Integrand size = 31, antiderivative size = 135 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {2 a \left (A b^2-a^2 C+2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
C*arctanh(sin(d*x+c))/b^2/d+2*a*(A*b^2-C*a^2+2*C*b^2)*arctanh((a-b)^(1/2)* tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^2/(a+b)^(3/2)/d-(A*b^2+C*a^2 )*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))
Result contains complex when optimal does not.
Time = 2.60 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.45 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (b+a \cos (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \left (-C (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 a \left (-A b^2+\left (a^2-2 b^2\right ) C\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x)) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^{3/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b \left (A b^2+a^2 C\right ) (b \sin (c)-a \sin (d x))}{a (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{b^2 d (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \]
(2*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*(-(C*(b + a*Cos[c + d*x])*L og[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + C*(b + a*Cos[c + d*x])*Log[Cos[ (c + d*x)/2] + Sin[(c + d*x)/2]] + (2*a*(-(A*b^2) + (a^2 - 2*b^2)*C)*ArcTa n[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^ 2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])*(I*Cos[c] + Si n[c]))/((a^2 - b^2)^(3/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*(A*b^2 + a^2*C )*(b*Sin[c] - a*Sin[d*x]))/(a*(a - b)*(a + b)*(Cos[c/2] - Sin[c/2])*(Cos[c /2] + Sin[c/2]))))/(b^2*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d* x])^2)
Time = 0.82 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4569, 25, 3042, 4486, 3042, 4257, 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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4569 |
| \(\displaystyle -\frac {\int -\frac {\sec (c+d x) \left (a b (A+C)+\left (a^2-b^2\right ) C \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (a b (A+C)+\left (a^2-b^2\right ) C \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b (A+C)+\left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}+\frac {C \left (a^2-b^2\right ) \int \sec (c+d x)dx}{b}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}+\frac {C \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}+\frac {C \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}+\frac {C \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {C \left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
(((a^2 - b^2)*C*ArcTanh[Sin[c + d*x]])/(b*d) + (2*a*(A*b^2 - a^2*C + 2*b^2 *C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sq rt[a + b]*d))/(b*(a^2 - b^2)) - ((A*b^2 + a^2*C)*Tan[c + d*x])/(b*(a^2 - b ^2)*d*(a + b*Sec[c + d*x]))
3.7.86.3.1 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[((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[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ (e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(A*b^2 + a^2*C) )*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*Simp[a*b*(A + C)*(m + 1) - (A*b^2 + a^2*C + b*(A*b + b*C)*(m + 1))*C sc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && LtQ[m, -1] && Ne Q[a^2 - b^2, 0]
Time = 0.41 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \left (-\frac {b \left (A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (A \,b^{2}-C \,a^{2}+2 C \,b^{2}\right ) a \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\) | \(188\) |
| default | \(\frac {-\frac {2 \left (-\frac {b \left (A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (A \,b^{2}-C \,a^{2}+2 C \,b^{2}\right ) a \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\) | \(188\) |
| risch | \(-\frac {2 i \left (A \,b^{2}+C \,a^{2}\right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{\left (a^{2}-b^{2}\right ) d b a \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d \,b^{2}}\) | \(619\) |
1/d*(-2/b^2*(-b*(A*b^2+C*a^2)/(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)-(A*b^2-C*a^2+2*C*b^2)*a/(a+b)/(a-b)/( (a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))+ C/b^2*ln(tan(1/2*d*x+1/2*c)+1)-C/b^2*ln(tan(1/2*d*x+1/2*c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (126) = 252\).
Time = 1.42 (sec) , antiderivative size = 676, normalized size of antiderivative = 5.01 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2}\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 ) + {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}, -\frac {2 \, {\left (C a^{3} b - {\left (A + 2 \, C\right )} a b^{3} + {\left (C a^{4} - {\left (A + 2 \, C\right )} a^{2} b^{2}\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 - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{4} b - 2 \, C a^{2} b^{3} + C b^{5} + {\left (C a^{5} - 2 \, C a^{3} b^{2} + C a b^{4}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{4} b + {\left (A - C\right )} a^{2} b^{3} - A b^{5}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} b^{2} - 2 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} d\right )}}\right ] \]
[1/2*((C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*C)*a^2*b^2)*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)) + (C*a^4*b - 2*C*a^2*b^3 + C*b ^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(C*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d *x + c))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2* b^5 + b^7)*d), -1/2*(2*(C*a^3*b - (A + 2*C)*a*b^3 + (C*a^4 - (A + 2*C)*a^2 *b^2)*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))) - (C*a^4*b - 2*C*a^2*b^3 + C*b^5 + (C *a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*log(sin(d*x + c) + 1) + (C*a^4 *b - 2*C*a^2*b^3 + C*b^5 + (C*a^5 - 2*C*a^3*b^2 + C*a*b^4)*cos(d*x + c))*l og(-sin(d*x + c) + 1) + 2*(C*a^4*b + (A - C)*a^2*b^3 - A*b^5)*sin(d*x + c) )/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^4*b^3 - 2*a^2*b^5 + b ^7)*d)]
\[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.71 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (C a^{3} - A a b^{2} - 2 \, C a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]
(2*(C*a^3 - A*a*b^2 - 2*C*a*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^2*b^2 - b^4)*sqrt(-a^2 + b^2)) + C*log(abs(tan(1/2*d*x + 1 /2*c) + 1))/b^2 - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*(C*a^2*tan( 1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^2*b - b^3)*(a*tan(1/2*d *x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d
Time = 24.33 (sec) , antiderivative size = 3838, normalized size of antiderivative = 28.43 \[ \int \frac {\sec (c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
- (C*atan(((C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2* b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b ^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^ 2))))/b^2 - (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2* C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2 - (C*((C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3* C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3 *b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4* b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^ 2 + (32*tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5* b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^ 2*b^4 - 2*A*C*a^4*b^2))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))*1i)/b^2)/((C*(( C*((32*(A*a^4*b^5 - A*a^2*b^7 - A*a^3*b^6 - C*b^9 + C*a^2*b^7 - 3*C*a^3*b^ 6 + C*a^5*b^4 + A*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2* a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^3*b^2))))/b^2 - (32* tan(c/2 + (d*x)/2)*(2*C^2*a^6 + C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2* a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C^2*a^4*b^2 + 4*A*C*a^2*b^4...