Integrand size = 40, antiderivative size = 220 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{b^3 d}+\frac {\left (a^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-6 a b^4 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {a \left (a^2 b B-4 b^3 B-3 a^3 C+6 a b^2 C\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \]
C*arctanh(sin(d*x+c))/b^3/d+(B*a^2*b^3+2*B*b^5-2*C*a^5+5*C*a^3*b^2-6*C*a*b ^4)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/b^3/(a +b)^(5/2)/d-1/2*a^2*(B*b-C*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^ 2+1/2*a*(B*a^2*b-4*B*b^3-3*C*a^3+6*C*a*b^2)*tan(d*x+c)/b^2/(a^2-b^2)^2/d/( a+b*sec(d*x+c))
Time = 1.56 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {\cos (c+d x) (B+C \sec (c+d x)) \left (\frac {2 \left (-a^2 b^3 B-2 b^5 B+2 a^5 C-5 a^3 b^2 C+6 a b^4 C\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^2 (-b B+a C) \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))^2}+\frac {a b \left (-3 b^3 B-2 a^3 C+5 a b^2 C\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (b+a \cos (c+d x))}\right )}{2 b^3 d (C+B \cos (c+d x))} \]
(Cos[c + d*x]*(B + C*Sec[c + d*x])*((2*(-(a^2*b^3*B) - 2*b^5*B + 2*a^5*C - 5*a^3*b^2*C + 6*a*b^4*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b ^2]])/(a^2 - b^2)^(5/2) - 2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2 *C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a*b^2*(-(b*B) + a*C)*Sin[c + d*x])/((-a + b)*(a + b)*(b + a*Cos[c + d*x])^2) + (a*b*(-3*b^3*B - 2*a^3 *C + 5*a*b^2*C)*Sin[c + d*x])/((a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])))) /(2*b^3*d*(C + B*Cos[c + d*x]))
Time = 1.52 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.16, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 3042, 4516, 25, 3042, 4568, 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.
| \(\displaystyle \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4516 |
| \(\displaystyle -\frac {\int -\frac {\sec (c+d x) \left (2 b \left (a^2-b^2\right ) C \sec ^2(c+d x)+\left (a^2-2 b^2\right ) (b B-a C) \sec (c+d x)+2 a b (b B-a C)\right )}{(a+b \sec (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) \left (2 b \left (a^2-b^2\right ) C \sec ^2(c+d x)+\left (a^2-2 b^2\right ) (b B-a C) \sec (c+d x)+2 a b (b B-a C)\right )}{(a+b \sec (c+d x))^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 b \left (a^2-b^2\right ) C \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (a^2-2 b^2\right ) (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a b (b B-a C)\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4568 |
| \(\displaystyle \frac {\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {\sec (c+d x) \left (\left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 C \sec (c+d x) b\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (\left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 C \sec (c+d x) b\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (C a^3+b B a^2-4 b^2 C a+2 b^3 B\right ) b^2+2 \left (a^2-b^2\right )^2 C \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {2 C \left (a^2-b^2\right )^2 \int \sec (c+d x)dx+\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 C \left (a^2-b^2\right )^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\frac {\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {2 \left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}+\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {a \left (-3 a^3 C+a^2 b B+6 a b^2 C-4 b^3 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {2 \left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-6 a b^4 C+2 b^5 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b \left (a^2-b^2\right )}}{2 b^2 \left (a^2-b^2\right )}-\frac {a^2 (b B-a C) \tan (c+d x)}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
-1/2*(a^2*(b*B - a*C)*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x] )^2) + (((2*(a^2 - b^2)^2*C*ArcTanh[Sin[c + d*x]])/d + (2*(a^2*b^3*B + 2*b ^5*B - 2*a^5*C + 5*a^3*b^2*C - 6*a*b^4*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d* x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/(b*(a^2 - b^2)) + (a*(a^ 2*b*B - 4*b^3*B - 3*a^3*C + 6*a*b^2*C)*Tan[c + d*x])/((a^2 - b^2)*d*(a + b *Sec[c + d*x])))/(2*b^2*(a^2 - b^2))
3.9.9.3.1 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[(-(A*b^2 - a*b*B + 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[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 2, 0]
Time = 0.69 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {2 \left (\frac {-\frac {\left (B a \,b^{2}+4 B \,b^{3}+2 a^{3} C -a^{2} b C -6 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (B a \,b^{2}-4 B \,b^{3}-2 a^{3} C -a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (B \,a^{2} b^{3}+2 B \,b^{5}-2 a^{5} C +5 C \,a^{3} b^{2}-6 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(305\) |
| default | \(\frac {-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}-\frac {2 \left (\frac {-\frac {\left (B a \,b^{2}+4 B \,b^{3}+2 a^{3} C -a^{2} b C -6 C a \,b^{2}\right ) a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {b a \left (B a \,b^{2}-4 B \,b^{3}-2 a^{3} C -a^{2} b C +6 C a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (B \,a^{2} b^{3}+2 B \,b^{5}-2 a^{5} C +5 C \,a^{3} b^{2}-6 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\) | \(305\) |
| risch | \(\text {Expression too large to display}\) | \(1191\) |
1/d*(-C/b^3*ln(tan(1/2*d*x+1/2*c)-1)-2/b^3*((-1/2*(B*a*b^2+4*B*b^3+2*C*a^3 -C*a^2*b-6*C*a*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*a *(B*a*b^2-4*B*b^3-2*C*a^3-C*a^2*b+6*C*a*b^2)/(a+b)/(a-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)^2-1/2*(B*a^2*b^3+ 2*B*b^5-2*C*a^5+5*C*a^3*b^2-6*C*a*b^4)/(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)))+C/b^3*ln(tan(1 /2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 680 vs. \(2 (210) = 420\).
Time = 10.23 (sec) , antiderivative size = 1419, normalized size of antiderivative = 6.45 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[-1/4*((2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 6*C*a*b^6 - 2*B*b^7 + (2*C *a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 6*C*a^3*b^4 - 2*B*a^2*b^5)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 6*C*a^2*b^5 - 2*B*a*b^6)*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^2 - 3*C*a^4 *b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^ 6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*cos( d*x + c))*log(sin(d*x + c) + 1) + 2*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C*a^2*b^6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7)*cos(d*x + c))*log(-sin (d*x + c) + 1) + 2*(3*C*a^6*b^2 - B*a^5*b^3 - 9*C*a^4*b^4 + 5*B*a^3*b^5 + 6*C*a^2*b^6 - 4*B*a*b^7 + (2*C*a^7*b - 7*C*a^5*b^3 + 3*B*a^4*b^4 + 5*C*a^3 *b^5 - 3*B*a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3* a^4*b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d), -1/ 2*((2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 6*C*a*b^6 - 2*B*b^7 + (2*C*a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 6*C*a^3*b^4 - 2*B*a^2*b^5)*cos(d*x + c)^2 + 2 *(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 6*C*a^2*b^5 - 2*B*a*b^6)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a...
\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]
integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^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
Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (210) = 420\).
Time = 0.38 (sec) , antiderivative size = 486, normalized size of antiderivative = 2.21 \[ \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))^3} \, dx=-\frac {\frac {{\left (2 \, C a^{5} - 5 \, C a^{3} b^{2} - B a^{2} b^{3} + 6 \, C a b^{4} - 2 \, B b^{5}\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} b^{3} - 2 \, a^{2} b^{5} + b^{7}\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^{3}} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac {2 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, B a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\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*C*a^5 - 5*C*a^3*b^2 - B*a^2*b^3 + 6*C*a*b^4 - 2*B*b^5)*(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*b^3 - 2*a^2*b^5 + b^7)*sqrt (-a^2 + b^2)) - C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + C*log(abs(tan(1 /2*d*x + 1/2*c) - 1))/b^3 - (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 3*C*a^4*b*ta n(1/2*d*x + 1/2*c)^3 + B*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^3*b^2*tan( 1/2*d*x + 1/2*c)^3 + 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*b^3*tan( 1/2*d*x + 1/2*c)^3 - 4*B*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^5*tan(1/2*d* x + 1/2*c) - 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + B*a^3*b^2*tan(1/2*d*x + 1/2* c) + 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*C*a^2*b^3*tan(1/2*d*x + 1/2*c) - 4*B*a*b^4*tan(1/2*d*x + 1/2*c))/((a^4* b^2 - 2*a^2*b^4 + b^6)*(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 = 27.14 (sec) , antiderivative size = 6899, normalized size of antiderivative = 31.36 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \]
((tan(c/2 + (d*x)/2)^3*(2*C*a^4 + B*a^2*b^2 - 6*C*a^2*b^2 + 4*B*a*b^3 - C* a^3*b))/((a*b^2 - b^3)*(a + b)^2) - (tan(c/2 + (d*x)/2)*(2*C*a^4 - B*a^2*b ^2 - 6*C*a^2*b^2 + 4*B*a*b^3 + C*a^3*b))/((a + b)*(b^4 - 2*a*b^3 + a^2*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)) - (C*atan(-((C*((C*((8*(4*B*b^15 + 4*C *b^15 - 6*B*a^2*b^13 + 6*B*a^3*b^12 + 2*B*a^6*b^9 - 2*B*a^7*b^8 - 8*C*a^2* b^13 + 34*C*a^3*b^12 + 6*C*a^4*b^11 - 36*C*a^5*b^10 - 4*C*a^6*b^9 + 18*C*a ^7*b^8 + 2*C*a^8*b^7 - 4*C*a^9*b^6 - 4*B*a*b^14 - 12*C*a*b^14))/(a*b^12 + b^13 - 3*a^2*b^11 - 3*a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6 ) - (8*C*tan(c/2 + (d*x)/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4* b^12 + 48*a^5*b^11 - 48*a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8 *a^10*b^6))/(b^3*(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^ 5*b^6 - a^6*b^5 - a^7*b^4))))/b^3 - (8*tan(c/2 + (d*x)/2)*(4*B^2*b^10 + 8* C^2*a^10 + 4*C^2*b^10 - 8*C^2*a*b^9 - 8*C^2*a^9*b + 4*B^2*a^2*b^8 + B^2*a^ 4*b^6 + 24*C^2*a^2*b^8 + 32*C^2*a^3*b^7 - 52*C^2*a^4*b^6 - 48*C^2*a^5*b^5 + 57*C^2*a^6*b^4 + 32*C^2*a^7*b^3 - 32*C^2*a^8*b^2 - 24*B*C*a*b^9 + 8*B*C* a^3*b^7 + 2*B*C*a^5*b^5 - 4*B*C*a^7*b^3))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a ^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^5 - a^7*b^4))*1i)/b^3 - (C*((C*((8* (4*B*b^15 + 4*C*b^15 - 6*B*a^2*b^13 + 6*B*a^3*b^12 + 2*B*a^6*b^9 - 2*B*a^7 *b^8 - 8*C*a^2*b^13 + 34*C*a^3*b^12 + 6*C*a^4*b^11 - 36*C*a^5*b^10 - 4*...