Integrand size = 38, antiderivative size = 217 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {5 a^3 (A-15 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 (A-15 B) c \cos ^5(e+f x)}{48 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (A-15 B) \cos ^3(e+f x)}{192 c f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 (A-15 B) \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}} \] Output:
-5/256*a^3*(A-15*B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e) )^(1/2))*2^(1/2)/c^(9/2)/f+1/8*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e ))^(15/2)+1/48*a^3*(A-15*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(11/2)-5/192 *a^3*(A-15*B)*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^(7/2)+5/128*a^3*(A-15*B)*c os(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 15.46 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.64 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (1765 A \cos \left (\frac {1}{2} (e+f x)\right )+405 B \cos \left (\frac {1}{2} (e+f x)\right )-895 A \cos \left (\frac {3}{2} (e+f x)\right )-2703 B \cos \left (\frac {3}{2} (e+f x)\right )-397 A \cos \left (\frac {5}{2} (e+f x)\right )+579 B \cos \left (\frac {5}{2} (e+f x)\right )+15 A \cos \left (\frac {7}{2} (e+f x)\right )+543 B \cos \left (\frac {7}{2} (e+f x)\right )+(120+120 i) \sqrt [4]{-1} (A-15 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+1765 A \sin \left (\frac {1}{2} (e+f x)\right )+405 B \sin \left (\frac {1}{2} (e+f x)\right )+895 A \sin \left (\frac {3}{2} (e+f x)\right )+2703 B \sin \left (\frac {3}{2} (e+f x)\right )-397 A \sin \left (\frac {5}{2} (e+f x)\right )+579 B \sin \left (\frac {5}{2} (e+f x)\right )-15 A \sin \left (\frac {7}{2} (e+f x)\right )-543 B \sin \left (\frac {7}{2} (e+f x)\right )\right )}{3072 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{9/2}} \] Input:
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^(9/2),x]
Output:
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(1765*A*Co s[(e + f*x)/2] + 405*B*Cos[(e + f*x)/2] - 895*A*Cos[(3*(e + f*x))/2] - 270 3*B*Cos[(3*(e + f*x))/2] - 397*A*Cos[(5*(e + f*x))/2] + 579*B*Cos[(5*(e + f*x))/2] + 15*A*Cos[(7*(e + f*x))/2] + 543*B*Cos[(7*(e + f*x))/2] + (120 + 120*I)*(-1)^(1/4)*(A - 15*B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8 + 1765*A*Sin[(e + f*x)/2 ] + 405*B*Sin[(e + f*x)/2] + 895*A*Sin[(3*(e + f*x))/2] + 2703*B*Sin[(3*(e + f*x))/2] - 397*A*Sin[(5*(e + f*x))/2] + 579*B*Sin[(5*(e + f*x))/2] - 15 *A*Sin[(7*(e + f*x))/2] - 543*B*Sin[(7*(e + f*x))/2]))/(3072*f*(Cos[(e + f *x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(9/2))
Time = 1.17 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 3042, 3159, 3042, 3159, 3042, 3128, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{15/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{15/2}}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{13/2}}dx}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{13/2}}dx}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{9/2}}dx}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c^2 f}+\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}\right )}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^3 c^3 \left (\frac {(A-15 B) \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{5/2} f}\right )}{4 c^2}\right )}{6 c^2}\right )}{16 c}+\frac {(A+B) \cos ^7(e+f x)}{8 f (c-c \sin (e+f x))^{15/2}}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(9/ 2),x]
Output:
a^3*c^3*(((A + B)*Cos[e + f*x]^7)/(8*f*(c - c*Sin[e + f*x])^(15/2)) + ((A - 15*B)*(Cos[e + f*x]^5/(3*c*f*(c - c*Sin[e + f*x])^(11/2)) - (5*(Cos[e + f*x]^3/(2*c*f*(c - c*Sin[e + f*x])^(7/2)) - (3*(-(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(Sqrt[2]*c^(5/2)*f)) + Cos[e + f*x]/(c*f*(c - c*Sin[e + f*x])^(3/2))))/(4*c^2)))/(6*c^2)))/(16*c))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(190)=380\).
Time = 1.89 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.99
| method | result | size |
| default | \(-\frac {a^{3} \left (-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} \left (A -15 B \right ) \cos \left (f x +e \right )^{4}-60 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} \left (A -15 B \right ) \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+120 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} \left (A -15 B \right ) \cos \left (f x +e \right )^{2}+120 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} \left (A -15 B \right ) \sin \left (f x +e \right )+240 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {7}{2}}-440 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {5}{2}}+292 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {3}{2}}+30 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} \sqrt {c}-3600 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {7}{2}}+6600 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {5}{2}}-4380 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{\frac {3}{2}}+1086 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {7}{2}} \sqrt {c}-120 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}+1800 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{768 c^{\frac {17}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(432\) |
| parts | \(\text {Expression too large to display}\) | \(1529\) |
Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x,method=_R ETURNVERBOSE)
Output:
-1/768/c^(17/2)*a^3*(-15*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2 )/c^(1/2))*c^4*(A-15*B)*cos(f*x+e)^4-60*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e ))^(1/2)*2^(1/2)/c^(1/2))*c^4*(A-15*B)*cos(f*x+e)^2*sin(f*x+e)+120*2^(1/2) *arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4*(A-15*B)*cos(f*x+ e)^2+120*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^4*( A-15*B)*sin(f*x+e)+240*A*(c+c*sin(f*x+e))^(1/2)*c^(7/2)-440*A*(c+c*sin(f*x +e))^(3/2)*c^(5/2)+292*A*(c+c*sin(f*x+e))^(5/2)*c^(3/2)+30*A*(c+c*sin(f*x+ e))^(7/2)*c^(1/2)-3600*B*(c+c*sin(f*x+e))^(1/2)*c^(7/2)+6600*B*(c+c*sin(f* x+e))^(3/2)*c^(5/2)-4380*B*(c+c*sin(f*x+e))^(5/2)*c^(3/2)+1086*B*(c+c*sin( f*x+e))^(7/2)*c^(1/2)-120*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^( 1/2)/c^(1/2))*c^4+1800*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2 )/c^(1/2))*c^4)*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)^3/cos(f*x+e)/(c-c* sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 633 vs. \(2 (190) = 380\).
Time = 0.14 (sec) , antiderivative size = 633, normalized size of antiderivative = 2.92 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, al gorithm="fricas")
Output:
-1/1536*(15*sqrt(2)*((A - 15*B)*a^3*cos(f*x + e)^5 + 5*(A - 15*B)*a^3*cos( f*x + e)^4 - 8*(A - 15*B)*a^3*cos(f*x + e)^3 - 20*(A - 15*B)*a^3*cos(f*x + e)^2 + 8*(A - 15*B)*a^3*cos(f*x + e) + 16*(A - 15*B)*a^3 - ((A - 15*B)*a^ 3*cos(f*x + e)^4 - 4*(A - 15*B)*a^3*cos(f*x + e)^3 - 12*(A - 15*B)*a^3*cos (f*x + e)^2 + 8*(A - 15*B)*a^3*cos(f*x + e) + 16*(A - 15*B)*a^3)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)* sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f *x + e) - cos(f*x + e) - 2)) - 4*(3*(5*A + 181*B)*a^3*cos(f*x + e)^4 - (19 1*A - 561*B)*a^3*cos(f*x + e)^3 - 2*(169*A + 537*B)*a^3*cos(f*x + e)^2 + 1 2*(21*A - 59*B)*a^3*cos(f*x + e) + 384*(A + B)*a^3 - (3*(5*A + 181*B)*a^3* cos(f*x + e)^3 + 2*(103*A - 9*B)*a^3*cos(f*x + e)^2 - 12*(11*A + 91*B)*a^3 *cos(f*x + e) - 384*(A + B)*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/ (c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f - (c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f *x + e) + 16*c^5*f)*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, al gorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c )^(9/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x, al gorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(9/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(9/ 2), x)
\[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx =\text {Too large to display} \] Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x)
Output:
(sqrt(c)*a**3*( - int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1) ,x)*a - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**4)/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f *x) - 1),x)*b - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f *x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*s in(e + f*x) - 1),x)*a - 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/ (sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x )**2 + 5*sin(e + f*x) - 1),x)*b - 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*s in(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*a - 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x) **3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*b - 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*a - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10* sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*b))/c**5