Integrand size = 48, antiderivative size = 140 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {(b B-a C) x}{a^2}-\frac {2 b \left (2 a^2 b B-b^3 B-3 a^3 C+a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 (b B-2 a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
(B*b-C*a)*x/a^2-2*b*(2*B*a^2*b-B*b^3-3*C*a^3+C*a*b^2)*arctanh((a-b)^(1/2)* tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+b^2*(B*b-2*C *a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))
Time = 1.82 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.51 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) (b B-a C+b C \sec (c+d x)) \left ((b B-a C) (c+d x) (b+a \cos (c+d x))-\frac {2 b \left (-2 a^2 b B+b^3 B+3 a^3 C-a b^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))}{\left (a^2-b^2\right )^{3/2}}+\frac {a b^2 (b B-2 a C) \sin (c+d x)}{(a-b) (a+b)}\right )}{a^2 d (b C+(b B-a C) \cos (c+d x)) (a+b \sec (c+d x))^2} \]
Integrate[(a*b*B - a^2*C + b^2*B*Sec[c + d*x] + b^2*C*Sec[c + d*x]^2)/(a + b*Sec[c + d*x])^3,x]
((b + a*Cos[c + d*x])*Sec[c + d*x]*(b*B - a*C + b*C*Sec[c + d*x])*((b*B - a*C)*(c + d*x)*(b + a*Cos[c + d*x]) - (2*b*(-2*a^2*b*B + b^3*B + 3*a^3*C - a*b^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[ c + d*x]))/(a^2 - b^2)^(3/2) + (a*b^2*(b*B - 2*a*C)*Sin[c + d*x])/((a - b) *(a + b))))/(a^2*d*(b*C + (b*B - a*C)*Cos[c + d*x])*(a + b*Sec[c + d*x])^2 )
Time = 0.79 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.24, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {2014, 3042, 4411, 25, 3042, 4407, 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.
| \(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 2014 |
| \(\displaystyle \frac {\int \frac {C \sec (c+d x) b^3+(b B-a C) b^2}{(a+b \sec (c+d x))^2}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{b^2}\) |
| \(\Big \downarrow \) 4411 |
| \(\displaystyle \frac {\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {b^2 \left (a^2-b^2\right ) (b B-a C)-a b^3 (b B-2 a C) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}}{b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {b^2 \left (a^2-b^2\right ) (b B-a C)-a b^3 (b B-2 a C) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {b^2 \left (a^2-b^2\right ) (b B-a C)-a b^3 (b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {b^3 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {b^3 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {b^2 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {b^2 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {2 b^2 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{b^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {b^4 (b B-2 a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {b^2 x \left (a^2-b^2\right ) (b B-a C)}{a}-\frac {2 b^3 \left (-3 a^3 C+2 a^2 b B+a b^2 C-b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}}{b^2}\) |
(((b^2*(a^2 - b^2)*(b*B - a*C)*x)/a - (2*b^3*(2*a^2*b*B - b^3*B - 3*a^3*C + a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2)) + (b^4*(b*B - 2*a*C)*Tan[c + d*x])/(a *(a^2 - b^2)*d*(a + b*Sec[c + d*x])))/b^2
3.10.32.3.1 Defintions of rubi rules used
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[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S ymbol] :> Simp[1/b^2 Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ [m, -1]
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_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[b*(b*c - a*d)*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[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && N eQ[a^2 - b^2, 0] && IntegerQ[2*m]
Time = 0.33 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.30
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \left (-\frac {b \left (B b -2 C a \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 B \,a^{2} b -B \,b^{3}-3 a^{3} C +C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(182\) |
| default | \(\frac {\frac {2 \left (B b -C a \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \left (-\frac {b \left (B b -2 C a \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 B \,a^{2} b -B \,b^{3}-3 a^{3} C +C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}}{d}\) | \(182\) |
| risch | \(\frac {x B b}{a^{2}}-\frac {x C}{a}-\frac {2 i b^{2} \left (-B b +2 C a \right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{4} \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {3 b a \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {2 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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{4} \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 b a \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \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}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}\) | \(768\) |
int((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x ,method=_RETURNVERBOSE)
1/d*(2*(B*b-C*a)/a^2*arctan(tan(1/2*d*x+1/2*c))+2*b/a^2*(-b*(B*b-2*C*a)*a/ (a^2-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*B*a^2*b-B*b^3-3*C*a^3+C*a*b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*a rctanh((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. 327 vs. \(2 (134) = 268\).
Time = 0.34 (sec) , antiderivative size = 714, normalized size of antiderivative = 5.10 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\left [-\frac {2 \, {\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x - {\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} + {\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\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 (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, -\frac {{\left (C a^{6} - B a^{5} b - 2 \, C a^{4} b^{2} + 2 \, B a^{3} b^{3} + C a^{2} b^{4} - B a b^{5}\right )} d x \cos \left (d x + c\right ) + {\left (C a^{5} b - B a^{4} b^{2} - 2 \, C a^{3} b^{3} + 2 \, B a^{2} b^{4} + C a b^{5} - B b^{6}\right )} d x - {\left (3 \, C a^{3} b^{2} - 2 \, B a^{2} b^{3} - C a b^{4} + B b^{5} + {\left (3 \, C a^{4} b - 2 \, B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\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 (2 \, C a^{4} b^{2} - B a^{3} b^{3} - 2 \, C a^{2} b^{4} + B a b^{5}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^3,x, algorithm="fricas")
[-1/2*(2*(C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3*b^3 + C*a^2*b^4 - B*a*b^ 5)*d*x*cos(d*x + c) + 2*(C*a^5*b - B*a^4*b^2 - 2*C*a^3*b^3 + 2*B*a^2*b^4 + C*a*b^5 - B*b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a*b^4 + B*b^5 + (3* C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*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*(2*C*a^4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a *b^5)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), -((C*a^6 - B*a^5*b - 2*C*a^4*b^2 + 2*B*a^3*b^3 + C*a^2*b^4 - B*a*b^5)*d*x*cos(d*x + c) + (C*a^5*b - B*a^4*b^2 - 2*C*a^3*b^ 3 + 2*B*a^2*b^4 + C*a*b^5 - B*b^6)*d*x - (3*C*a^3*b^2 - 2*B*a^2*b^3 - C*a* b^4 + B*b^5 + (3*C*a^4*b - 2*B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*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))) + (2*C*a^4*b^2 - B*a^3*b^3 - 2*C*a^2*b^4 + B*a*b^5)*sin (d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^ 3 + a^2*b^5)*d)]
\[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=- \int \left (- \frac {B b}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\right )\, dx - \int \frac {C a}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\, dx - \int \left (- \frac {C b \sec {\left (c + d x \right )}}{a^{2} + 2 a b \sec {\left (c + d x \right )} + b^{2} \sec ^{2}{\left (c + d x \right )}}\right )\, dx \]
-Integral(-B*b/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2), x) - In tegral(C*a/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2), x) - Integr al(-C*b*sec(c + d*x)/(a**2 + 2*a*b*sec(c + d*x) + b**2*sec(c + d*x)**2), x )
Exception generated. \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^3,x, algorithm="maxima")
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
Time = 0.38 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.59 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, C a^{3} b - 2 \, B a^{2} b^{2} - C a b^{3} + B b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (C a - B b\right )} {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (2 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]
integrate((B*a*b-C*a^2+b^2*B*sec(d*x+c)+b^2*C*sec(d*x+c)^2)/(a+b*sec(d*x+c ))^3,x, algorithm="giac")
(2*(3*C*a^3*b - 2*B*a^2*b^2 - C*a*b^3 + B*b^4)*(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 - a^2*b^2)*sqrt(-a^2 + b^2)) - (C*a - B* b)*(d*x + c)/a^2 + 2*(2*C*a*b^2*tan(1/2*d*x + 1/2*c) - B*b^3*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) ^2 - a - b)))/d
Time = 26.83 (sec) , antiderivative size = 5260, normalized size of antiderivative = 37.57 \[ \int \frac {a b B-a^2 C+b^2 B \sec (c+d x)+b^2 C \sec ^2(c+d x)}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
int(((B*b^2)/cos(c + d*x) - C*a^2 + (C*b^2)/cos(c + d*x)^2 + B*a*b)/(a + b /cos(c + d*x))^3,x)
(2*atan(((((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C ^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b ^3 + 8*C^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^ 3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^ 2*b^3 - a^3*b^2) + (((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2 *B*a^8*b^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/( a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a^9* b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*( a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*(B*b - C*a)*1i)/a^2)*(B*b - C*a))/a^2 + (((32*tan(c/2 + (d*x)/2)*(2*B^2*b^8 + C^2*a^8 - 2*B^2*a*b^7 - 2*C^2*a^7*b - 5*B^2*a^2*b^6 + 4*B^2*a^3*b^5 + 3*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + B^2*a^6 *b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 7*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 8*C ^2*a^6*b^2 - 4*B*C*a*b^7 - 2*B*C*a^7*b + 4*B*C*a^2*b^6 + 12*B*C*a^3*b^5 - 8*B*C*a^4*b^4 - 10*B*C*a^5*b^3 + 4*B*C*a^6*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (((32*(C*a^10 + B*a^4*b^6 - 3*B*a^6*b^4 + B*a^7*b^3 + 2*B*a^8*b ^2 - C*a^5*b^5 - C*a^6*b^4 + 4*C*a^7*b^3 - B*a^9*b - 3*C*a^9*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a^9*b - 2*a^ 4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*(B*b - C*a)*1i)/a^2)*(B*b - C*a))/a^2)/((64*...