Integrand size = 12, antiderivative size = 184 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=\frac {a \left (2 a^2+3 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{7/2} (a+b)^{7/2} d}-\frac {b \sin (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^3}-\frac {5 a b \sin (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))^2}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \cos (c+d x))} \] Output:
a*(2*a^2+3*b^2)*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*sin(d*x+c)/(a^2-b^2)/d/(a+b*cos(d*x+c))^3-5/6*a*b *sin(d*x+c)/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^2-1/6*b*(11*a^2+4*b^2)*sin(d*x+ c)/(a^2-b^2)^3/d/(a+b*cos(d*x+c))
Time = 1.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{7/2}}-\frac {b \left (18 a^4-5 a^2 b^2+2 b^4+3 a b \left (9 a^2+b^2\right ) \cos (c+d x)+b^2 \left (11 a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sin (c+d x)}{(a-b)^3 (a+b)^3 (a+b \cos (c+d x))^3}}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^(-4),x]
Output:
((6*a*(2*a^2 + 3*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]] )/(-a^2 + b^2)^(7/2) - (b*(18*a^4 - 5*a^2*b^2 + 2*b^4 + 3*a*b*(9*a^2 + b^2 )*Cos[c + d*x] + b^2*(11*a^2 + 4*b^2)*Cos[c + d*x]^2)*Sin[c + d*x])/((a - b)^3*(a + b)^3*(a + b*Cos[c + d*x])^3))/(6*d)
Time = 0.74 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.23, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 3143, 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.
| \(\displaystyle \int \frac {1}{(a+b \cos (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4}dx\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle -\frac {\int -\frac {3 a-2 b \cos (c+d x)}{(a+b \cos (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 a-2 b \cos (c+d x)}{(a+b \cos (c+d x))^3}dx}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a-2 b \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 \left (3 a^2+2 b^2\right )-5 a b \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (3 a^2+2 b^2\right )-5 a b \cos (c+d x)}{(a+b \cos (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (3 a^2+2 b^2\right )-5 a b \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {3 a \left (2 a^2+3 b^2\right )}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 a \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a^2-b^2}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 a \left (2 a^2+3 b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {6 a \left (2 a^2+3 b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))}}{2 \left (a^2-b^2\right )}-\frac {5 a b \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}}{3 \left (a^2-b^2\right )}-\frac {b \sin (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^3}\) |
Input:
Int[(a + b*Cos[c + d*x])^(-4),x]
Output:
-1/3*(b*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*Cos[c + d*x])^3) + ((-5*a*b*Si n[c + d*x])/(2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^2) + ((6*a*(2*a^2 + 3*b^ 2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)*d) - (b*(11*a^2 + 4*b^2)*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))
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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 = 1.14 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.52
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (6 a^{2}+3 a b +2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (9 a^{2}+b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-3 a b +2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {a \left (2 a^{2}+3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(280\) |
| default | \(\frac {\frac {-\frac {\left (6 a^{2}+3 a b +2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{\left (a -b \right ) \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {4 \left (9 a^{2}+b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 \left (a^{2}-2 a b +b^{2}\right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-3 a b +2 b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a +b \right ) \left (a^{3}-3 a^{2} b +3 b^{2} a -b^{3}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{3}}+\frac {a \left (2 a^{2}+3 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a^{6}-3 a^{4} b^{2}+3 b^{4} a^{2}-b^{6}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(280\) |
| risch | \(-\frac {i \left (6 a^{3} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+9 a \,b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+30 a^{4} b \,{\mathrm e}^{4 i \left (d x +c \right )}+45 a^{2} b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+44 a^{5} {\mathrm e}^{3 i \left (d x +c \right )}+82 a^{3} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+24 a \,b^{4} {\mathrm e}^{3 i \left (d x +c \right )}+102 a^{4} b \,{\mathrm e}^{2 i \left (d x +c \right )}+36 a^{2} b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{5} {\mathrm e}^{2 i \left (d x +c \right )}+60 a^{3} b^{2} {\mathrm e}^{i \left (d x +c \right )}+15 a \,b^{4} {\mathrm e}^{i \left (d x +c \right )}+11 a^{2} b^{3}+4 b^{5}\right )}{3 \left (a^{2}-b^{2}\right )^{3} d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}-\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}+\frac {3 a \,b^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{3} \left (a -b \right )^{3} d}\) | \(585\) |
Input:
int(1/(a+cos(d*x+c)*b)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(2*(-1/2*(6*a^2+3*a*b+2*b^2)*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*a^2+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*a^2-3*a*b+2*b^2)*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+a*(2* a^2+3*b^2)/(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. 413 vs. \(2 (169) = 338\).
Time = 0.14 (sec) , antiderivative size = 895, normalized size of antiderivative = 4.86 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+b*cos(d*x+c))^4,x, algorithm="fricas")
Output:
[1/12*(3*(2*a^6 + 3*a^4*b^2 + (2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 + 3*(2* a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + 3*(2*a^5*b + 3*a^3*b^3)*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*(18*a^6*b - 23*a^4*b^3 + 7*a^2*b^5 - 2*b^7 + (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2 + 3*(9 *a^5*b^2 - 8*a^3*b^4 - a*b^6)*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*a^6 + 3*a^4*b^2 + (2*a ^3*b^3 + 3*a*b^5)*cos(d*x + c)^3 + 3*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^ 2 + 3*(2*a^5*b + 3*a^3*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d *x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - (18*a^6*b - 23*a^4*b^3 + 7* a^2*b^5 - 2*b^7 + (11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^2 + 3*(9*a ^5*b^2 - 8*a^3*b^4 - a*b^6)*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)]
Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*cos(d*x+c))**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+b*cos(d*x+c))^4,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. 399 vs. \(2 (169) = 338\).
Time = 0.47 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.17 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=-\frac {\frac {3 \, {\left (2 \, a^{3} + 3 \, a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} + \frac {18 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 32 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, a^{4} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, b^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{3}}}{3 \, d} \] Input:
integrate(1/(a+b*cos(d*x+c))^4,x, algorithm="giac")
Output:
-1/3*(3*(2*a^3 + 3*a*b^2)*(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)) + (18*a^4*b*ta n(1/2*d*x + 1/2*c)^5 - 27*a^3*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b^3*tan(1 /2*d*x + 1/2*c)^5 - 3*a*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*b^5*tan(1/2*d*x + 1 /2*c)^5 + 36*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 32*a^2*b^3*tan(1/2*d*x + 1/2*c )^3 - 4*b^5*tan(1/2*d*x + 1/2*c)^3 + 18*a^4*b*tan(1/2*d*x + 1/2*c) + 27*a^ 3*b^2*tan(1/2*d*x + 1/2*c) + 6*a^2*b^3*tan(1/2*d*x + 1/2*c) + 3*a*b^4*tan( 1/2*d*x + 1/2*c) + 6*b^5*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 = 48.93 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.05 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a^2+3\,b^2\right )\,\left (2\,a-2\,b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{2\,\left (2\,a^3+3\,a\,b^2\right )\,\sqrt {a+b}\,{\left (a-b\right )}^{7/2}}\right )\,\left (2\,a^2+3\,b^2\right )}{d\,{\left (a+b\right )}^{7/2}\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (9\,a^2\,b+b^3\right )}{3\,{\left (a+b\right )}^2\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (6\,a^2\,b+3\,a\,b^2+2\,b^3\right )}{{\left (a+b\right )}^3\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a^2\,b-3\,a\,b^2+2\,b^3\right )}{\left (a+b\right )\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}}{d\,\left (3\,a\,b^2-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (-3\,a^3+3\,a^2\,b+3\,a\,b^2-3\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (-3\,a^3-3\,a^2\,b+3\,a\,b^2+3\,b^3\right )+3\,a^2\,b+a^3+b^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )\right )} \] Input:
int(1/(a + b*cos(c + d*x))^4,x)
Output:
(a*atan((a*tan(c/2 + (d*x)/2)*(2*a^2 + 3*b^2)*(2*a - 2*b)*(3*a*b^2 - 3*a^2 *b + a^3 - b^3))/(2*(3*a*b^2 + 2*a^3)*(a + b)^(1/2)*(a - b)^(7/2)))*(2*a^2 + 3*b^2))/(d*(a + b)^(7/2)*(a - b)^(7/2)) - ((4*tan(c/2 + (d*x)/2)^3*(9*a ^2*b + b^3))/(3*(a + b)^2*(a^2 - 2*a*b + b^2)) + (tan(c/2 + (d*x)/2)^5*(3* a*b^2 + 6*a^2*b + 2*b^3))/((a + b)^3*(a - b)) + (tan(c/2 + (d*x)/2)*(6*a^2 *b - 3*a*b^2 + 2*b^3))/((a + b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(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 + ta n(c/2 + (d*x)/2)^6*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))
Time = 0.20 (sec) , antiderivative size = 1065, normalized size of antiderivative = 5.79 \[ \int \frac {1}{(a+b \cos (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(1/(a+b*cos(d*x+c))^4,x)
Output:
(12*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**3*b**3 + 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**5 - 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 - 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**3 - 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**5 + 36*sqrt( a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b* *2))*sin(c + d*x)**2*a**4*b**2 + 54*sqrt(a**2 - b**2)*atan((tan((c + d*x)/ 2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**4 - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a **2 - b**2))*a**6 - 54*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**4*b**2 - 54*sqrt(a**2 - b**2)*atan((ta n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt(a**2 - b**2))*a**2*b**4 + 27*c os(c + d*x)*sin(c + d*x)*a**5*b**2 - 24*cos(c + d*x)*sin(c + d*x)*a**3*b** 4 - 3*cos(c + d*x)*sin(c + d*x)*a*b**6 - 11*sin(c + d*x)**3*a**4*b**3 + 7* sin(c + d*x)**3*a**2*b**5 + 4*sin(c + d*x)**3*b**7 + 18*sin(c + d*x)*a**6* b - 12*sin(c + d*x)*a**4*b**3 - 6*sin(c + d*x)*b**7)/(6*d*(cos(c + d*x)*si n(c + d*x)**2*a**8*b**3 - 4*cos(c + d*x)*sin(c + d*x)**2*a**6*b**5 + 6*...