Integrand size = 40, antiderivative size = 164 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b B-2 a C) \text {arctanh}(\sin (c+d x))}{b^3 d}-\frac {2 a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {C \tan (c+d x)}{b^2 d}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
(B*b-2*C*a)*arctanh(sin(d*x+c))/b^3/d-2*a*(B*a^2*b-2*B*b^3-2*C*a^3+3*C*a*b ^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(3/2)/b^3/(a +b)^(3/2)/d+C*tan(d*x+c)/b^2/d-a^2*(B*b-C*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a +b*sec(d*x+c))
Time = 4.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {-\frac {2 a \left (-a^2 b B+2 b^3 B+2 a^3 C-3 a b^2 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 b (-b B+a C) \sin (c+d x)}{(a-b) (a+b) (b+a \cos (c+d x))}+b C \tan (c+d x)}{b^3 d} \]
((-2*a*(-(a^2*b*B) + 2*b^3*B + 2*a^3*C - 3*a*b^2*C)*ArcTanh[((-a + b)*Tan[ (c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - b*B*Log[Cos[(c + d*x)/ 2] - Sin[(c + d*x)/2]] + 2*a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + b*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 2*a*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*b*(-(b*B) + a*C)*Sin[c + d*x])/((a - b)*(a + b) *(b + a*Cos[c + d*x])) + b*C*Tan[c + d*x])/(b^3*d)
Time = 1.33 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.18, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4560, 3042, 4516, 25, 3042, 4570, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4560 |
\(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4516 |
\(\displaystyle -\frac {\int -\frac {\sec (c+d x) \left (b \left (a^2-b^2\right ) C \sec ^2(c+d x)+\left (a^2-b^2\right ) (b B-a C) \sec (c+d x)+a b (b B-a C)\right )}{a+b \sec (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sec (c+d x) \left (b \left (a^2-b^2\right ) C \sec ^2(c+d x)+\left (a^2-b^2\right ) (b B-a C) \sec (c+d x)+a b (b B-a C)\right )}{a+b \sec (c+d x)}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (a^2-b^2\right ) (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+a b (b B-a C)\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (a (b B-a C) b^2+\left (a^2-b^2\right ) (b B-2 a C) \sec (c+d x) b\right )}{a+b \sec (c+d x)}dx}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a (b B-a C) b^2+\left (a^2-b^2\right ) (b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4486 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \sec (c+d x)dx-a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}-a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {\frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}+\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac {\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}}{b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
-((a^2*(b*B - a*C)*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))) + ((((a^2 - b^2)*(b*B - 2*a*C)*ArcTanh[Sin[c + d*x]])/d - (2*a*(a^2*b*B - 2*b^3*B - 2*a^3*C + 3*a*b^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqr t[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/b + ((a^2 - b^2)*C*Tan[c + d*x])/d )/(b^2*(a^2 - b^2))
3.9.2.3.1 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_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B) *Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x ] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x] )^(m + 1)*Simp[a*b*(A*b - a*B)*(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Cs c[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 ]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.55 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.49
method | result | size |
derivativedivides | \(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {2 a \left (\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (B \,a^{2} b -2 B \,b^{3}-2 a^{3} C +3 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\) | \(245\) |
default | \(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {2 a \left (\frac {a b \left (B b -C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (B \,a^{2} b -2 B \,b^{3}-2 a^{3} C +3 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}}{d}\) | \(245\) |
risch | \(\frac {2 i \left (B a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+B a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-3 C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+2 C \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+B \,a^{2} b -2 a^{3} C +C a \,b^{2}\right )}{d \,b^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (-a^{2}+b^{2}\right ) \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^{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 ) B}{\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 ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 a^{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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}+\frac {3 a^{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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\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 ) B}{\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 ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {2 a^{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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{3}}-\frac {3 a^{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 ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{b^{2} d}+\frac {2 C a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{b^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{b^{2} d}-\frac {2 C a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{b^{3} d}\) | \(977\) |
1/d*(-C/b^2/(tan(1/2*d*x+1/2*c)+1)+(B*b-2*C*a)/b^3*ln(tan(1/2*d*x+1/2*c)+1 )-C/b^2/(tan(1/2*d*x+1/2*c)-1)+1/b^3*(-B*b+2*C*a)*ln(tan(1/2*d*x+1/2*c)-1) +2*a/b^3*(a*b*(B*b-C*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)-(B*a^2*b-2*B*b^3-2*C*a^3+3*C*a*b^2)/(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))))
Leaf count of result is larger than twice the leaf count of optimal. 529 vs. \(2 (157) = 314\).
Time = 7.36 (sec) , antiderivative size = 1114, normalized size of antiderivative = 6.79 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
[1/2*(((2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a^2*b^3)*cos(d*x + c)^2 + (2 *C*a^4*b - B*a^3*b^2 - 3*C*a^2*b^3 + 2*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*C*a^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3 *b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4* C*a^3*b^3 + 2*B*a^2*b^4 + 2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(sin(d*x + c ) + 1) + ((2*C*a^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B *a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*b^ 4 + 2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(C*a^4*b^2 - 2*C*a^2*b^4 + C*b^6 + (2*C*a^5*b - B*a^4*b^2 - 3*C*a^3*b^3 + B*a^2*b^4 + C*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^3 - 2*a^3*b^5 + a*b^7)*d*co s(d*x + c)^2 + (a^4*b^4 - 2*a^2*b^6 + b^8)*d*cos(d*x + c)), 1/2*(2*((2*C*a ^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a^2*b^3)*cos(d*x + c)^2 + (2*C*a^4*b - B* a^3*b^2 - 3*C*a^2*b^3 + 2*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^6 - B*a^5*b - 4*C*a^4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*b - B*a^4*b^2 - 4*C*a^3*b^3 + 2*B*a^2*b^4 + 2*C*a*b^5 - B*b^6)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((2*C*a^6 - B*a^5*b - 4*C*a^ 4*b^2 + 2*B*a^3*b^3 + 2*C*a^2*b^4 - B*a*b^5)*cos(d*x + c)^2 + (2*C*a^5*...
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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
Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (157) = 314\).
Time = 0.33 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.46 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, C a^{4} - B a^{3} b - 3 \, C a^{2} b^{2} + 2 \, B a b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} + \frac {{\left (2 \, C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}}}{d} \]
(2*(2*C*a^4 - B*a^3*b - 3*C*a^2*b^2 + 2*B*a*b^3)*(pi*floor(1/2*(d*x + c)/p i + 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^3 - b^5)*sqrt(-a^2 + b^2)) - 2*(2*C* a^3*tan(1/2*d*x + 1/2*c)^3 - 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 - C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + C*b^3*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^3*tan(1/2*d*x + 1/2*c) + B*a^2*b*tan(1/2*d*x + 1/2*c) - C *a^2*b*tan(1/2*d*x + 1/2*c) + C*a*b^2*tan(1/2*d*x + 1/2*c) + C*b^3*tan(1/2 *d*x + 1/2*c))/((a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a *tan(1/2*d*x + 1/2*c)^2 + a + b)*(a^2*b^2 - b^4)) - (2*C*a - B*b)*log(abs( tan(1/2*d*x + 1/2*c) + 1))/b^3 + (2*C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c ) - 1))/b^3)/d
Time = 26.22 (sec) , antiderivative size = 5436, normalized size of antiderivative = 33.15 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
((2*tan(c/2 + (d*x)/2)^3*(B*a^2*b - C*b^3 - 2*C*a^3 + C*a*b^2 + C*a^2*b))/ (b^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(C*b^3 - 2*C*a^3 + B*a^2*b + C*a*b^2 - C*a^2*b))/(b^2*(a + b)*(a - b)))/(d*(a + b + tan(c/2 + (d*x)/2) ^4*(a - b) - 2*a*tan(c/2 + (d*x)/2)^2)) + (atan((((B*b - 2*C*a)*((32*tan(c /2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2 *b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5*b^3 + 2*B^2*a^6*b^2 + 4*C ^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C^2*a^5*b^3 - 16*C^2*a^6*b ^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^ 4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) + (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + B*a^5*b^7 - 3*C*a^2*b^10 - 3 *C*a^3*b^9 + 5*C*a^4*b^8 + C*a^5*b^7 - 2*C*a^6*b^6 + 2*B*a*b^11 + 2*C*a*b^ 11))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + (32*tan(c/2 + (d*x)/2)*(B*b - 2*C *a)*(2*a*b^11 - 2*a^2*b^10 - 4*a^3*b^9 + 4*a^4*b^8 + 2*a^5*b^7 - 2*a^6*b^6 ))/(b^3*(a*b^6 + b^7 - a^2*b^5 - a^3*b^4)))*(B*b - 2*C*a))/b^3)*1i)/b^3 + ((B*b - 2*C*a)*((32*tan(c/2 + (d*x)/2)*(B^2*b^8 + 8*C^2*a^8 - 2*B^2*a*b^7 - 8*C^2*a^7*b + 3*B^2*a^2*b^6 + 4*B^2*a^3*b^5 - 5*B^2*a^4*b^4 - 2*B^2*a^5* b^3 + 2*B^2*a^6*b^2 + 4*C^2*a^2*b^6 - 8*C^2*a^3*b^5 + 5*C^2*a^4*b^4 + 16*C ^2*a^5*b^3 - 16*C^2*a^6*b^2 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 8*B*C*a^2*b^6 - 8*B*C*a^3*b^5 - 16*B*C*a^4*b^4 + 18*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/(a*b^6 + b^7 - a^2*b^5 - a^3*b^4) - (((32*(B*a^2*b^10 - B*b^12 - 3*B*a^3*b^9 + ...