Integrand size = 41, antiderivative size = 177 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b B-2 a C) \text {arctanh}(\sin (c+d x))}{b^3 d}-\frac {2 \left (A b^4+a^3 b B-2 a b^3 B-2 a^4 C+3 a^2 b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^3 (a+b)^{3/2} d}+\frac {C \tan (c+d x)}{b^2 d}+\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:
(B*b-2*C*a)*arctanh(sin(d*x+c))/b^3/d-2*(A*b^4+B*a^3*b-2*B*a*b^3-2*C*a^4+3 *C*a^2*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*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b^ 2/(a^2-b^2)/d/(a+b*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(382\) vs. \(2(177)=354\).
Time = 4.23 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.16 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (A b^4+a \left (a^2 b B-2 b^3 B-2 a^3 C+3 a b^2 C\right )\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cos (c+d x))}{\left (a^2-b^2\right )^{3/2}}-(b B-2 a C) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+(b B-2 a C) (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b C (b+a \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {b C (b+a \cos (c+d x)) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {a b \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{(a-b) (a+b)}\right )}{b^3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \] Input:
Integrate[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b* Sec[c + d*x])^2,x]
Output:
(2*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(A*b^4 + 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]]*(b + a*Cos[c + d*x]))/(a^2 - b^2)^(3/2) - (b*B - 2*a*C)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + ( b*B - 2*a*C)*(b + a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b*C*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (b*C*(b + a*Cos[c + d*x])*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (a*b*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a - b)*(a + b))))/(b^3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
Time = 1.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4578, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4578 |
| \(\displaystyle \frac {\int -\frac {\sec (c+d x) \left (-b \left (a^2-b^2\right ) C \sec ^2(c+d x)-\left (a^2-b^2\right ) (b B-a C) \sec (c+d x)+b \left (A b^2-a (b B-a C)\right )\right )}{a+b \sec (c+d x)}dx}{b^2 \left (a^2-b^2\right )}+\frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {\sec (c+d x) \left (-b \left (a^2-b^2\right ) C \sec ^2(c+d x)-\left (a^2-b^2\right ) (b B-a C) \sec (c+d x)+b \left (A b^2-a (b B-a C)\right )\right )}{a+b \sec (c+d x)}dx}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-b \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2-\left (a^2-b^2\right ) (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left (A b^2-a (b B-a C)\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\int \frac {\sec (c+d x) \left (b^2 \left (A b^2-a (b B-a C)\right )-b \left (a^2-b^2\right ) (b B-2 a C) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b^2 \left (A b^2-a (b B-a C)\right )-b \left (a^2-b^2\right ) (b B-2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx-\left (a^2-b^2\right ) (b B-2 a C) \int \sec (c+d x)dx}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx-\left (a^2-b^2\right ) (b B-2 a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {\left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {\left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {2 \left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\frac {\frac {2 \left (a \left (-2 a^3 C+a^2 b B+3 a b^2 C-2 b^3 B\right )+A b^4\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}-\frac {\left (a^2-b^2\right ) (b B-2 a C) \text {arctanh}(\sin (c+d x))}{d}}{b}-\frac {C \left (a^2-b^2\right ) \tan (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}\) |
Input:
Int[(Sec[c + d*x]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
Output:
(a*(A*b^2 - a*(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 *b^4 + a*(a^2*b*B - 2*b^3*B - 2*a^3*C + 3*a*b^2*C))*ArcTanh[(Sqrt[a - b]*T an[(c + d*x)/2])/Sqrt[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))
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_)]*(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]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[a*(A*b^2 - a*b*B + a^2*C)*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[b*(m + 1)*((-a)*(b *B - a*C) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^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 = 0.71 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.47
| method | result | size |
| derivativedivides | \(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}+\frac {-\frac {2 a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (A \,b^{4}+B \,a^{3} b -2 B a \,b^{3}-2 a^{4} C +3 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{3}}}{d}\) | \(261\) |
| default | \(\frac {-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {C}{b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -2 C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}+\frac {-\frac {2 a b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (A \,b^{4}+B \,a^{3} b -2 B a \,b^{3}-2 a^{4} C +3 C \,a^{2} b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{3}}}{d}\) | \(261\) |
| risch | \(\text {Expression too large to display}\) | \(1204\) |
Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x,meth od=_RETURNVERBOSE)
Output:
1/d*(-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)-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) +2/b^3*(-a*b*(A*b^2-B*a*b+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^4+B*a^3*b-2*B*a*b^3-2*C*a^4+3 *C*a^2*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. 550 vs. \(2 (171) = 342\).
Time = 12.59 (sec) , antiderivative size = 1156, normalized size of antiderivative = 6.53 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2, x, algorithm="fricas")
Output:
[1/2*(((2*C*a^5 - B*a^4*b - 3*C*a^3*b^2 + 2*B*a^2*b^3 - A*a*b^4)*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 - A*b^5)*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 + (A - 3 *C)*a^3*b^3 + B*a^2*b^4 - (A - 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*cos(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 - A *a*b^4)*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 - A*b^5)*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...
\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))* *2,x)
Output:
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a + b*s ec(c + d*x))**2, x)
Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2, x, algorithm="maxima")
Output:
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. 443 vs. \(2 (171) = 342\).
Time = 0.25 (sec) , antiderivative size = 443, normalized size of antiderivative = 2.50 \[ \int \frac {\sec ^2(c+d x) \left (A+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} - A 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^{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} + A 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} + 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 ) - A 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 ) + 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} \] Input:
integrate(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2, x, algorithm="giac")
Output:
(2*(2*C*a^4 - B*a^3*b - 3*C*a^2*b^2 + 2*B*a*b^3 - A*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^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 + A*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 + 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) - A*a *b^2*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*t an(1/2*d*x + 1/2*c)^2 + a + b)*(a^2*b^2 - b^4)) - (2*C*a - B*b)*log(abs(ta n(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 = 21.53 (sec) , antiderivative size = 6421, normalized size of antiderivative = 36.28 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b/cos(c + d*x))^2),x)
Output:
((2*tan(c/2 + (d*x)/2)*(2*C*a^3 - C*b^3 + A*a*b^2 - B*a^2*b - C*a*b^2 + C* a^2*b))/(b^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)^3*(2*C*a^3 + C*b^3 + A*a*b^2 - 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)*(((B*b - 2*C*a)*((32*(A*a^2*b^10 - B*b^12 - A*b^12 - A*a^3*b^9 + B*a^2*b^10 - 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 + A*a*b^11 + 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^3 - (32*tan(c/2 + (d*x)/2)*(A^2*b^ 8 + 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*A*B* a*b^7 - 4*B*C*a*b^7 - 8*B*C*a^7*b + 2*A*B*a^3*b^5 + 6*A*C*a^2*b^6 - 4*A*C* a^4*b^4 + 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))*1i)/b^3 - ((B*b - 2*C *a)*(((B*b - 2*C*a)*((32*(A*a^2*b^10 - B*b^12 - A*b^12 - A*a^3*b^9 + B*a^2 *b^10 - 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 + A*a*b^11 + 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...
Time = 0.16 (sec) , antiderivative size = 1793, normalized size of antiderivative = 10.13 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x)
Output:
(4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr t( - a**2 + b**2))*cos(c + d*x)*a**4*b*c - 2*sqrt( - a**2 + b**2)*atan((ta n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)* a**3*b**3 - 6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x )/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**2*b**3*c + 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2)) *cos(c + d*x)*a*b**5 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - t an((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**5*c + 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**4*b**2 + 6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a* *3*b**2*c - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x )/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**2*b**4 + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2) )*a**5*c - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x) /2)*b)/sqrt( - a**2 + b**2))*a**4*b**2 - 6*sqrt( - a**2 + b**2)*atan((tan( (c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**3*b**2*c + 2 *sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**2*b**4 + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**5 *b*c - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b**3 - 4*cos(c + d*x...