Integrand size = 19, antiderivative size = 133 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {b \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {3 a b \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
(2*a^2+b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2 )/(a+b)^(5/2)/d-1/2*b*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-3/2*a*b*ta n(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
Time = 0.59 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b \left (-3 a b+\left (-4 a^2+b^2\right ) \cos (c+d x)\right ) \sin (c+d x)}{(b+a \cos (c+d x))^2}}{2 (a-b)^2 (a+b)^2 d} \]
((-2*(2*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/S qrt[a^2 - b^2] + (b*(-3*a*b + (-4*a^2 + b^2)*Cos[c + d*x])*Sin[c + d*x])/( b + a*Cos[c + d*x])^2)/(2*(a - b)^2*(a + b)^2*d)
Time = 0.68 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.20, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4320, 25, 3042, 4491, 25, 27, 3042, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4320 |
| \(\displaystyle -\frac {\int -\frac {\sec (c+d x) (2 a-b \sec (c+d x))}{(a+b \sec (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) (2 a-b \sec (c+d x))}{(a+b \sec (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a-b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4491 |
| \(\displaystyle \frac {-\frac {\int -\frac {\left (2 a^2+b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 a^2+b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (2 a^2+b^2\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2+b^2\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d \left (a^2-b^2\right )}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 \left (2 a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {3 a b \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
-1/2*(b*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ((2*(2*a^2 + b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*S qrt[a + b]*(a^2 - b^2)*d) - (3*a*b*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec [c + d*x])))/(2*(a^2 - b^2))
3.6.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)* (a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b* Csc[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + 2)*Csc[e + f*x]), x], x] /; FreeQ [{a, b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 /((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp [(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m , -1]
Time = 0.51 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.40
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {\left (4 a +b \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(186\) |
| default | \(\frac {-\frac {2 \left (-\frac {\left (4 a +b \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(186\) |
| risch | \(\frac {i b \left (-5 a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-4 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-7 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-11 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+2 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}-4 a^{4}+a^{2} b^{2}\right )}{a^{2} \left (-a^{2}+b^{2}\right )^{2} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right ) b^{2}}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(503\) |
1/d*(-2*(-1/2*(4*a+b)*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(4* a-b)*b/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-t an(1/2*d*x+1/2*c)^2*b-a-b)^2+(2*a^2+b^2)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b)) ^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (120) = 240\).
Time = 0.32 (sec) , antiderivative size = 595, normalized size of antiderivative = 4.47 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\left [\frac {{\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (3 \, a^{3} b^{2} - 3 \, a b^{4} + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\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 ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}, \frac {{\left (2 \, a^{2} b^{2} + b^{4} + {\left (2 \, a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (3 \, a^{3} b^{2} - 3 \, a b^{4} + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\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 ) + {\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d\right )}}\right ] \]
[1/4*((2*a^2*b^2 + b^4 + (2*a^4 + a^2*b^2)*cos(d*x + c)^2 + 2*(2*a^3*b + a *b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2 )*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2 *a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(3*a^3*b^ 2 - 3*a*b^4 + (4*a^4*b - 5*a^2*b^3 + b^5)*cos(d*x + c))*sin(d*x + c))/((a^ 8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b ^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d), 1/2*((2*a^2*b^2 + b^4 + (2*a^4 + a^2*b^2)*cos(d*x + c)^2 + 2*(2 *a^3*b + a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b *cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (3*a^3*b^2 - 3*a*b^4 + (4 *a^4*b - 5*a^2*b^3 + b^5)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d)]
\[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (120) = 240\).
Time = 0.37 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.91 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} {\left (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \]
-((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*(2*a^2 + b^2)/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - (4*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 3*a*b^2*tan(1/2*d*x + 1/2*c)^3 - b^3*tan(1/2*d*x + 1/2*c)^3 - 4*a^2*b*tan( 1/2*d*x + 1/2*c) - 3*a*b^2*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) )/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2 *c)^2 - a - b)^2))/d
Time = 16.17 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.53 \[ \int \frac {\sec (c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (b^2+4\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,a\,b-b^2\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,a^2+b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
((tan(c/2 + (d*x)/2)^3*(4*a*b + b^2))/((a + b)^2*(a - b)) - (tan(c/2 + (d* x)/2)*(4*a*b - b^2))/((a + b)*(a^2 - 2*a*b + b^2)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a^2 - 2*a*b + b^2) + a^ 2 + b^2)) + (atanh((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(a^2 - 2*a*b + b^2))/(2 *(a + b)^(1/2)*(a - b)^(5/2)))*(2*a^2 + b^2))/(d*(a + b)^(5/2)*(a - b)^(5/ 2))