Integrand size = 25, antiderivative size = 298 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=-\frac {2 d^3 (4 c+3 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^4 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 c-9 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))} \] Output:
-2*d^3*(4*c+3*d)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a^3/(c-d )^4/(c+d)/(c^2-d^2)^(1/2)/f-1/15*d*(2*c^3-12*c^2*d+43*c*d^2+72*d^3)*cos(f* x+e)/a^3/(c-d)^4/(c+d)/f/(c+d*sin(f*x+e))-1/5*cos(f*x+e)/(c-d)/f/(a+a*sin( f*x+e))^3/(c+d*sin(f*x+e))-1/15*(2*c-9*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin( f*x+e))^2/(c+d*sin(f*x+e))-1/15*(2*c^2-12*c*d+45*d^2)*cos(f*x+e)/(c-d)^3/f /(a^3+a^3*sin(f*x+e))/(c+d*sin(f*x+e))
Time = 7.57 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 (c-6 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (c-6 d) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 \left (2 c^2-14 c d+57 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-\frac {30 d^3 (4 c+3 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d) \sqrt {c^2-d^2}}-\frac {15 d^4 \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{(c+d) (c+d \sin (e+f x))}\right )}{15 a^3 (c-d)^4 f (1+\sin (e+f x))^3} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2),x]
Output:
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(6*(c - d)^2*Sin[(e + f*x)/2] - 3*( c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 4*(c - 6*d)*(c - d)*Sin[( e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*(c - 6*d)*(c - d)* (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 2*(2*c^2 - 14*c*d + 57*d^2)*Sin[ (e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (30*d^3*(4*c + 3*d) *ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[ (e + f*x)/2])^5)/((c + d)*Sqrt[c^2 - d^2]) - (15*d^4*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)/((c + d)*(c + d*Sin[e + f*x]))))/(15*a^3* (c - d)^4*f*(1 + Sin[e + f*x])^3)
Time = 1.40 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.09, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 3245, 25, 3042, 3457, 25, 3042, 3457, 25, 3042, 3233, 27, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c+d \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c+d \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3245 |
| \(\displaystyle -\frac {\int -\frac {2 a (c-3 d)+3 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))^2}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 a (c-3 d)+3 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))^2}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (c-3 d)+3 a d \sin (e+f x)}{(\sin (e+f x) a+a)^2 (c+d \sin (e+f x))^2}dx}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {-\frac {\int -\frac {\left (2 c^2-8 d c+27 d^2\right ) a^2+2 (2 c-9 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 c^2-8 d c+27 d^2\right ) a^2+2 (2 c-9 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\left (2 c^2-8 d c+27 d^2\right ) a^2+2 (2 c-9 d) d \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {-\frac {\int -\frac {2 (c-36 d) d^2 a^3+d \left (2 c^2-12 d c+45 d^2\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 (c-36 d) d^2 a^3+d \left (2 c^2-12 d c+45 d^2\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 (c-36 d) d^2 a^3+d \left (2 c^2-12 d c+45 d^2\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {15 a^3 d^3 (4 c+3 d)}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {-\frac {15 a^3 d^3 (4 c+3 d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {-\frac {15 a^3 d^3 (4 c+3 d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {\frac {-\frac {30 a^3 d^3 (4 c+3 d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {\frac {\frac {60 a^3 d^3 (4 c+3 d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {\frac {-\frac {30 a^3 d^3 (4 c+3 d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a^3 d \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (2 c^2-12 c d+45 d^2\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 c-9 d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2),x]
Output:
-1/5*Cos[e + f*x]/((c - d)*f*(a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])) + (-1/3*(a*(2*c - 9*d)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])) + (-((a^2*(2*c^2 - 12*c*d + 45*d^2)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])*(c + d*Sin[e + f*x]))) + ((-30*a^3*d^3*(4*c + 3 *d)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c^2 - d^2) ^(3/2)*f) - (a^3*d*(2*c^3 - 12*c^2*d + 43*c*d^2 + 72*d^3)*Cos[e + f*x])/(( c^2 - d^2)*f*(c + d*Sin[e + f*x])))/(a^2*(c - d)))/(3*a^2*(c - d)))/(5*a^2 *(c - d))
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, 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_)])^(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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Time = 2.68 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 d^{3} \left (\frac {\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (4 c +3 d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{4}}-\frac {2 \left (8 c -12 d \right )}{3 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8 d -4 c}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (c^{2}-4 c d +6 d^{2}\right )}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{a^{3} f}\) | \(273\) |
| default | \(\frac {-\frac {2 d^{3} \left (\frac {\frac {d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (4 c +3 d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{4}}-\frac {2 \left (8 c -12 d \right )}{3 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8 d -4 c}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (c^{2}-4 c d +6 d^{2}\right )}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {8}{5 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}}{a^{3} f}\) | \(273\) |
| risch | \(-\frac {2 \left (-72 d^{4}-2 c^{3} d +12 c^{2} d^{2}-43 c \,d^{3}-410 i c^{2} d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+26 i c^{2} d^{2} {\mathrm e}^{i \left (f x +e \right )}-14 i c^{3} d \,{\mathrm e}^{i \left (f x +e \right )}+299 i c \,d^{3} {\mathrm e}^{i \left (f x +e \right )}-1220 i c \,d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+170 i c^{3} d \,{\mathrm e}^{3 i \left (f x +e \right )}+45 d^{4} {\mathrm e}^{6 i \left (f x +e \right )}-40 i c^{4} {\mathrm e}^{3 i \left (f x +e \right )}+40 c^{3} {\mathrm e}^{4 i \left (f x +e \right )} d +60 c \,d^{3} {\mathrm e}^{6 i \left (f x +e \right )}+225 i d^{4} {\mathrm e}^{5 i \left (f x +e \right )}-270 c^{2} d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+315 i d^{4} {\mathrm e}^{i \left (f x +e \right )}+238 d^{2} c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+4 i c^{4} {\mathrm e}^{i \left (f x +e \right )}-865 c \,d^{3} {\mathrm e}^{4 i \left (f x +e \right )}-98 d \,c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+848 d^{3} c \,{\mathrm e}^{2 i \left (f x +e \right )}-600 i d^{4} {\mathrm e}^{3 i \left (f x +e \right )}-480 d^{4} {\mathrm e}^{4 i \left (f x +e \right )}+567 d^{4} {\mathrm e}^{2 i \left (f x +e \right )}+20 c^{4} {\mathrm e}^{2 i \left (f x +e \right )}+60 i c^{2} d^{2} {\mathrm e}^{5 i \left (f x +e \right )}+345 i c \,d^{3} {\mathrm e}^{5 i \left (f x +e \right )}\right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left (c +d \right ) \left (d \,{\mathrm e}^{2 i \left (f x +e \right )}-d +2 i c \,{\mathrm e}^{i \left (f x +e \right )}\right ) \left (c -d \right )^{4} f \,a^{3}}-\frac {4 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{4} f \,a^{3}}-\frac {3 d^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{4} f \,a^{3}}+\frac {4 d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{4} f \,a^{3}}+\frac {3 d^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{4} f \,a^{3}}\) | \(804\) |
Input:
int(1/(a+sin(f*x+e)*a)^3/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
2/f/a^3*(-d^3/(c-d)^4*((d^2/(c+d)/c*tan(1/2*f*x+1/2*e)+d/(c+d))/(tan(1/2*f *x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(4*c+3*d)/(c+d)/(c^2-d^2)^(1/2)*ar ctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2)))-1/3*(8*c-12*d)/(c- d)^3/(tan(1/2*f*x+1/2*e)+1)^3-1/2*(8*d-4*c)/(c-d)^3/(tan(1/2*f*x+1/2*e)+1) ^2-(c^2-4*c*d+6*d^2)/(c-d)^4/(tan(1/2*f*x+1/2*e)+1)-4/5/(c-d)^2/(tan(1/2*f *x+1/2*e)+1)^5+2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^4)
Leaf count of result is larger than twice the leaf count of optimal. 1575 vs. \(2 (285) = 570\).
Time = 0.19 (sec) , antiderivative size = 3235, normalized size of antiderivative = 10.86 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,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*d^2-4*c^2>0)', see `assume?` f or more de
Time = 0.19 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {15 \, {\left (4 \, c d^{3} + 3 \, d^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a^{3} c^{5} - 3 \, a^{3} c^{4} d + 2 \, a^{3} c^{3} d^{2} + 2 \, a^{3} c^{2} d^{3} - 3 \, a^{3} c d^{4} + a^{3} d^{5}\right )} \sqrt {c^{2} - d^{2}}} + \frac {15 \, {\left (d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c d^{4}\right )}}{{\left (a^{3} c^{6} - 3 \, a^{3} c^{5} d + 2 \, a^{3} c^{4} d^{2} + 2 \, a^{3} c^{3} d^{3} - 3 \, a^{3} c^{2} d^{4} + a^{3} c d^{5}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} + \frac {15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 90 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 150 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 300 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 190 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 420 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 110 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 270 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} - 34 \, c d + 72 \, d^{2}}{{\left (a^{3} c^{4} - 4 \, a^{3} c^{3} d + 6 \, a^{3} c^{2} d^{2} - 4 \, a^{3} c d^{3} + a^{3} d^{4}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algorithm="giac")
Output:
-2/15*(15*(4*c*d^3 + 3*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arc tan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*sqrt(c^2 - d^2)) + 15*(d^5*tan(1/2*f*x + 1/2*e) + c*d^4)/((a^3*c^6 - 3*a^3*c^5*d + 2*a^3*c ^4*d^2 + 2*a^3*c^3*d^3 - 3*a^3*c^2*d^4 + a^3*c*d^5)*(c*tan(1/2*f*x + 1/2*e )^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) + (15*c^2*tan(1/2*f*x + 1/2*e)^4 - 60 *c*d*tan(1/2*f*x + 1/2*e)^4 + 90*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1 /2*f*x + 1/2*e)^3 - 150*c*d*tan(1/2*f*x + 1/2*e)^3 + 300*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*c^2*tan(1/2*f*x + 1/2*e)^2 - 190*c*d*tan(1/2*f*x + 1/2*e)^2 + 420*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x + 1/2*e) - 110*c*d* tan(1/2*f*x + 1/2*e) + 270*d^2*tan(1/2*f*x + 1/2*e) + 7*c^2 - 34*c*d + 72* d^2)/((a^3*c^4 - 4*a^3*c^3*d + 6*a^3*c^2*d^2 - 4*a^3*c*d^3 + a^3*d^4)*(tan (1/2*f*x + 1/2*e) + 1)^5))/f
Time = 18.98 (sec) , antiderivative size = 987, normalized size of antiderivative = 3.31 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:
int(1/((a + a*sin(e + f*x))^3*(c + d*sin(e + f*x))^2),x)
Output:
- ((2*(72*c*d^3 - 27*c^3*d + 7*c^4 + 15*d^4 + 38*c^2*d^2))/(15*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (4*tan(e/2 + (f*x)/2)^3*(84*c*d^3 - 18*c^3*d + 5*c^4 + 15*d^4 + 19*c^2*d^2))/(3*c*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)*(219*c*d^4 - 76*c^4*d + 20*c^5 + 15 *d^5 + 346*c^2*d^3 + 106*c^3*d^2))/(15*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2* d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^6*(c^5 - 3*c^4*d + d^5 + 6*c^2*d^3 + 2*c^3*d^2))/(c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*ta n(e/2 + (f*x)/2)^5*(13*c*d^4 - 6*c^4*d + 2*c^5 + 5*d^5 + 24*c^2*d^3 + 4*c^ 3*d^2))/(c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^4*(135*c*d^4 - 27*c^4*d + 11*c^5 + 30*d^5 + 162*c^2*d^3 + 4*c^3* d^2))/(3*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (2*tan(e/2 + (f*x)/2)^2*(690*c*d^4 - 137*c^4*d + 47*c^5 + 75*d^5 + 812*c^2*d^3 + 88*c^ 3*d^2))/(15*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)))/(f*(a^3*c + tan(e/2 + (f*x)/2)*(5*a^3*c + 2*a^3*d) + tan(e/2 + (f*x)/2)^6*(5*a^3*c + 2*a^3*d) + tan(e/2 + (f*x)/2)^2*(11*a^3*c + 10*a^3*d) + tan(e/2 + (f*x)/2 )^5*(11*a^3*c + 10*a^3*d) + tan(e/2 + (f*x)/2)^3*(15*a^3*c + 20*a^3*d) + t an(e/2 + (f*x)/2)^4*(15*a^3*c + 20*a^3*d) + a^3*c*tan(e/2 + (f*x)/2)^7)) - (2*d^3*atan(((d^3*(4*c + 3*d)*(2*a^3*d^6 - 6*a^3*c*d^5 + 2*a^3*c^5*d + 4* a^3*c^2*d^4 + 4*a^3*c^3*d^3 - 6*a^3*c^4*d^2))/(a^3*(c + d)^(3/2)*(c - d)^( 9/2)) + (2*c*d^3*tan(e/2 + (f*x)/2)*(4*c + 3*d)*(a^3*c^5 + a^3*d^5 - 3*...
Time = 0.21 (sec) , antiderivative size = 3777, normalized size of antiderivative = 12.67 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:
int(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x)
Output:
(2*( - 300*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d** 2))*tan((e + f*x)/2)**7*c**3*d**3 - 345*sqrt(c**2 - d**2)*atan((tan((e + f *x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**7*c**2*d**4 - 90*sqrt(c **2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x) /2)**7*c*d**5 - 1500*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt( c**2 - d**2))*tan((e + f*x)/2)**6*c**3*d**3 - 2325*sqrt(c**2 - d**2)*atan( (tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**6*c**2*d**4 - 1140*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))* tan((e + f*x)/2)**6*c*d**5 - 180*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)* c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**6*d**6 - 3300*sqrt(c**2 - d**2 )*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*c** 3*d**3 - 6795*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*c**2*d**4 - 4440*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*c*d**5 - 900*sqrt (c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f* x)/2)**5*d**6 - 4500*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt( c**2 - d**2))*tan((e + f*x)/2)**4*c**3*d**3 - 11175*sqrt(c**2 - d**2)*atan ((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**4*c**2*d**4 - 8250*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2)) *tan((e + f*x)/2)**4*c*d**5 - 1800*sqrt(c**2 - d**2)*atan((tan((e + f*x...