Integrand size = 40, antiderivative size = 189 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {2 b \left (3 a^2 b B-2 b^3 B-2 a^3 C+a b^2 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {(2 b B-a C) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {\left (a^2 B-2 b^2 B+a b C\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
2*b*(3*B*a^2*b-2*B*b^3-2*C*a^3+C*a*b^2)*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2 *c)/(a+b)^(1/2))/a^3/(a-b)^(3/2)/(a+b)^(3/2)/d-(2*B*b-C*a)*arctanh(sin(d*x +c))/a^3/d+(B*a^2-2*B*b^2+C*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d+b*(B*b-C*a)*ta n(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))
Time = 2.63 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.27 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {2 b \left (-3 a^2 b B+2 b^3 B+2 a^3 C-a b^2 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}+2 b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+a C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^2 (-b B+a C) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+a B \tan (c+d x)}{a^3 d} \]
((-2*b*(-3*a^2*b*B + 2*b^3*B + 2*a^3*C - a*b^2*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 2*b*B*Log[Cos[(c + d*x) /2] - Sin[(c + d*x)/2]] - a*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2 *b*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + a*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)* (a + b*Cos[c + d*x])) + a*B*Tan[c + d*x])/(a^3*d)
Time = 1.32 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.325, Rules used = {3042, 3508, 3042, 3479, 3042, 3534, 25, 3042, 3480, 3042, 3138, 218, 4257}
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.
\(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3508 |
\(\displaystyle \int \frac {\sec ^2(c+d x) (B+C \cos (c+d x))}{(a+b \cos (c+d x))^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {\int \frac {\left (B a^2+b C a-(b B-a C) \cos (c+d x) a+b (b B-a C) \cos ^2(c+d x)-2 b^2 B\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {B a^2+b C a-(b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 b^2 B}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {\int -\frac {\left (\left (a^2-b^2\right ) (2 b B-a C)-a b (b B-a C) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (\left (a^2-b^2\right ) (2 b B-a C)-a b (b B-a C) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\int \frac {\left (a^2-b^2\right ) (2 b B-a C)-a b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) (2 b B-a C) \int \sec (c+d x)dx}{a}-\frac {b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) (2 b B-a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) (2 b B-a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) (2 b B-a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {\left (a^2 B+a b C-2 b^2 B\right ) \tan (c+d x)}{a d}-\frac {\frac {\left (a^2-b^2\right ) (2 b B-a C) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \left (-2 a^3 C+3 a^2 b B+a b^2 C-2 b^3 B\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a}}{a \left (a^2-b^2\right )}\) |
(b*(b*B - a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x])) + (-(( (-2*b*(3*a^2*b*B - 2*b^3*B - 2*a^3*C + a*b^2*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 - b^2)*(2* b*B - a*C)*ArcTanh[Sin[c + d*x]])/(a*d))/a) + ((a^2*B - 2*b^2*B + a*b*C)*T an[c + d*x])/(a*d))/(a*(a^2 - b^2))
3.9.6.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.78 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.28
method | result | size |
derivativedivides | \(\frac {-\frac {B}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 B b +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}+\frac {2 b \left (-\frac {a b \left (B b -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (3 B \,a^{2} b -2 B \,b^{3}-2 C \,a^{3}+C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {B}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 B b -a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(241\) |
default | \(\frac {-\frac {B}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-2 B b +a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}+\frac {2 b \left (-\frac {a b \left (B b -a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (3 B \,a^{2} b -2 B \,b^{3}-2 C \,a^{3}+C a \,b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}-\frac {B}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (2 B b -a C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}}{d}\) | \(241\) |
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 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b -2 B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 B a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+B \,a^{2} b -2 B \,b^{3}+C a \,b^{2}\right )}{a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \left (a^{2}-b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}-\frac {3 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}+\frac {2 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C b}{\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}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {3 b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d a}-\frac {2 b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{3}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C b}{\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}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}\) | \(985\) |
1/d*(-B/a^2/(tan(1/2*d*x+1/2*c)+1)+1/a^3*(-2*B*b+C*a)*ln(tan(1/2*d*x+1/2*c )+1)+2*b/a^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-b*tan(1/2*d*x+1/2*c)^2+a+b)+(3*B*a^2*b-2*B*b^3-2*C*a^3+C*a*b^2)/(a -b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b) )^(1/2)))-B/a^2/(tan(1/2*d*x+1/2*c)-1)+(2*B*b-C*a)/a^3*ln(tan(1/2*d*x+1/2* c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (180) = 360\).
Time = 7.23 (sec) , antiderivative size = 1088, normalized size of antiderivative = 5.76 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(d*x+c))^2,x, algorithm="fricas")
[-1/2*(((2*C*a^3*b^2 - 3*B*a^2*b^3 - C*a*b^4 + 2*B*b^5)*cos(d*x + c)^2 + ( 2*C*a^4*b - 3*B*a^3*b^2 - 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) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c) ^2 + 2*a*b*cos(d*x + c) + a^2)) - ((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C *a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C*a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(B*a^6 - 2*B*a^4*b^2 + B*a^2*b^4 + (B*a^5*b + C*a^4*b^2 - 3*B*a^3*b^3 - C*a^2*b^4 + 2*B*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^3*b^5)*d *cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 + a^4*b^4)*d*cos(d*x + c)), -1/2*(2*((2 *C*a^3*b^2 - 3*B*a^2*b^3 - C*a*b^4 + 2*B*b^5)*cos(d*x + c)^2 + (2*C*a^4*b - 3*B*a^3*b^2 - C*a^2*b^3 + 2*B*a*b^4)*cos(d*x + c))*sqrt(a^2 - b^2)*arcta n(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - ((C*a^5*b - 2*B* a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b - 2*C*a^4*b^2 + 4*B*a^3*b^3 + C*a^2*b^4 - 2*B*a*b^5)*co s(d*x + c))*log(sin(d*x + c) + 1) + ((C*a^5*b - 2*B*a^4*b^2 - 2*C*a^3*b^3 + 4*B*a^2*b^4 + C*a*b^5 - 2*B*b^6)*cos(d*x + c)^2 + (C*a^6 - 2*B*a^5*b ...
\[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(a+b*cos(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*b^2-4*a^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (180) = 360\).
Time = 0.37 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.14 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, C a^{3} b - 3 \, B a^{2} b^{2} - C a b^{3} + 2 \, B b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}} + \frac {{\left (C a - 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (C a - 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}}{d} \]
(2*(2*C*a^3*b - 3*B*a^2*b^2 - C*a*b^3 + 2*B*b^4)*(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^5 - a^3*b^2)*sqrt(a^2 - b^2)) - 2*(B*a^3* tan(1/2*d*x + 1/2*c)^3 - B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - B*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 + 2*B*b^3*tan(1/2*d*x + 1/ 2*c)^3 + B*a^3*tan(1/2*d*x + 1/2*c) + B*a^2*b*tan(1/2*d*x + 1/2*c) - B*a*b ^2*tan(1/2*d*x + 1/2*c) + C*a*b^2*tan(1/2*d*x + 1/2*c) - 2*B*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*b*t an(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)) + (C*a - 2*B*b)*log(abs(ta n(1/2*d*x + 1/2*c) + 1))/a^3 - (C*a - 2*B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3)/d
Time = 10.57 (sec) , antiderivative size = 5464, normalized size of antiderivative = 28.91 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
(atan((((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8*B^2*b^8 + C^2*a^8 - 8*B^2 *a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^3*b^5 + 5*B^2*a^4*b^4 - 8 *B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2*C^2*a^3*b^5 - 5*C^2*a^4*b ^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2 *b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^5*b^3 + 8*B*C*a^6*b^2))/( a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (((32*(B*a^7*b^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a^7*b^5 - 3*C*a^9*b^3 + C* a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + ( 32*tan(c/2 + (d*x)/2)*(2*B*b - C*a)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4* a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)) )*(2*B*b - C*a))/a^3)*1i)/a^3 + ((2*B*b - C*a)*((32*tan(c/2 + (d*x)/2)*(8* B^2*b^8 + C^2*a^8 - 8*B^2*a*b^7 - 2*C^2*a^7*b - 16*B^2*a^2*b^6 + 16*B^2*a^ 3*b^5 + 5*B^2*a^4*b^4 - 8*B^2*a^5*b^3 + 4*B^2*a^6*b^2 + 2*C^2*a^2*b^6 - 2* C^2*a^3*b^5 - 5*C^2*a^4*b^4 + 4*C^2*a^5*b^3 + 3*C^2*a^6*b^2 - 8*B*C*a*b^7 - 4*B*C*a^7*b + 8*B*C*a^2*b^6 + 18*B*C*a^3*b^5 - 16*B*C*a^4*b^4 - 8*B*C*a^ 5*b^3 + 8*B*C*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (((32*(B*a^7*b ^5 - 2*B*a^6*b^6 - C*a^12 + 5*B*a^8*b^4 - 3*B*a^9*b^3 - 3*B*a^10*b^2 + C*a ^7*b^5 - 3*C*a^9*b^3 + C*a^10*b^2 + 2*B*a^11*b + 2*C*a^11*b))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*tan(c/2 + (d*x)/2)*(2*B*b - C*a)*(2*a^11*b - 2 *a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2))/(a^3*(a^6*b...