Integrand size = 41, antiderivative size = 214 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b} d}-\frac {\left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{2 a^4 d}+\frac {\left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 a^3 d}-\frac {(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 a d} \]
-1/2*(2*A*b^3-B*a^3-2*B*a*b^2+a^2*b*(A+2*C))*arctanh(sin(d*x+c))/a^4/d+2*b ^2*(A*b^2-a*(B*b-C*a))*arctan((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/ a^4/d/(a-b)^(1/2)/(a+b)^(1/2)+1/3*(3*A*b^2-3*B*a*b+a^2*(2*A+3*C))*tan(d*x+ c)/a^3/d-1/2*(A*b-B*a)*sec(d*x+c)*tan(d*x+c)/a^2/d+1/3*A*sec(d*x+c)^2*tan( d*x+c)/a/d
Leaf count is larger than twice the leaf count of optimal. \(466\) vs. \(2(214)=428\).
Time = 4.01 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.18 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {-\frac {24 b^2 \left (A b^2+a (-b B+a C)\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+6 \left (2 A b^3-a^3 B-2 a b^2 B+a^2 b (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 \left (-2 A b^3+a^3 B+2 a b^2 B-a^2 b (A+2 C)\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a^2 (-3 A b+a (A+3 B))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 a \left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {2 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-\frac {a^2 (-3 A b+a (A+3 B))}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a \left (3 A b^2-3 a b B+a^2 (2 A+3 C)\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}}{12 a^4 d} \]
((-24*b^2*(A*b^2 + a*(-(b*B) + a*C))*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sq rt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 6*(2*A*b^3 - a^3*B - 2*a*b^2*B + a^2*b *(A + 2*C))*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(-2*A*b^3 + a^3*B + 2*a*b^2*B - a^2*b*(A + 2*C))*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (a^2*(-3*A*b + a*(A + 3*B)))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2 *a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*a*(3 *A*b^2 - 3*a*b*B + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (a^2*(-3*A*b + a*(A + 3*B)))/(Cos[(c + d*x)/2] + Sin[(c + d *x)/2])^2 + (4*a*(3*A*b^2 - 3*a*b*B + a^2*(2*A + 3*C))*Sin[(c + d*x)/2])/( Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(12*a^4*d)
Time = 1.58 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3534, 25, 3042, 3534, 25, 3042, 3534, 27, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {\int -\frac {\left (-2 A b \cos ^2(c+d x)-a (2 A+3 C) \cos (c+d x)+3 (A b-a B)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {\left (-2 A b \cos ^2(c+d x)-a (2 A+3 C) \cos (c+d x)+3 (A b-a B)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\int \frac {-2 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {\int -\frac {\left (-3 b (A b-a B) \cos ^2(c+d x)+a (A b+3 a B) \cos (c+d x)+2 \left ((2 A+3 C) a^2-3 b B a+3 A b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}+\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}}{3 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {\left (-3 b (A b-a B) \cos ^2(c+d x)+a (A b+3 a B) \cos (c+d x)+2 \left (\frac {1}{2} (4 A+6 C) a^2-3 b B a+3 A b^2\right )\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\int \frac {-3 b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (A b+3 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (\frac {1}{2} (4 A+6 C) a^2-3 b B a+3 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3534 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {\int -\frac {3 \left (-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \cos (c+d x) a+2 A b^3\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}+\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \int \frac {\left (-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \cos (c+d x) a+2 A b^3\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \int \frac {-B a^3+b (A+2 C) a^2-2 b^2 B a+b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^3}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3480 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {\left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right ) \int \sec (c+d x)dx}{a}-\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {\left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {\left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {\left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {4 b^2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {A \tan (c+d x) \sec ^2(c+d x)}{3 a d}-\frac {\frac {3 (A b-a B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {\frac {2 \tan (c+d x) \left (a^2 (2 A+3 C)-3 a b B+3 A b^2\right )}{a d}-\frac {3 \left (\frac {\left (a^3 (-B)+a^2 b (A+2 C)-2 a b^2 B+2 A b^3\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {4 b^2 \left (A b^2-a (b B-a C)\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}\right )}{a}}{2 a}}{3 a}\) |
(A*Sec[c + d*x]^2*Tan[c + d*x])/(3*a*d) - ((3*(A*b - a*B)*Sec[c + d*x]*Tan [c + d*x])/(2*a*d) - ((-3*((-4*b^2*(A*b^2 - a*(b*B - a*C))*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d) + ((2*A *b^3 - a^3*B - 2*a*b^2*B + a^2*b*(A + 2*C))*ArcTanh[Sin[c + d*x]])/(a*d))) /a + (2*(3*A*b^2 - 3*a*b*B + a^2*(2*A + 3*C))*Tan[c + d*x])/(a*d))/(2*a))/ (3*a)
3.10.84.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_)])/(((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[(-(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 = 0.60 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.72
| method | result | size |
| derivativedivides | \(\frac {-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a A +A b -B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}+2 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}-B \,a^{2}-2 B a b +2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a A -A b +B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-A \,a^{2} b -2 A \,b^{3}+B \,a^{3}+2 B a \,b^{2}-2 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}-B \,a^{2}-2 B a b +2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{2} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(369\) |
| default | \(\frac {-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a A +A b -B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {\left (A \,a^{2} b +2 A \,b^{3}-B \,a^{3}-2 B a \,b^{2}+2 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}-B \,a^{2}-2 B a b +2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {A}{3 a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a A -A b +B a}{2 a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {\left (-A \,a^{2} b -2 A \,b^{3}+B \,a^{3}+2 B a \,b^{2}-2 a^{2} b C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{4}}-\frac {2 A \,a^{2}+a A b +2 A \,b^{2}-B \,a^{2}-2 B a b +2 a^{2} C}{2 a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 b^{2} \left (A \,b^{2}-B a b +a^{2} C \right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(369\) |
| risch | \(\frac {i \left (3 A a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+6 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-12 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 A a b \,{\mathrm e}^{i \left (d x +c \right )}+3 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}+4 A \,a^{2}+6 A \,b^{2}-6 B a b +6 a^{2} C \right )}{3 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{2 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b C}{a^{2} d}-\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}+\frac {b^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) A}{\sqrt {-a^{2}+b^{2}}\, d \,a^{4}}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) B}{\sqrt {-a^{2}+b^{2}}\, d \,a^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {-a^{2}+b^{2}}\, a}{b \sqrt {-a^{2}+b^{2}}}\right ) C}{\sqrt {-a^{2}+b^{2}}\, d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{2 a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b C}{a^{2} d}\) | \(895\) |
1/d*(-1/3*A/a/(tan(1/2*d*x+1/2*c)-1)^3-1/2*(A*a+A*b-B*a)/a^2/(tan(1/2*d*x+ 1/2*c)-1)^2+1/2*(A*a^2*b+2*A*b^3-B*a^3-2*B*a*b^2+2*C*a^2*b)/a^4*ln(tan(1/2 *d*x+1/2*c)-1)-1/2*(2*A*a^2+A*a*b+2*A*b^2-B*a^2-2*B*a*b+2*C*a^2)/a^3/(tan( 1/2*d*x+1/2*c)-1)-1/3*A/a/(tan(1/2*d*x+1/2*c)+1)^3-1/2*(-A*a-A*b+B*a)/a^2/ (tan(1/2*d*x+1/2*c)+1)^2+1/2/a^4*(-A*a^2*b-2*A*b^3+B*a^3+2*B*a*b^2-2*C*a^2 *b)*ln(tan(1/2*d*x+1/2*c)+1)-1/2*(2*A*a^2+A*a*b+2*A*b^2-B*a^2-2*B*a*b+2*C* a^2)/a^3/(tan(1/2*d*x+1/2*c)+1)+2*b^2*(A*b^2-B*a*b+C*a^2)/a^4/((a-b)*(a+b) )^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 14.20 (sec) , antiderivative size = 795, normalized size of antiderivative = 3.71 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\left [-\frac {6 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right )^{3} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, A a^{5} - 2 \, A a^{3} b^{2} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{5} - 3 \, B a^{4} b + {\left (A - 3 \, C\right )} a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )^{3}}, \frac {12 \, {\left (C a^{2} b^{2} - B a b^{3} + A b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{5} - {\left (A + 2 \, C\right )} a^{4} b + B a^{3} b^{2} - {\left (A - 2 \, C\right )} a^{2} b^{3} - 2 \, B a b^{4} + 2 \, A b^{5}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{5} - 2 \, A a^{3} b^{2} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{5} - 3 \, B a^{4} b + {\left (A - 3 \, C\right )} a^{3} b^{2} + 3 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (B a^{5} - A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{6} - a^{4} b^{2}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),x, algorithm="fricas")
[-1/12*(6*(C*a^2*b^2 - B*a*b^3 + A*b^4)*sqrt(-a^2 + b^2)*cos(d*x + c)^3*lo g((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)) - 3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2* C)*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) + 3 *(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b^4 + 2* A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) - 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*((2*A + 3*C)*a^5 - 3*B*a^4*b + (A - 3*C)*a^3*b^2 + 3*B*a^2*b^3 - 3*A*a*b ^4)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3), 1/12*(12*(C*a^2*b^2 - B*a*b^3 + A*b^4)*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c)^3 + 3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3 *b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a^5 - (A + 2*C)*a^4*b + B*a^3*b^2 - (A - 2*C)*a^2*b^3 - 2*B*a*b^4 + 2*A*b^5)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*A*a^5 - 2*A*a^3*b^2 + 2*((2*A + 3*C)*a^5 - 3*B*a^4*b + (A - 3*C)*a^3*b^2 + 3*B*a^2 *b^3 - 3*A*a*b^4)*cos(d*x + c)^2 + 3*(B*a^5 - A*a^4*b - B*a^3*b^2 + A*a^2* b^3)*cos(d*x + c))*sin(d*x + c))/((a^6 - a^4*b^2)*d*cos(d*x + c)^3)]
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Exception raised: ValueError} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+b*cos(d*x+c)),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. 483 vs. \(2 (196) = 392\).
Time = 0.35 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.26 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {\frac {3 \, {\left (B a^{3} - A a^{2} b - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3 \, {\left (B a^{3} - A a^{2} b - 2 \, C a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {12 \, {\left (C a^{2} b^{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 )}}{\sqrt {a^{2} - b^{2}} a^{4}} - \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3}}}{6 \, d} \]
1/6*(3*(B*a^3 - A*a^2*b - 2*C*a^2*b + 2*B*a*b^2 - 2*A*b^3)*log(abs(tan(1/2 *d*x + 1/2*c) + 1))/a^4 - 3*(B*a^3 - A*a^2*b - 2*C*a^2*b + 2*B*a*b^2 - 2*A *b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - 12*(C*a^2*b^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)))/(sqrt(a^2 - b^ 2)*a^4) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c) ^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*B *a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan (1/2*d*x + 1/2*c)^3 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*B*a*b*tan(1/2*d *x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/ 2*c) + 3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d*x + 1/2*c) - 3*A*a *b*tan(1/2*d*x + 1/2*c) - 6*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*d *x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^3))/d
Time = 11.09 (sec) , antiderivative size = 7033, normalized size of antiderivative = 32.86 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Too large to display} \]
(b^2*atan(((b^2*(-(a + b)*(a - b))^(1/2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2*a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2 *b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4*b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^2*a^4*b^5 - 16*C^2*a^5*b^4 + 12*C^2*a^6*b^3 - 4*C^2*a ^7*b^2 - 16*A*B*a*b^8 + 2*A*B*a^8*b + 4*B*C*a^8*b + 32*A*B*a^2*b^7 - 32*A* B*a^3*b^6 + 32*A*B*a^4*b^5 - 26*A*B*a^5*b^4 + 14*A*B*a^6*b^3 - 6*A*B*a^7*b ^2 + 16*A*C*a^2*b^7 - 32*A*C*a^3*b^6 + 28*A*C*a^4*b^5 - 20*A*C*a^5*b^4 + 1 2*A*C*a^6*b^3 - 4*A*C*a^7*b^2 - 16*B*C*a^3*b^6 + 32*B*C*a^4*b^5 - 28*B*C*a ^5*b^4 + 20*B*C*a^6*b^3 - 12*B*C*a^7*b^2))/a^6 + (b^2*(-(a + b)*(a - b))^( 1/2)*((8*(4*A*a^8*b^5 - 2*B*a^13 - 6*A*a^9*b^4 + 2*A*a^10*b^3 - 2*A*a^11*b ^2 - 4*B*a^9*b^4 + 6*B*a^10*b^3 - 2*B*a^11*b^2 + 4*C*a^10*b^3 - 8*C*a^11*b ^2 + 2*A*a^12*b + 2*B*a^12*b + 4*C*a^12*b))/a^9 - (8*b^2*tan(c/2 + (d*x)/2 )*(-(a + b)*(a - b))^(1/2)*(A*b^2 + C*a^2 - B*a*b)*(8*a^10*b + 8*a^8*b^3 - 16*a^9*b^2))/(a^6*(a^6 - a^4*b^2)))*(A*b^2 + C*a^2 - B*a*b))/(a^6 - a^4*b ^2))*(A*b^2 + C*a^2 - B*a*b)*1i)/(a^6 - a^4*b^2) + (b^2*(-(a + b)*(a - b)) ^(1/2)*((8*tan(c/2 + (d*x)/2)*(8*A^2*b^9 - B^2*a^9 - 16*A^2*a*b^8 + 3*B^2* a^8*b + 16*A^2*a^2*b^7 - 16*A^2*a^3*b^6 + 13*A^2*a^4*b^5 - 7*A^2*a^5*b^4 + 3*A^2*a^6*b^3 - A^2*a^7*b^2 + 8*B^2*a^2*b^7 - 16*B^2*a^3*b^6 + 16*B^2*a^4 *b^5 - 16*B^2*a^5*b^4 + 13*B^2*a^6*b^3 - 7*B^2*a^7*b^2 + 8*C^2*a^4*b^5 ...