Integrand size = 29, antiderivative size = 180 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 a x}{b^4}-\frac {\left (6 a^4-9 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{3/2} d}+\frac {3 \cos (c+d x)}{2 b^3 d}-\frac {\cos (c+d x) \sin ^2(c+d x)}{2 b d (a+b \sin (c+d x))^2}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{2 b^3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:
3*a*x/b^4-(6*a^4-9*a^2*b^2+2*b^4)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2 )^(1/2))/b^4/(a^2-b^2)^(3/2)/d+3/2*cos(d*x+c)/b^3/d-1/2*cos(d*x+c)*sin(d*x +c)^2/b/d/(a+b*sin(d*x+c))^2+1/2*a*(3*a^2-2*b^2)*cos(d*x+c)/b^3/(a^2-b^2)/ d/(a+b*sin(d*x+c))
Time = 1.65 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {6 a (c+d x)-\frac {2 \left (6 a^4-9 a^2 b^2+2 b^4\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+2 b \cos (c+d x)-\frac {a^2 b \cos (c+d x)}{(a+b \sin (c+d x))^2}+\frac {a b \left (5 a^2-4 b^2\right ) \cos (c+d x)}{(a-b) (a+b) (a+b \sin (c+d x))}}{2 b^4 d} \] Input:
Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
(6*a*(c + d*x) - (2*(6*a^4 - 9*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[(c + d*x )/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 2*b*Cos[c + d*x] - (a^2*b*Cos[ c + d*x])/(a + b*Sin[c + d*x])^2 + (a*b*(5*a^2 - 4*b^2)*Cos[c + d*x])/((a - b)*(a + b)*(a + b*Sin[c + d*x])))/(2*b^4*d)
Time = 1.27 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.28, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.552, Rules used = {3042, 3368, 3042, 3527, 25, 3042, 3511, 25, 3042, 3502, 3042, 3214, 3042, 3139, 1083, 217}
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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^2}{(a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3368 |
\(\displaystyle \int \frac {\sin ^2(c+d x) \left (1-\sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \left (1-\sin (c+d x)^2\right )}{(a+b \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3527 |
\(\displaystyle -\frac {\int -\frac {\sin (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\sin (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin (c+d x) \left (2 \left (a^2-b^2\right )-3 \left (a^2-b^2\right ) \sin (c+d x)^2\right )}{(a+b \sin (c+d x))^2}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3511 |
\(\displaystyle \frac {\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}-\frac {\int -\frac {-3 b \sin ^2(c+d x) \left (a^2-b^2\right )^2+3 a \sin (c+d x) \left (a^2-b^2\right )^2+b \left (3 a^4-5 b^2 a^2+2 b^4\right )}{a+b \sin (c+d x)}dx}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {-3 b \sin ^2(c+d x) \left (a^2-b^2\right )^2+3 a \sin (c+d x) \left (a^2-b^2\right )^2+b \left (3 a^4-5 b^2 a^2+2 b^4\right )}{a+b \sin (c+d x)}dx}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {-3 b \sin (c+d x)^2 \left (a^2-b^2\right )^2+3 a \sin (c+d x) \left (a^2-b^2\right )^2+b \left (3 a^4-5 b^2 a^2+2 b^4\right )}{a+b \sin (c+d x)}dx}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\left (3 a^4-5 b^2 a^2+2 b^4\right ) b^2+6 a \left (a^2-b^2\right )^2 \sin (c+d x) b}{a+b \sin (c+d x)}dx}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {\left (3 a^4-5 b^2 a^2+2 b^4\right ) b^2+6 a \left (a^2-b^2\right )^2 \sin (c+d x) b}{a+b \sin (c+d x)}dx}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {\frac {6 a x \left (a^2-b^2\right )^2-\left (6 a^6-15 a^4 b^2+11 a^2 b^4-2 b^6\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {6 a x \left (a^2-b^2\right )^2-\left (6 a^6-15 a^4 b^2+11 a^2 b^4-2 b^6\right ) \int \frac {1}{a+b \sin (c+d x)}dx}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {\frac {\frac {6 a x \left (a^2-b^2\right )^2-\frac {2 \left (6 a^6-15 a^4 b^2+11 a^2 b^4-2 b^6\right ) \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {\frac {\frac {4 \left (6 a^6-15 a^4 b^2+11 a^2 b^4-2 b^6\right ) \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{d}+6 a x \left (a^2-b^2\right )^2}{b}+\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}}{b^2 \left (a^2-b^2\right )}+\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {a \left (3 a^2-2 b^2\right ) \cos (c+d x)}{b^2 d (a+b \sin (c+d x))}+\frac {\frac {3 \left (a^2-b^2\right )^2 \cos (c+d x)}{d}+\frac {6 a x \left (a^2-b^2\right )^2-\frac {2 \left (6 a^6-15 a^4 b^2+11 a^2 b^4-2 b^6\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \sqrt {a^2-b^2}}}{b}}{b^2 \left (a^2-b^2\right )}}{2 b \left (a^2-b^2\right )}-\frac {\sin ^2(c+d x) \cos (c+d x)}{2 b d (a+b \sin (c+d x))^2}\) |
Input:
Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^3,x]
Output:
-1/2*(Cos[c + d*x]*Sin[c + d*x]^2)/(b*d*(a + b*Sin[c + d*x])^2) + (((6*a*( a^2 - b^2)^2*x - (2*(6*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - 2*b^6)*ArcTan[(2*b + 2*a*Tan[(c + d*x)/2])/(2*Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*d))/b + (3* (a^2 - b^2)^2*Cos[c + d*x])/d)/(b^2*(a^2 - b^2)) + (a*(3*a^2 - 2*b^2)*Cos[ c + d*x])/(b^2*d*(a + b*Sin[c + d*x])))/(2*b*(a^2 - b^2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *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_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n }, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d )) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b *C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e , f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Time = 1.37 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.61
method | result | size |
derivativedivides | \(\frac {\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {a \,b^{2} \left (3 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {b \left (4 a^{4}+5 a^{2} b^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{2}-b^{2}\right )}-\frac {a \,b^{2} \left (13 a^{2}-10 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a^{2} b \left (4 a^{2}-3 b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (6 a^{4}-9 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}}{d}\) | \(289\) |
default | \(\frac {\frac {\frac {2 b}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}-\frac {2 \left (\frac {-\frac {a \,b^{2} \left (3 a^{2}-2 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a^{2}-b^{2}\right )}-\frac {b \left (4 a^{4}+5 a^{2} b^{2}-6 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 \left (a^{2}-b^{2}\right )}-\frac {a \,b^{2} \left (13 a^{2}-10 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a^{2}-b^{2}\right )}-\frac {a^{2} b \left (4 a^{2}-3 b^{2}\right )}{2 \left (a^{2}-b^{2}\right )}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )^{2}}+\frac {\left (6 a^{4}-9 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}}{d}\) | \(289\) |
risch | \(\frac {3 a x}{b^{4}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b^{3} d}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{3} d}-\frac {i a \left (-6 i a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}+5 i a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+14 i a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-11 i a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+10 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-3 b^{2} a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-5 a^{2} b^{2}+4 b^{4}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} \left (a^{2}-b^{2}\right ) d \,b^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{4}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right ) a^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(710\) |
Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(2/b^4*(b/(1+tan(1/2*d*x+1/2*c)^2)+3*a*arctan(tan(1/2*d*x+1/2*c)))-2/b ^4*((-1/2*a*b^2*(3*a^2-2*b^2)/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3-1/2*b*(4*a^4+ 5*a^2*b^2-6*b^4)/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2-1/2*a*b^2*(13*a^2-10*b^2)/ (a^2-b^2)*tan(1/2*d*x+1/2*c)-1/2*a^2*b*(4*a^2-3*b^2)/(a^2-b^2))/(tan(1/2*d *x+1/2*c)^2*a+2*b*tan(1/2*d*x+1/2*c)+a)^2+1/2*(6*a^4-9*a^2*b^2+2*b^4)/(a^2 -b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (169) = 338\).
Time = 0.13 (sec) , antiderivative size = 919, normalized size of antiderivative = 5.11 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="frica s")
Output:
[1/4*(12*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 4*(a^4*b^3 - 2 *a^2*b^5 + b^7)*cos(d*x + c)^3 - 12*(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b^2 - 7*a^2*b^4 + 2*b^6 - (6*a^4*b^2 - 9*a^2*b^4 + 2*b^6) *cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(-a^ 2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^ 2 + 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^ 2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(6*a^6*b - 9*a^4*b ^3 + a^2*b^5 + 2*b^7)*cos(d*x + c) - 2*(12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d *x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4* b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9) *d*sin(d*x + c) - (a^6*b^4 - a^4*b^6 - a^2*b^8 + b^10)*d), 1/2*(6*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*d*x*cos(d*x + c)^2 + 2*(a^4*b^3 - 2*a^2*b^5 + b^7)*c os(d*x + c)^3 - 6*(a^7 - a^5*b^2 - a^3*b^4 + a*b^6)*d*x - (6*a^6 - 3*a^4*b ^2 - 7*a^2*b^4 + 2*b^6 - (6*a^4*b^2 - 9*a^2*b^4 + 2*b^6)*cos(d*x + c)^2 + 2*(6*a^5*b - 9*a^3*b^3 + 2*a*b^5)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a *sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (6*a^6*b - 9*a^4*b^3 + a^2*b^5 + 2*b^7)*cos(d*x + c) - (12*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*x + (9*a^5*b^2 - 17*a^3*b^4 + 8*a*b^6)*cos(d*x + c))*sin(d*x + c))/((a^4*b^6 - 2*a^2*b^8 + b^10)*d*cos(d*x + c)^2 - 2*(a^5*b^5 - 2*a^3*b^7 + a*b^9)*d*si n(d*x + c) - (a^6*b^4 - a^4*b^6 - a^2*b^8 + b^10)*d)]
Timed out. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+b*sin(d*x+c))**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="maxim a")
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
Time = 0.15 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {\frac {{\left (6 \, a^{4} - 9 \, a^{2} b^{2} + 2 \, b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {a^{2} - b^{2}}} - \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} - 3 \, a^{2} b^{2}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2}} - \frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}}}{d} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-((6*a^4 - 9*a^2*b^2 + 2*b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + a rctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/((a^2*b^4 - b^6)*sqrt (a^2 - b^2)) - (3*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^3*tan(1/2*d*x + 1/2 *c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 + 5*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 - 6*b^4*tan(1/2*d*x + 1/2*c)^2 + 13*a^3*b*tan(1/2*d*x + 1/2*c) - 10*a*b^3*ta n(1/2*d*x + 1/2*c) + 4*a^4 - 3*a^2*b^2)/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2) - 3*(d*x + c)*a/b^4 - 2/((tan( 1/2*d*x + 1/2*c)^2 + 1)*b^3))/d
Time = 38.51 (sec) , antiderivative size = 3031, normalized size of antiderivative = 16.84 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + b*sin(c + d*x))^3,x)
Output:
((6*a^4 - 5*a^2*b^2)/(b^3*(a^2 - b^2)) + (3*tan(c/2 + (d*x)/2)^4*(2*a^4 - 2*b^4 + a^2*b^2))/(b^3*(a^2 - b^2)) + (2*tan(c/2 + (d*x)/2)^2*(6*a^4 - 7*b ^4 + 3*a^2*b^2))/(b^3*(a^2 - b^2)) - (3*tan(c/2 + (d*x)/2)*(6*a*b^2 - 7*a^ 3))/(b^2*(a^2 - b^2)) - (tan(c/2 + (d*x)/2)^5*(2*a*b^2 - 3*a^3))/(b^2*(a^2 - b^2)) - (4*tan(c/2 + (d*x)/2)^3*(5*a*b^2 - 6*a^3))/(b^2*(a^2 - b^2)))/( d*(tan(c/2 + (d*x)/2)^2*(3*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^4*(3*a^2 + 4* b^2) + a^2*tan(c/2 + (d*x)/2)^6 + a^2 + 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*a*b *tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) + (6*a*atan((192*a^2*b^ 7*tan(c/2 + (d*x)/2))/((192*a^2*b^19)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (384 *a^4*b^17)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (48*a^6*b^15)/(b^12 - 2*a^2*b^1 0 + a^4*b^8) + (288*a^8*b^13)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (144*a^10*b^ 11)/(b^12 - 2*a^2*b^10 + a^4*b^8)) - (144*a^6*b^3*tan(c/2 + (d*x)/2))/((19 2*a^2*b^19)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (384*a^4*b^17)/(b^12 - 2*a^2*b ^10 + a^4*b^8) + (48*a^6*b^15)/(b^12 - 2*a^2*b^10 + a^4*b^8) + (288*a^8*b^ 13)/(b^12 - 2*a^2*b^10 + a^4*b^8) - (144*a^10*b^11)/(b^12 - 2*a^2*b^10 + a ^4*b^8))))/(b^4*d) + (atan((((-(a + b)^3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - ( 9*a^2*b^2)/2)*((8*(36*a^4*b^7 - 72*a^6*b^5 + 36*a^8*b^3))/(b^12 - 2*a^2*b^ 10 + a^4*b^8) - (8*tan(c/2 + (d*x)/2)*(4*a*b^11 - 108*a^3*b^9 + 285*a^5*b^ 7 - 252*a^7*b^5 + 72*a^9*b^3))/(b^13 - 2*a^2*b^11 + a^4*b^9) + ((-(a + b)^ 3*(a - b)^3)^(1/2)*(3*a^4 + b^4 - (9*a^2*b^2)/2)*((8*tan(c/2 + (d*x)/2)...
Time = 0.18 (sec) , antiderivative size = 941, normalized size of antiderivative = 5.23 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+b*sin(d*x+c))^3,x)
Output:
( - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))* sin(c + d*x)**2*a**4*b**2 + 36*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*a**2*b**4 - 8*sqrt(a**2 - b**2)*at an((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)**2*b**6 - 48*s qrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**5*b + 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3*b**3 - 16*sqrt(a**2 - b**2)*atan((tan((c + d*x )/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b**5 - 24*sqrt(a**2 - b**2)* atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**6 + 36*sqrt(a**2 - b** 2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*b**2 - 8*sqrt(a** 2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**4 + 4*c os(c + d*x)*sin(c + d*x)**2*a**4*b**3 - 8*cos(c + d*x)*sin(c + d*x)**2*a** 2*b**5 + 4*cos(c + d*x)*sin(c + d*x)**2*b**7 + 18*cos(c + d*x)*sin(c + d*x )*a**5*b**2 - 34*cos(c + d*x)*sin(c + d*x)*a**3*b**4 + 16*cos(c + d*x)*sin (c + d*x)*a*b**6 + 12*cos(c + d*x)*a**6*b - 22*cos(c + d*x)*a**4*b**3 + 10 *cos(c + d*x)*a**2*b**5 + 12*sin(c + d*x)**2*a**5*b**2*d*x + 9*sin(c + d*x )**2*a**4*b**3 - 24*sin(c + d*x)**2*a**3*b**4*d*x - 17*sin(c + d*x)**2*a** 2*b**5 + 12*sin(c + d*x)**2*a*b**6*d*x + 8*sin(c + d*x)**2*b**7 + 24*sin(c + d*x)*a**6*b*d*x + 18*sin(c + d*x)*a**5*b**2 - 48*sin(c + d*x)*a**4*b**3 *d*x - 34*sin(c + d*x)*a**3*b**4 + 24*sin(c + d*x)*a**2*b**5*d*x + 16*s...