Integrand size = 48, antiderivative size = 249 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=\frac {\left (2 a^3 b B+3 a b^3 B-2 a^4 C-7 a^2 b^2 C-b^4 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {b (b B-2 a C) \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {b \left (5 a b B-7 a^2 C-3 b^2 C\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {b \left (11 a^2 b B+4 b^3 B-13 a^3 C-17 a b^2 C\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:
(2*B*a^3*b+3*B*a*b^3-2*C*a^4-7*C*a^2*b^2-C*b^4)*arctan((a-b)^(1/2)*tan(1/2 *d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(7/2)/(a+b)^(7/2)/d-1/3*b*(B*b-2*C*a)*sin(d *x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^3-1/6*b*(5*B*a*b-7*C*a^2-3*C*b^2)*sin(d *x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2-1/6*b*(11*B*a^2*b+4*B*b^3-13*C*a^3- 17*C*a*b^2)*sin(d*x+c)/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
Time = 1.31 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.99 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=\frac {\frac {24 \left (-2 a^3 b B-3 a b^3 B+2 a^4 C+7 a^2 b^2 C+b^4 C\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-\frac {2 b \left (36 a^4 b B+a^2 b^3 B+8 b^5 B-48 a^5 C-23 a^3 b^2 C-19 a b^4 C+6 b \left (9 a^3 b B+a b^3 B-11 a^4 C-10 a^2 b^2 C+b^4 C\right ) \cos (c+d x)+b^2 \left (11 a^2 b B+4 b^3 B-13 a^3 C-17 a b^2 C\right ) \cos (2 (c+d x))\right ) \sin (c+d x)}{(a+b \cos (c+d x))^3}}{24 \left (a^2-b^2\right )^3 d} \] Input:
Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos[c + d*x])^5,x]
Output:
((24*(-2*a^3*b*B - 3*a*b^3*B + 2*a^4*C + 7*a^2*b^2*C + b^4*C)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - (2*b*(36*a^4* b*B + a^2*b^3*B + 8*b^5*B - 48*a^5*C - 23*a^3*b^2*C - 19*a*b^4*C + 6*b*(9* a^3*b*B + a*b^3*B - 11*a^4*C - 10*a^2*b^2*C + b^4*C)*Cos[c + d*x] + b^2*(1 1*a^2*b*B + 4*b^3*B - 13*a^3*C - 17*a*b^2*C)*Cos[2*(c + d*x)])*Sin[c + d*x ])/(a + b*Cos[c + d*x])^3)/(24*(a^2 - b^2)^3*d)
Time = 1.05 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.271, Rules used = {2014, 3042, 3233, 25, 3042, 3233, 25, 3042, 3233, 27, 3042, 3138, 218}
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.
| \(\displaystyle \int \frac {a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx\) |
| \(\Big \downarrow \) 2014 |
| \(\displaystyle \frac {\int \frac {C \cos (c+d x) b^3+(b B-a C) b^2}{(a+b \cos (c+d x))^4}dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx}{b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int -\frac {3 b^2 \left (-C a^2+b B a-b^2 C\right )-2 b^3 (b B-2 a C) \cos (c+d x)}{(a+b \cos (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {3 b^2 \left (-C a^2+b B a-b^2 C\right )-2 b^3 (b B-2 a C) \cos (c+d x)}{(a+b \cos (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 b^2 \left (-C a^2+b B a-b^2 C\right )-2 b^3 (b B-2 a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {2 b^2 \left (-3 C a^3+3 b B a^2-7 b^2 C a+2 b^3 B\right )-b^3 \left (-7 C a^2+5 b B a-3 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 b^2 \left (-3 C a^3+3 b B a^2-7 b^2 C a+2 b^3 B\right )-b^3 \left (-7 C a^2+5 b B a-3 b^2 C\right ) \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 b^2 \left (-3 C a^3+3 b B a^2-7 b^2 C a+2 b^3 B\right )-b^3 \left (-7 C a^2+5 b B a-3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int -\frac {3 b^2 \left (-2 C a^4+2 b B a^3-7 b^2 C a^2+3 b^3 B a-b^4 C\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b^3 \left (-13 a^3 C+11 a^2 b B-17 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 b^2 \left (-2 a^4 C+2 a^3 b B-7 a^2 b^2 C+3 a b^3 B-b^4 C\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b^3 \left (-13 a^3 C+11 a^2 b B-17 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 b^2 \left (-2 a^4 C+2 a^3 b B-7 a^2 b^2 C+3 a b^3 B-b^4 C\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {b^3 \left (-13 a^3 C+11 a^2 b B-17 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {\frac {6 b^2 \left (-2 a^4 C+2 a^3 b B-7 a^2 b^2 C+3 a b^3 B-b^4 C\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {b^3 \left (-13 a^3 C+11 a^2 b B-17 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {6 b^2 \left (-2 a^4 C+2 a^3 b B-7 a^2 b^2 C+3 a b^3 B-b^4 C\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b^3 \left (-13 a^3 C+11 a^2 b B-17 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {b^3 \left (-7 a^2 C+5 a b B-3 b^2 C\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b^3 (b B-2 a C) \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}}{b^2}\) |
Input:
Int[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/(a + b*Cos [c + d*x])^5,x]
Output:
(-1/3*(b^3*(b*B - 2*a*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x]) ^3) + (-1/2*(b^3*(5*a*b*B - 7*a^2*C - 3*b^2*C)*Sin[c + d*x])/((a^2 - b^2)* d*(a + b*Cos[c + d*x])^2) + ((6*b^2*(2*a^3*b*B + 3*a*b^3*B - 2*a^4*C - 7*a ^2*b^2*C - b^4*C)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqr t[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - (b^3*(11*a^2*b*B + 4*b^3*B - 13*a^3* C - 17*a*b^2*C)*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])))/(2*(a^ 2 - b^2)))/(3*(a^2 - b^2)))/b^2
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[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_S ymbol] :> Simp[1/b^2 Int[u*(a + b*v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] && LeQ [m, -1]
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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 1.20 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.54
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (6 B \,a^{2} b +3 B a \,b^{2}+2 B \,b^{3}-8 a^{3} C -5 a^{2} b C -8 C a \,b^{2}-C \,b^{3}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (9 B \,a^{2} b +B \,b^{3}-12 a^{3} C -8 C a \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 B \,a^{2} b -3 B a \,b^{2}+2 B \,b^{3}-8 a^{3} C +5 a^{2} b C -8 C a \,b^{2}+C \,b^{3}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (2 B \,a^{3} b +3 B a \,b^{3}-2 a^{4} C -7 C \,a^{2} b^{2}-C \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(384\) |
| default | \(\frac {\frac {-\frac {\left (6 B \,a^{2} b +3 B a \,b^{2}+2 B \,b^{3}-8 a^{3} C -5 a^{2} b C -8 C a \,b^{2}-C \,b^{3}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {4 \left (9 B \,a^{2} b +B \,b^{3}-12 a^{3} C -8 C a \,b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 B \,a^{2} b -3 B a \,b^{2}+2 B \,b^{3}-8 a^{3} C +5 a^{2} b C -8 C a \,b^{2}+C \,b^{3}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {\left (2 B \,a^{3} b +3 B a \,b^{3}-2 a^{4} C -7 C \,a^{2} b^{2}-C \,b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(384\) |
| risch | \(\text {Expression too large to display}\) | \(1394\) |
Input:
int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^5,x ,method=_RETURNVERBOSE)
Output:
1/d*(2*(-1/2*(6*B*a^2*b+3*B*a*b^2+2*B*b^3-8*C*a^3-5*C*a^2*b-8*C*a*b^2-C*b^ 3)*b/(a-b)/(a^3+3*a^2*b+3*a*b^2+b^3)*tan(1/2*d*x+1/2*c)^5-2/3*(9*B*a^2*b+B *b^3-12*C*a^3-8*C*a*b^2)*b/(a^2-2*a*b+b^2)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2 *c)^3-1/2*(6*B*a^2*b-3*B*a*b^2+2*B*b^3-8*C*a^3+5*C*a^2*b-8*C*a*b^2+C*b^3)* b/(a+b)/(a^3-3*a^2*b+3*a*b^2-b^3)*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)^3+(2*B*a^3*b+3*B*a*b^3-2*C*a^4-7*C*a^2*b^2 -C*b^4)/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/((a-b)*(a+b))^(1/2)*arctan((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. 618 vs. \(2 (235) = 470\).
Time = 0.17 (sec) , antiderivative size = 1305, normalized size of antiderivative = 5.24 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=\text {Too large to display} \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^5,x, algorithm="fricas")
Output:
[1/12*(3*(2*C*a^7 - 2*B*a^6*b + 7*C*a^5*b^2 - 3*B*a^4*b^3 + C*a^3*b^4 + (2 *C*a^4*b^3 - 2*B*a^3*b^4 + 7*C*a^2*b^5 - 3*B*a*b^6 + C*b^7)*cos(d*x + c)^3 + 3*(2*C*a^5*b^2 - 2*B*a^4*b^3 + 7*C*a^3*b^4 - 3*B*a^2*b^5 + C*a*b^6)*cos (d*x + c)^2 + 3*(2*C*a^6*b - 2*B*a^5*b^2 + 7*C*a^4*b^3 - 3*B*a^3*b^4 + C*a ^2*b^5)*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)) + 2*(24*C *a^7*b - 18*B*a^6*b^2 - 19*C*a^5*b^3 + 23*B*a^4*b^4 - 4*C*a^3*b^5 - 7*B*a^ 2*b^6 - C*a*b^7 + 2*B*b^8 + (13*C*a^5*b^3 - 11*B*a^4*b^4 + 4*C*a^3*b^5 + 7 *B*a^2*b^6 - 17*C*a*b^7 + 4*B*b^8)*cos(d*x + c)^2 + 3*(11*C*a^6*b^2 - 9*B* a^5*b^3 - C*a^4*b^4 + 8*B*a^3*b^5 - 11*C*a^2*b^6 + B*a*b^7 + C*b^8)*cos(d* x + c))*sin(d*x + c))/((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11 )*d*cos(d*x + c)^3 + 3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^ 10)*d*cos(d*x + c)^2 + 3*(a^10*b - 4*a^8*b^3 + 6*a^6*b^5 - 4*a^4*b^7 + a^2 *b^9)*d*cos(d*x + c) + (a^11 - 4*a^9*b^2 + 6*a^7*b^4 - 4*a^5*b^6 + a^3*b^8 )*d), -1/6*(3*(2*C*a^7 - 2*B*a^6*b + 7*C*a^5*b^2 - 3*B*a^4*b^3 + C*a^3*b^4 + (2*C*a^4*b^3 - 2*B*a^3*b^4 + 7*C*a^2*b^5 - 3*B*a*b^6 + C*b^7)*cos(d*x + c)^3 + 3*(2*C*a^5*b^2 - 2*B*a^4*b^3 + 7*C*a^3*b^4 - 3*B*a^2*b^5 + C*a*b^6 )*cos(d*x + c)^2 + 3*(2*C*a^6*b - 2*B*a^5*b^2 + 7*C*a^4*b^3 - 3*B*a^3*b^4 + C*a^2*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)...
Timed out. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=\text {Timed out} \] Input:
integrate((B*a*b-a**2*C+b**2*B*cos(d*x+c)+b**2*C*cos(d*x+c)**2)/(a+b*cos(d *x+c))**5,x)
Output:
Timed out
Exception generated. \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^5,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*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. 711 vs. \(2 (235) = 470\).
Time = 0.24 (sec) , antiderivative size = 711, normalized size of antiderivative = 2.86 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx =\text {Too large to display} \] Input:
integrate((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c ))^5,x, algorithm="giac")
Output:
-1/3*(3*(2*C*a^4 - 2*B*a^3*b + 7*C*a^2*b^2 - 3*B*a*b^3 + C*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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) - (24*C*a^5*b*tan(1/2*d*x + 1/2*c)^5 - 18*B*a^4*b^2* tan(1/2*d*x + 1/2*c)^5 - 33*C*a^4*b^2*tan(1/2*d*x + 1/2*c)^5 + 27*B*a^3*b^ 3*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^3*b^3*tan(1/2*d*x + 1/2*c)^5 - 6*B*a^2*b ^4*tan(1/2*d*x + 1/2*c)^5 - 30*C*a^2*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*b^ 5*tan(1/2*d*x + 1/2*c)^5 + 18*C*a*b^5*tan(1/2*d*x + 1/2*c)^5 - 6*B*b^6*tan (1/2*d*x + 1/2*c)^5 + 3*C*b^6*tan(1/2*d*x + 1/2*c)^5 + 48*C*a^5*b*tan(1/2* d*x + 1/2*c)^3 - 36*B*a^4*b^2*tan(1/2*d*x + 1/2*c)^3 - 16*C*a^3*b^3*tan(1/ 2*d*x + 1/2*c)^3 + 32*B*a^2*b^4*tan(1/2*d*x + 1/2*c)^3 - 32*C*a*b^5*tan(1/ 2*d*x + 1/2*c)^3 + 4*B*b^6*tan(1/2*d*x + 1/2*c)^3 + 24*C*a^5*b*tan(1/2*d*x + 1/2*c) - 18*B*a^4*b^2*tan(1/2*d*x + 1/2*c) + 33*C*a^4*b^2*tan(1/2*d*x + 1/2*c) - 27*B*a^3*b^3*tan(1/2*d*x + 1/2*c) + 18*C*a^3*b^3*tan(1/2*d*x + 1 /2*c) - 6*B*a^2*b^4*tan(1/2*d*x + 1/2*c) + 30*C*a^2*b^4*tan(1/2*d*x + 1/2* c) - 3*B*a*b^5*tan(1/2*d*x + 1/2*c) + 18*C*a*b^5*tan(1/2*d*x + 1/2*c) - 6* B*b^6*tan(1/2*d*x + 1/2*c) - 3*C*b^6*tan(1/2*d*x + 1/2*c))/((a^6 - 3*a^4*b ^2 + 3*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)^3))/d
Time = 2.93 (sec) , antiderivative size = 462, normalized size of antiderivative = 1.86 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx=-\frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B\,b^4+C\,b^4+6\,B\,a^2\,b^2+5\,C\,a^2\,b^2-3\,B\,a\,b^3-8\,C\,a\,b^3-8\,C\,a^3\,b\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (C\,b^4-2\,B\,b^4-6\,B\,a^2\,b^2+5\,C\,a^2\,b^2-3\,B\,a\,b^3+8\,C\,a\,b^3+8\,C\,a^3\,b\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (-12\,C\,a^3\,b+9\,B\,a^2\,b^2-8\,C\,a\,b^3+B\,b^4\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )}-\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,C\,a^4-2\,B\,a^3\,b+7\,C\,a^2\,b^2-3\,B\,a\,b^3+C\,b^4\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}} \] Input:
int((C*b^2*cos(c + d*x)^2 - C*a^2 + B*a*b + B*b^2*cos(c + d*x))/(a + b*cos (c + d*x))^5,x)
Output:
- ((tan(c/2 + (d*x)/2)*(2*B*b^4 + C*b^4 + 6*B*a^2*b^2 + 5*C*a^2*b^2 - 3*B* a*b^3 - 8*C*a*b^3 - 8*C*a^3*b))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) - (tan(c/2 + (d*x)/2)^5*(C*b^4 - 2*B*b^4 - 6*B*a^2*b^2 + 5*C*a^2*b^2 - 3*B *a*b^3 + 8*C*a*b^3 + 8*C*a^3*b))/((a + b)^3*(a - b)) + (4*tan(c/2 + (d*x)/ 2)^3*(B*b^4 + 9*B*a^2*b^2 - 8*C*a*b^3 - 12*C*a^3*b))/(3*(a + b)^2*(a^2 - 2 *a*b + b^2)))/(d*(3*a*b^2 - tan(c/2 + (d*x)/2)^4*(3*a*b^2 + 3*a^2*b - 3*a^ 3 - 3*b^3) - tan(c/2 + (d*x)/2)^2*(3*a*b^2 - 3*a^2*b - 3*a^3 + 3*b^3) + 3* a^2*b + a^3 + b^3 + tan(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3))) - (atan((tan(c/2 + (d*x)/2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3))/ (2*(a + b)^(1/2)*(a - b)^(7/2)))*(2*C*a^4 + C*b^4 + 7*C*a^2*b^2 - 3*B*a*b^ 3 - 2*B*a^3*b))/(d*(a + b)^(7/2)*(a - b)^(7/2))
Time = 0.18 (sec) , antiderivative size = 2116, normalized size of antiderivative = 8.50 \[ \int \frac {a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{(a+b \cos (c+d x))^5} \, dx =\text {Too large to display} \] Input:
int((B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^5,x )
Output:
( - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq rt(a**2 - b**2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**3*c + 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*c os(c + d*x)*sin(c + d*x)**2*a**3*b**5 - 42*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)*sin(c + d*x)**2*a**2*b**5*c + 18*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)*sin(c + d*x)**2*a*b**7 - 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)*sin(c + d*x)**2*b**7*c + 36*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**6*b*c - 36*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**5*b**3 + 138*sqrt(a**2 - b**2)*at an((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*cos(c + d* x)*a**4*b**3*c - 66*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**3*b**5 + 60*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**5*c - 18*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**7 + 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)*b**7*c - 36*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c ...