Integrand size = 24, antiderivative size = 207 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=-\frac {a \left (5 a^2+3 c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{32 c^7 e}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a+a \cos (d+e x)+c \sin (d+e x))^3}+\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{96 c^4 e (a+a \cos (d+e x)+c \sin (d+e x))^2}-\frac {c \left (15 a^2+4 c^2\right ) \cos (d+e x)-a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{96 c^6 e (a+a \cos (d+e x)+c \sin (d+e x))} \] Output:
-1/32*a*(5*a^2+3*c^2)*ln(a+c*tan(1/2*e*x+1/2*d))/c^7/e-1/48*(c*cos(e*x+d)- a*sin(e*x+d))/c^2/e/(a+a*cos(e*x+d)+c*sin(e*x+d))^3+5/96*(a*c*cos(e*x+d)-a ^2*sin(e*x+d))/c^4/e/(a+a*cos(e*x+d)+c*sin(e*x+d))^2-1/96*(c*(15*a^2+4*c^2 )*cos(e*x+d)-a*(15*a^2+4*c^2)*sin(e*x+d))/c^6/e/(a+a*cos(e*x+d)+c*sin(e*x+ d))
Time = 4.15 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.92 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=\frac {\cos ^8\left (\frac {1}{2} (d+e x)\right ) \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right ) \left (3 c^4 \left (a^2+2 c^2\right ) \sec ^4\left (\frac {1}{2} (d+e x)\right )+c^6 \sec ^6\left (\frac {1}{2} (d+e x)\right )-2 \left (37 a^6+36 a^4 c^2+3 a^2 c^4+4 c^6+30 a^6 \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )+18 a^4 c^2 \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )+48 a c^5 \csc ^5(d+e x) \sin ^6\left (\frac {1}{2} (d+e x)\right )+3 a c \left (27 a^4+30 a^2 c^2+c^4+6 a^2 \left (5 a^2+3 c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )\right ) \tan \left (\frac {1}{2} (d+e x)\right )+6 c^2 \left (6 a^4+11 a^2 c^2+2 c^4+3 a^2 \left (5 a^2+3 c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (d+e x)\right )+6 a c^3 \left (-3 a^2+\left (5 a^2+3 c^2\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )\right ) \tan ^3\left (\frac {1}{2} (d+e x)\right )-6 a^2 c^4 \tan ^4\left (\frac {1}{2} (d+e x)\right )\right )\right )}{24 c^7 e (a+a \cos (d+e x)+c \sin (d+e x))^4} \] Input:
Integrate[(2*a + 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^(-4),x]
Output:
(Cos[(d + e*x)/2]^8*(a + c*Tan[(d + e*x)/2])*(3*c^4*(a^2 + 2*c^2)*Sec[(d + e*x)/2]^4 + c^6*Sec[(d + e*x)/2]^6 - 2*(37*a^6 + 36*a^4*c^2 + 3*a^2*c^4 + 4*c^6 + 30*a^6*Log[a + c*Tan[(d + e*x)/2]] + 18*a^4*c^2*Log[a + c*Tan[(d + e*x)/2]] + 48*a*c^5*Csc[d + e*x]^5*Sin[(d + e*x)/2]^6 + 3*a*c*(27*a^4 + 30*a^2*c^2 + c^4 + 6*a^2*(5*a^2 + 3*c^2)*Log[a + c*Tan[(d + e*x)/2]])*Tan[ (d + e*x)/2] + 6*c^2*(6*a^4 + 11*a^2*c^2 + 2*c^4 + 3*a^2*(5*a^2 + 3*c^2)*L og[a + c*Tan[(d + e*x)/2]])*Tan[(d + e*x)/2]^2 + 6*a*c^3*(-3*a^2 + (5*a^2 + 3*c^2)*Log[a + c*Tan[(d + e*x)/2]])*Tan[(d + e*x)/2]^3 - 6*a^2*c^4*Tan[( d + e*x)/2]^4)))/(24*c^7*e*(a + a*Cos[d + e*x] + c*Sin[d + e*x])^4)
Time = 0.85 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {3042, 3608, 27, 3042, 3635, 25, 3042, 3632, 3042, 3603, 16}
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.
| \(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(2 a \cos (d+e x)+2 a+2 c \sin (d+e x))^4}dx\) |
| \(\Big \downarrow \) 3608 |
| \(\displaystyle \frac {\int -\frac {-2 \cos (d+e x) a+3 a-2 c \sin (d+e x)}{4 (\cos (d+e x) a+a+c \sin (d+e x))^3}dx}{12 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {-2 \cos (d+e x) a+3 a-2 c \sin (d+e x)}{(\cos (d+e x) a+a+c \sin (d+e x))^3}dx}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {-2 \cos (d+e x) a+3 a-2 c \sin (d+e x)}{(\cos (d+e x) a+a+c \sin (d+e x))^3}dx}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3635 |
| \(\displaystyle -\frac {\frac {\int -\frac {-5 \cos (d+e x) a^2-5 c \sin (d+e x) a+2 \left (5 a^2+2 c^2\right )}{(\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {-\frac {\int \frac {-5 \cos (d+e x) a^2-5 c \sin (d+e x) a+2 \left (5 a^2+2 c^2\right )}{(\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {-5 \cos (d+e x) a^2-5 c \sin (d+e x) a+2 \left (5 a^2+2 c^2\right )}{(\cos (d+e x) a+a+c \sin (d+e x))^2}dx}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3632 |
| \(\displaystyle -\frac {-\frac {-3 a \left (\frac {5 a^2}{c^2}+3\right ) \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {c \left (15 a^2+4 c^2\right ) \cos (d+e x)-a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-3 a \left (\frac {5 a^2}{c^2}+3\right ) \int \frac {1}{\cos (d+e x) a+a+c \sin (d+e x)}dx-\frac {c \left (15 a^2+4 c^2\right ) \cos (d+e x)-a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 3603 |
| \(\displaystyle -\frac {-\frac {-\frac {6 a \left (\frac {5 a^2}{c^2}+3\right ) \int \frac {1}{2 a+2 c \tan \left (\frac {1}{2} (d+e x)\right )}d\tan \left (\frac {1}{2} (d+e x)\right )}{e}-\frac {c \left (15 a^2+4 c^2\right ) \cos (d+e x)-a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{2 c^2}-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {-\frac {5 \left (a c \cos (d+e x)-a^2 \sin (d+e x)\right )}{2 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^2}-\frac {-\frac {3 a \left (\frac {5 a^2}{c^2}+3\right ) \log \left (a+c \tan \left (\frac {1}{2} (d+e x)\right )\right )}{c e}-\frac {c \left (15 a^2+4 c^2\right ) \cos (d+e x)-a \left (15 a^2+4 c^2\right ) \sin (d+e x)}{c^2 e (a \cos (d+e x)+a+c \sin (d+e x))}}{2 c^2}}{48 c^2}-\frac {c \cos (d+e x)-a \sin (d+e x)}{48 c^2 e (a \cos (d+e x)+a+c \sin (d+e x))^3}\) |
Input:
Int[(2*a + 2*a*Cos[d + e*x] + 2*c*Sin[d + e*x])^(-4),x]
Output:
-1/48*(c*Cos[d + e*x] - a*Sin[d + e*x])/(c^2*e*(a + a*Cos[d + e*x] + c*Sin [d + e*x])^3) - ((-5*(a*c*Cos[d + e*x] - a^2*Sin[d + e*x]))/(2*c^2*e*(a + a*Cos[d + e*x] + c*Sin[d + e*x])^2) - ((-3*a*(3 + (5*a^2)/c^2)*Log[a + c*T an[(d + e*x)/2]])/(c*e) - (c*(15*a^2 + 4*c^2)*Cos[d + e*x] - a*(15*a^2 + 4 *c^2)*Sin[d + e*x])/(c^2*e*(a + a*Cos[d + e*x] + c*Sin[d + e*x])))/(2*c^2) )/(48*c^2)
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f /e) Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) /2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ (n_), x_Symbol] :> Simp[((-c)*Cos[d + e*x] + b*Sin[d + e*x])*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*(a^2 - b^2 - c^2))), x] + Simp[ 1/((n + 1)*(a^2 - b^2 - c^2)) Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c *(n + 2)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x ] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1] && NeQ[n, -3/2]
Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)]) /((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)*Sin[ d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Simp[(a*A - b*B - c*C)/(a^2 - b^2 - c^2) Int[1/(a + b*Cos[d + e*x] + c*S in[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]
Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]) ^(n_)*((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_) ]), x_Symbol] :> Simp[(-(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A) *Sin[d + e*x]))*((a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*( a^2 - b^2 - c^2))), x] + Simp[1/((n + 1)*(a^2 - b^2 - c^2)) Int[(a + b*Co s[d + e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - b*B - c*C) + (n + 2)*(a*B - b*A)*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]
Time = 1.19 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {-60 a \left (a^{2}+\frac {3 c^{2}}{5}\right ) \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3} \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6} c^{6}-3 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5} c^{5}+\left (15 a^{2} c^{4}+9 c^{6}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\left (-180 c^{2} a^{4}-108 a^{2} c^{4}-9 c^{6}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\left (-270 a^{5} c -162 a^{3} c^{3}-9 a \,c^{5}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )-110 a^{6}-66 c^{2} a^{4}-3 a^{2} c^{4}-c^{6}}{384 c^{7} e \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}\) | \(215\) |
| derivativedivides | \(\frac {\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} c^{2}}{3}-2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} c +10 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c^{2}}{8 c^{6}}-\frac {a \left (5 a^{2}+3 c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 c^{7}}-\frac {a^{6}+3 c^{2} a^{4}+3 a^{2} c^{4}+c^{6}}{24 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}-\frac {15 a^{4}+18 a^{2} c^{2}+3 c^{4}}{8 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {3 a \left (a^{4}+2 a^{2} c^{2}+c^{4}\right )}{8 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}}{16 e}\) | \(221\) |
| default | \(\frac {\frac {\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} c^{2}}{3}-2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} c +10 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c^{2}}{8 c^{6}}-\frac {a \left (5 a^{2}+3 c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{2 c^{7}}-\frac {a^{6}+3 c^{2} a^{4}+3 a^{2} c^{4}+c^{6}}{24 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}-\frac {15 a^{4}+18 a^{2} c^{2}+3 c^{4}}{8 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}+\frac {3 a \left (a^{4}+2 a^{2} c^{2}+c^{4}\right )}{8 c^{7} \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{2}}}{16 e}\) | \(221\) |
| norman | \(\frac {-\frac {110 a^{6}+66 c^{2} a^{4}+3 a^{2} c^{4}+c^{6}}{384 c^{7} e}+\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}}{384 c e}-\frac {a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{128 c^{2} e}+\frac {\left (5 a^{2}+3 c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{128 c^{3} e}-\frac {3 \left (20 a^{4}+12 a^{2} c^{2}+c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{128 c^{5} e}-\frac {3 a \left (30 a^{4}+18 a^{2} c^{2}+c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{128 c^{6} e}}{\left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )^{3}}-\frac {a \left (5 a^{2}+3 c^{2}\right ) \ln \left (a +c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )\right )}{32 c^{7} e}\) | \(226\) |
| risch | \(\frac {i \left (-4 c^{5}-3 a^{2} c^{3}-75 i a^{5} {\mathrm e}^{i \left (e x +d \right )}+41 i a^{3} c^{2}+12 i a \,c^{4}-15 i a^{5} {\mathrm e}^{5 i \left (e x +d \right )}-30 a^{4} c \,{\mathrm e}^{5 i \left (e x +d \right )}-18 a^{2} c^{3} {\mathrm e}^{5 i \left (e x +d \right )}-75 a^{4} c \,{\mathrm e}^{4 i \left (e x +d \right )}+150 a^{4} c \,{\mathrm e}^{i \left (e x +d \right )}+30 a^{2} c^{3} {\mathrm e}^{i \left (e x +d \right )}-75 i a^{5} {\mathrm e}^{4 i \left (e x +d \right )}-150 i a^{5} {\mathrm e}^{3 i \left (e x +d \right )}-150 i a^{5} {\mathrm e}^{2 i \left (e x +d \right )}-45 a^{2} c^{3} {\mathrm e}^{4 i \left (e x +d \right )}+60 a^{2} c^{3} {\mathrm e}^{2 i \left (e x +d \right )}+150 a^{4} c \,{\mathrm e}^{2 i \left (e x +d \right )}+45 a^{4} c -60 i a^{3} c^{2} {\mathrm e}^{2 i \left (e x +d \right )}+6 i a^{3} c^{2} {\mathrm e}^{5 i \left (e x +d \right )}+9 i a \,c^{4} {\mathrm e}^{5 i \left (e x +d \right )}-45 i a^{3} c^{2} {\mathrm e}^{4 i \left (e x +d \right )}-130 i a^{3} c^{2} {\mathrm e}^{3 i \left (e x +d \right )}-24 i a \,c^{4} {\mathrm e}^{3 i \left (e x +d \right )}+60 i a^{3} c^{2} {\mathrm e}^{i \left (e x +d \right )}+15 i a \,c^{4} {\mathrm e}^{i \left (e x +d \right )}-12 i a \,c^{4} {\mathrm e}^{2 i \left (e x +d \right )}-15 i a^{5}+12 c^{5} {\mathrm e}^{2 i \left (e x +d \right )}\right )}{48 \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}+i a \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}+i a \right )^{3} c^{6} e}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{32 c^{7} e}+\frac {3 a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+1\right )}{32 c^{5} e}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (e x +d \right )}-\frac {i c +a}{i c -a}\right )}{32 c^{7} e}-\frac {3 a \ln \left ({\mathrm e}^{i \left (e x +d \right )}-\frac {i c +a}{i c -a}\right )}{32 c^{5} e}\) | \(593\) |
Input:
int(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))^4,x,method=_RETURNVERBOSE)
Output:
1/384*(-60*a*(a^2+3/5*c^2)*(a+c*tan(1/2*e*x+1/2*d))^3*ln(a+c*tan(1/2*e*x+1 /2*d))+tan(1/2*e*x+1/2*d)^6*c^6-3*a*tan(1/2*e*x+1/2*d)^5*c^5+(15*a^2*c^4+9 *c^6)*tan(1/2*e*x+1/2*d)^4+(-180*a^4*c^2-108*a^2*c^4-9*c^6)*tan(1/2*e*x+1/ 2*d)^2+(-270*a^5*c-162*a^3*c^3-9*a*c^5)*tan(1/2*e*x+1/2*d)-110*a^6-66*c^2* a^4-3*a^2*c^4-c^6)/c^7/e/(a+c*tan(1/2*e*x+1/2*d))^3
Leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (198) = 396\).
Time = 0.10 (sec) , antiderivative size = 791, normalized size of antiderivative = 3.82 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))^4,x, algorithm="fricas")
Output:
1/192*(60*a^4*c^2 + 6*a^2*c^4 - 2*(45*a^4*c^2 - 3*a^2*c^4 - 4*c^6)*cos(e*x + d)^3 - 12*(10*a^4*c^2 + a^2*c^4)*cos(e*x + d)^2 + 6*(5*a^4*c^2 - 2*a^2* c^4 - 2*c^6)*cos(e*x + d) - 3*(5*a^6 + 18*a^4*c^2 + 9*a^2*c^4 + (5*a^6 - 1 2*a^4*c^2 - 9*a^2*c^4)*cos(e*x + d)^3 + 3*(5*a^6 - 2*a^4*c^2 - 3*a^2*c^4)* cos(e*x + d)^2 + 3*(5*a^6 + 8*a^4*c^2 + 3*a^2*c^4)*cos(e*x + d) + (15*a^5* c + 14*a^3*c^3 + 3*a*c^5 + (15*a^5*c + 4*a^3*c^3 - 3*a*c^5)*cos(e*x + d)^2 + 6*(5*a^5*c + 3*a^3*c^3)*cos(e*x + d))*sin(e*x + d))*log(a*c*sin(e*x + d ) + 1/2*a^2 + 1/2*c^2 + 1/2*(a^2 - c^2)*cos(e*x + d)) + 3*(5*a^6 + 18*a^4* c^2 + 9*a^2*c^4 + (5*a^6 - 12*a^4*c^2 - 9*a^2*c^4)*cos(e*x + d)^3 + 3*(5*a ^6 - 2*a^4*c^2 - 3*a^2*c^4)*cos(e*x + d)^2 + 3*(5*a^6 + 8*a^4*c^2 + 3*a^2* c^4)*cos(e*x + d) + (15*a^5*c + 14*a^3*c^3 + 3*a*c^5 + (15*a^5*c + 4*a^3*c ^3 - 3*a*c^5)*cos(e*x + d)^2 + 6*(5*a^5*c + 3*a^3*c^3)*cos(e*x + d))*sin(e *x + d))*log(1/2*cos(e*x + d) + 1/2) + 2*(15*a^5*c + 14*a^3*c^3 + 6*a*c^5 + (15*a^5*c - 41*a^3*c^3 - 12*a*c^5)*cos(e*x + d)^2 + 3*(10*a^5*c - 9*a^3* c^3 - a*c^5)*cos(e*x + d))*sin(e*x + d))/((a^3*c^7 - 3*a*c^9)*e*cos(e*x + d)^3 + 3*(a^3*c^7 - a*c^9)*e*cos(e*x + d)^2 + 3*(a^3*c^7 + a*c^9)*e*cos(e* x + d) + (a^3*c^7 + 3*a*c^9)*e + (6*a^2*c^8*e*cos(e*x + d) + (3*a^2*c^8 - c^10)*e*cos(e*x + d)^2 + (3*a^2*c^8 + c^10)*e)*sin(e*x + d))
Timed out. \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=\text {Timed out} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))**4,x)
Output:
Timed out
Time = 0.06 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=-\frac {\frac {37 \, a^{6} + 39 \, a^{4} c^{2} + 3 \, a^{2} c^{4} + c^{6} + \frac {9 \, {\left (9 \, a^{5} c + 10 \, a^{3} c^{3} + a c^{5}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {9 \, {\left (5 \, a^{4} c^{2} + 6 \, a^{2} c^{4} + c^{6}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{a^{3} c^{7} + \frac {3 \, a^{2} c^{8} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {3 \, a c^{9} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {c^{10} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}}} + \frac {\frac {6 \, a c \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} - \frac {c^{2} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} - \frac {3 \, {\left (10 \, a^{2} + 3 \, c^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}}{c^{6}} + \frac {12 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} \log \left (a + \frac {c \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right )}{c^{7}}}{384 \, e} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))^4,x, algorithm="maxima")
Output:
-1/384*((37*a^6 + 39*a^4*c^2 + 3*a^2*c^4 + c^6 + 9*(9*a^5*c + 10*a^3*c^3 + a*c^5)*sin(e*x + d)/(cos(e*x + d) + 1) + 9*(5*a^4*c^2 + 6*a^2*c^4 + c^6)* sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/(a^3*c^7 + 3*a^2*c^8*sin(e*x + d)/(co s(e*x + d) + 1) + 3*a*c^9*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + c^10*sin(e *x + d)^3/(cos(e*x + d) + 1)^3) + (6*a*c*sin(e*x + d)^2/(cos(e*x + d) + 1) ^2 - c^2*sin(e*x + d)^3/(cos(e*x + d) + 1)^3 - 3*(10*a^2 + 3*c^2)*sin(e*x + d)/(cos(e*x + d) + 1))/c^6 + 12*(5*a^3 + 3*a*c^2)*log(a + c*sin(e*x + d) /(cos(e*x + d) + 1))/c^7)/e
Time = 0.14 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=-\frac {\frac {12 \, {\left (5 \, a^{3} + 3 \, a c^{2}\right )} \log \left ({\left | c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a \right |}\right )}{c^{7}} - \frac {110 \, a^{3} c^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 66 \, a c^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 285 \, a^{4} c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 144 \, a^{2} c^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 9 \, c^{6} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 249 \, a^{5} c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 108 \, a^{3} c^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 9 \, a c^{5} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 73 \, a^{6} + 27 \, a^{4} c^{2} - 3 \, a^{2} c^{4} - c^{6}}{{\left (c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a\right )}^{3} c^{7}} - \frac {c^{8} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 6 \, a c^{7} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 30 \, a^{2} c^{6} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + 9 \, c^{8} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )}{c^{12}}}{384 \, e} \] Input:
integrate(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))^4,x, algorithm="giac")
Output:
-1/384*(12*(5*a^3 + 3*a*c^2)*log(abs(c*tan(1/2*e*x + 1/2*d) + a))/c^7 - (1 10*a^3*c^3*tan(1/2*e*x + 1/2*d)^3 + 66*a*c^5*tan(1/2*e*x + 1/2*d)^3 + 285* a^4*c^2*tan(1/2*e*x + 1/2*d)^2 + 144*a^2*c^4*tan(1/2*e*x + 1/2*d)^2 - 9*c^ 6*tan(1/2*e*x + 1/2*d)^2 + 249*a^5*c*tan(1/2*e*x + 1/2*d) + 108*a^3*c^3*ta n(1/2*e*x + 1/2*d) - 9*a*c^5*tan(1/2*e*x + 1/2*d) + 73*a^6 + 27*a^4*c^2 - 3*a^2*c^4 - c^6)/((c*tan(1/2*e*x + 1/2*d) + a)^3*c^7) - (c^8*tan(1/2*e*x + 1/2*d)^3 - 6*a*c^7*tan(1/2*e*x + 1/2*d)^2 + 30*a^2*c^6*tan(1/2*e*x + 1/2* d) + 9*c^8*tan(1/2*e*x + 1/2*d))/c^12)/e
Time = 16.58 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3}{384\,c^4\,e}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (\frac {3}{128\,c^4}+\frac {5\,a^2}{64\,c^6}\right )}{e}-\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (27\,a^5+30\,a^3\,c^2+3\,a\,c^4\right )+\frac {37\,a^6+39\,a^4\,c^2+3\,a^2\,c^4+c^6}{3\,c}+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (15\,a^4\,c+18\,a^2\,c^3+3\,c^5\right )}{e\,\left (128\,a^3\,c^6+384\,a^2\,c^7\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+384\,a\,c^8\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+128\,c^9\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\right )}-\frac {a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2}{64\,c^5\,e}-\frac {\ln \left (a+c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )\,\left (5\,a^3+3\,a\,c^2\right )}{32\,c^7\,e} \] Input:
int(1/(2*a + 2*a*cos(d + e*x) + 2*c*sin(d + e*x))^4,x)
Output:
tan(d/2 + (e*x)/2)^3/(384*c^4*e) + (tan(d/2 + (e*x)/2)*(3/(128*c^4) + (5*a ^2)/(64*c^6)))/e - (tan(d/2 + (e*x)/2)*(3*a*c^4 + 27*a^5 + 30*a^3*c^2) + ( 37*a^6 + c^6 + 3*a^2*c^4 + 39*a^4*c^2)/(3*c) + tan(d/2 + (e*x)/2)^2*(15*a^ 4*c + 3*c^5 + 18*a^2*c^3))/(e*(128*c^9*tan(d/2 + (e*x)/2)^3 + 128*a^3*c^6 + 384*a^2*c^7*tan(d/2 + (e*x)/2) + 384*a*c^8*tan(d/2 + (e*x)/2)^2)) - (a*t an(d/2 + (e*x)/2)^2)/(64*c^5*e) - (log(a + c*tan(d/2 + (e*x)/2))*(3*a*c^2 + 5*a^3))/(32*c^7*e)
Time = 0.21 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(2 a+2 a \cos (d+e x)+2 c \sin (d+e x))^4} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} a^{4} c^{3}-36 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} a^{2} c^{5}-180 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a^{5} c^{2}-108 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a^{3} c^{4}-180 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{6} c -108 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{4} c^{3}-60 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) a^{7}-36 \,\mathrm {log}\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) c +a \right ) a^{5} c^{2}+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6} a \,c^{6}-3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5} a^{2} c^{5}+15 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4} a^{3} c^{4}+9 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4} a \,c^{6}+60 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} a^{4} c^{3}+36 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} a^{2} c^{5}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} c^{7}-90 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{6} c -54 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{4} c^{3}-50 a^{7}-30 a^{5} c^{2}-a \,c^{6}}{384 a \,c^{7} e \left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3} c^{3}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a \,c^{2}+3 \tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a^{2} c +a^{3}\right )} \] Input:
int(1/(2*a+2*a*cos(e*x+d)+2*c*sin(e*x+d))^4,x)
Output:
( - 60*log(tan((d + e*x)/2)*c + a)*tan((d + e*x)/2)**3*a**4*c**3 - 36*log( tan((d + e*x)/2)*c + a)*tan((d + e*x)/2)**3*a**2*c**5 - 180*log(tan((d + e *x)/2)*c + a)*tan((d + e*x)/2)**2*a**5*c**2 - 108*log(tan((d + e*x)/2)*c + a)*tan((d + e*x)/2)**2*a**3*c**4 - 180*log(tan((d + e*x)/2)*c + a)*tan((d + e*x)/2)*a**6*c - 108*log(tan((d + e*x)/2)*c + a)*tan((d + e*x)/2)*a**4* c**3 - 60*log(tan((d + e*x)/2)*c + a)*a**7 - 36*log(tan((d + e*x)/2)*c + a )*a**5*c**2 + tan((d + e*x)/2)**6*a*c**6 - 3*tan((d + e*x)/2)**5*a**2*c**5 + 15*tan((d + e*x)/2)**4*a**3*c**4 + 9*tan((d + e*x)/2)**4*a*c**6 + 60*ta n((d + e*x)/2)**3*a**4*c**3 + 36*tan((d + e*x)/2)**3*a**2*c**5 + 3*tan((d + e*x)/2)**3*c**7 - 90*tan((d + e*x)/2)*a**6*c - 54*tan((d + e*x)/2)*a**4* c**3 - 50*a**7 - 30*a**5*c**2 - a*c**6)/(384*a*c**7*e*(tan((d + e*x)/2)**3 *c**3 + 3*tan((d + e*x)/2)**2*a*c**2 + 3*tan((d + e*x)/2)*a**2*c + a**3))