + c) + 1) - 2*(a^8 - 3*a^4*b^4 - 2*a^2*b^6) *sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*(a^7*b + 2*a^5*b^3 + a^3*b^5)*sin (d*x + c)^3/(cos(d*x + c) + 1)^3 + (a^8 + 2*a^6*b^2 + a^4*b^4)*sin(d*x + c )^4/(cos(d*x + c) + 1)^4))/d
Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (143) = 286\).
Time = 0.54 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.89 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {{\left (2 \, a^{2} - b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, a^{4} b - a^{2} b^{3}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}^{2}}}{2 \, d} \]
-1/2*((2*a^2 - b^2)*log(abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*d*x + 1/2*c) - 2*b + 2*sqrt(a^2 + b^2)))/((a^4 + 2*a ^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2*(5*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*a *b^4*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*b*tan(1/2*d*x + 1/2*c)^2 - 7*a^2*b^3*t an(1/2*d*x + 1/2*c)^2 - 2*b^5*tan(1/2*d*x + 1/2*c)^2 - 11*a^3*b^2*tan(1/2* d*x + 1/2*c) - 2*a*b^4*tan(1/2*d*x + 1/2*c) - 4*a^4*b - a^2*b^3)/((a^6 + 2 *a^4*b^2 + a^2*b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - 2*b*tan(1/2*d*x + 1/2*c) - a)^2))/d
Time = 5.88 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.86 \[ \int \frac {\sec (c+d x)}{(a+b \tan (c+d x))^3} \, dx=\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}-a^4\,b-b^5-2\,a^2\,b^3+a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-\frac {b^2}{2}\right )}{d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\ln \left ({\left (a^2+b^2\right )}^{5/2}+a^4\,b+b^5+2\,a^2\,b^3-a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a^3\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,a^2-b^2\right )}{2\,d\,{\left (a^2+b^2\right )}^{5/2}}-\frac {\frac {4\,a^2\,b+b^3}{a^4+2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (a^2-2\,b^2\right )\,\left (4\,a^2\,b+b^3\right )}{a^2\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,a^2\,b+2\,b^3\right )}{a\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )} \]
(log((a^2 + b^2)^(5/2) - a^4*b - b^5 - 2*a^2*b^3 + a^5*tan(c/2 + (d*x)/2) + a*b^4*tan(c/2 + (d*x)/2) + 2*a^3*b^2*tan(c/2 + (d*x)/2))*(a^2 - b^2/2))/ (d*(a^2 + b^2)^(5/2)) - (log((a^2 + b^2)^(5/2) + a^4*b + b^5 + 2*a^2*b^3 - a^5*tan(c/2 + (d*x)/2) - a*b^4*tan(c/2 + (d*x)/2) - 2*a^3*b^2*tan(c/2 + ( d*x)/2))*(2*a^2 - b^2))/(2*d*(a^2 + b^2)^(5/2)) - ((4*a^2*b + b^3)/(a^4 + b^4 + 2*a^2*b^2) - (tan(c/2 + (d*x)/2)^2*(a^2 - 2*b^2)*(4*a^2*b + b^3))/(a ^2*(a^4 + b^4 + 2*a^2*b^2)) + (b*tan(c/2 + (d*x)/2)*(11*a^2*b + 2*b^3))/(a *(a^4 + b^4 + 2*a^2*b^2)) - (b*tan(c/2 + (d*x)/2)^3*(5*a^2*b + 2*b^3))/(a* (a^4 + b^4 + 2*a^2*b^2)))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2 )^2*(2*a^2 - 4*b^2) + a^2 - 4*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b*tan(c/2 + ( d*x)/2)))