Integrand size = 32, antiderivative size = 164 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\left (2 a^2 B+b^2 B-3 a b C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {(b B-a C) \tan (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (3 a b B-a^2 C-2 b^2 C\right ) \tan (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
(2*B*a^2+B*b^2-3*C*a*b)*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*b-C*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec( d*x+c))^2-1/2*(3*B*a*b-C*a^2-2*C*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d* x+c))
Time = 1.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.05 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 \left (2 a^2 B+b^2 B-3 a b C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {b (b B-a C) \sin (c+d x)}{a (a-b) (a+b) (b+a \cos (c+d x))^2}+\frac {\left (-4 a^2 b B+b^3 B+2 a^3 C+a b^2 C\right ) \sin (c+d x)}{a (a-b)^2 (a+b)^2 (b+a \cos (c+d x))}}{2 d} \]
((-2*(2*a^2*B + b^2*B - 3*a*b*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[ a^2 - b^2]])/(a^2 - b^2)^(5/2) + (b*(b*B - a*C)*Sin[c + d*x])/(a*(a - b)*( a + b)*(b + a*Cos[c + d*x])^2) + ((-4*a^2*b*B + b^3*B + 2*a^3*C + a*b^2*C) *Sin[c + d*x])/(a*(a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])))/(2*d)
Time = 0.77 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.19, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4548, 25, 3042, 4548, 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.
| \(\displaystyle \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle -\frac {\int -\frac {2 a (a B-b C) \sec (c+d x)-a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 a (a B-b C) \sec (c+d x)-a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^2}dx}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (a B-b C) \csc \left (c+d x+\frac {\pi }{2}\right )-a (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {-\frac {\int -\frac {a^2 \left (2 B a^2-3 b C a+b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 \left (2 B a^2-3 b C a+b^2 B\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a \left (2 a^2 B-3 a b C+b^2 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (2 a^2 B-3 a b C+b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {a \left (2 a^2 B-3 a b C+b^2 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (2 a^2 B-3 a b C+b^2 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 a \left (2 a^2 B-3 a b C+b^2 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a \left (2 a^2 B-3 a b C+b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {a \left (a^2 (-C)+3 a b B-2 b^2 C\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 a \left (a^2-b^2\right )}-\frac {(b B-a C) \tan (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
-1/2*((b*B - a*C)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])^2) + ( (2*a*(2*a^2*B + b^2*B - 3*a*b*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sq rt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - (a*(3*a*b*B - a^2*C - 2*b^2*C)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*a*(a^2 - b^2))
3.9.11.3.1 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[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^( m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x ] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Time = 0.36 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.44
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {\left (4 B a b +B \,b^{2}-2 C \,a^{2}-C a b -2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 B a b -B \,b^{2}-2 C \,a^{2}+C a b -2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 B \,a^{2}+B \,b^{2}-3 C a b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(236\) |
| default | \(\frac {-\frac {2 \left (-\frac {\left (4 B a b +B \,b^{2}-2 C \,a^{2}-C a b -2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 B a b -B \,b^{2}-2 C \,a^{2}+C a b -2 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 B \,a^{2}+B \,b^{2}-3 C a b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(236\) |
| risch | \(\frac {i \left (-5 B \,a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 B a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+3 C \,a^{4} b \,{\mathrm e}^{3 i \left (d x +c \right )}-4 B \,a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}-7 B \,a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 B \,b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+2 C \,a^{5} {\mathrm e}^{2 i \left (d x +c \right )}+5 C \,a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 C a \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-11 b^{2} B \,{\mathrm e}^{i \left (d x +c \right )} a^{3}+2 b^{4} B a \,{\mathrm e}^{i \left (d x +c \right )}+5 C \,a^{4} b \,{\mathrm e}^{i \left (d x +c \right )}+4 C \,a^{2} b^{3} {\mathrm e}^{i \left (d x +c \right )}-4 B \,a^{4} b +B \,a^{2} b^{3}+2 a^{5} C +C \,a^{3} 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 {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B \,a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{2 \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}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B \,a^{2}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}\) | \(802\) |
1/d*(-2*(-1/2*(4*B*a*b+B*b^2-2*C*a^2-C*a*b-2*C*b^2)/(a-b)/(a^2+2*a*b+b^2)* tan(1/2*d*x+1/2*c)^3+1/2*(4*B*a*b-B*b^2-2*C*a^2+C*a*b-2*C*b^2)/(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-tan(1/2*d*x+1/2*c)^ 2*b-a-b)^2+(2*B*a^2+B*b^2-3*C*a*b)/(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. 347 vs. \(2 (150) = 300\).
Time = 0.32 (sec) , antiderivative size = 752, normalized size of antiderivative = 4.59 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\left [\frac {{\left (2 \, B a^{2} b^{2} - 3 \, C a b^{3} + B b^{4} + {\left (2 \, B a^{4} - 3 \, C a^{3} b + B a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, B a^{3} b - 3 \, C a^{2} b^{2} + 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 (C a^{4} b - 3 \, B a^{3} b^{2} + C a^{2} b^{3} + 3 \, B a b^{4} - 2 \, C b^{5} + {\left (2 \, C a^{5} - 4 \, B a^{4} b - C a^{3} b^{2} + 5 \, B a^{2} b^{3} - C a b^{4} - B 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 \, B a^{2} b^{2} - 3 \, C a b^{3} + B b^{4} + {\left (2 \, B a^{4} - 3 \, C a^{3} b + B a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, B a^{3} b - 3 \, C a^{2} b^{2} + 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 (C a^{4} b - 3 \, B a^{3} b^{2} + C a^{2} b^{3} + 3 \, B a b^{4} - 2 \, C b^{5} + {\left (2 \, C a^{5} - 4 \, B a^{4} b - C a^{3} b^{2} + 5 \, B a^{2} b^{3} - C a b^{4} - B 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*B*a^2*b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2) *cos(d*x + c)^2 + 2*(2*B*a^3*b - 3*C*a^2*b^2 + 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*sqr t(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*(C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^ 3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B*a^4*b - C*a^3*b^2 + 5*B*a^2*b^3 - C*a*b^4 - B*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*B*a^2 *b^2 - 3*C*a*b^3 + B*b^4 + (2*B*a^4 - 3*C*a^3*b + B*a^2*b^2)*cos(d*x + c)^ 2 + 2*(2*B*a^3*b - 3*C*a^2*b^2 + B*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*a rctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (C*a^4*b - 3*B*a^3*b^2 + C*a^2*b^3 + 3*B*a*b^4 - 2*C*b^5 + (2*C*a^5 - 4*B *a^4*b - C*a^3*b^2 + 5*B*a^2*b^3 - C*a*b^4 - B*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)]
\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
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. 399 vs. \(2 (150) = 300\).
Time = 0.35 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.43 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\left (2 \, B a^{2} - 3 \, C a b + B b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}^{2}}}{d} \]
((2*B*a^2 - 3*C*a*b + B*b^2)*(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)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - (2*C*a^3*tan(1/2*d* x + 1/2*c)^3 - 4*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - C*a^2*b*tan(1/2*d*x + 1/ 2*c)^3 + 3*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + B*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^3 *tan(1/2*d*x + 1/2*c) + 4*B*a^2*b*tan(1/2*d*x + 1/2*c) - C*a^2*b*tan(1/2*d *x + 1/2*c) + 3*B*a*b^2*tan(1/2*d*x + 1/2*c) - C*a*b^2*tan(1/2*d*x + 1/2*c ) - B*b^3*tan(1/2*d*x + 1/2*c) - 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 = 20.76 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.53 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,B\,a^2-3\,C\,a\,b+B\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,C\,a^2-B\,b^2+2\,C\,b^2-4\,B\,a\,b+C\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B\,b^2+2\,C\,a^2+2\,C\,b^2-4\,B\,a\,b-C\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )} \]
(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*B*a^2 + B*b^2 - 3*C*a*b))/(d*(a + b)^(5/2)*(a - b)^( 5/2)) - ((tan(c/2 + (d*x)/2)^3*(2*C*a^2 - B*b^2 + 2*C*b^2 - 4*B*a*b + C*a* b))/((a + b)^2*(a - b)) - (tan(c/2 + (d*x)/2)*(B*b^2 + 2*C*a^2 + 2*C*b^2 - 4*B*a*b - C*a*b))/((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))