Integrand size = 38, antiderivative size = 218 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \sqrt {2} a^3 (5 A+9 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}+\frac {a^3 (5 A+9 B) c \cos ^5(e+f x)}{10 f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 (5 A+9 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}+\frac {2 a^3 (5 A+9 B) \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \] Output:
-2*2^(1/2)*a^3*(5*A+9*B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f *x+e))^(1/2))/c^(3/2)/f+1/2*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^ (9/2)+1/10*a^3*(5*A+9*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+1/3*a^3*( 5*A+9*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+2*a^3*(5*A+9*B)*cos(f*x+e)/ c/f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 12.39 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.04 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (120 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(120+120 i) \sqrt [4]{-1} (5 A+9 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+30 (9 A+20 B) \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-5 (2 A+9 B) \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-3 B \cos \left (\frac {5}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+240 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )+30 (9 A+20 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+5 (2 A+9 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {3}{2} (e+f x)\right )-3 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{30 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{3/2}} \] Input:
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^(3/2),x]
Output:
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(120*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) + (120 + 120*I)*(-1)^(1/4)*(5*A + 9*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])^2 + 30*(9*A + 20*B)*Cos[(e + f*x)/2]*(Cos[(e + f*x )/2] - Sin[(e + f*x)/2])^2 - 5*(2*A + 9*B)*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 - 3*B*Cos[(5*(e + f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + 240*(A + B)*Sin[(e + f*x)/2] + 30*(9*A + 20*B)*(C os[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 5*(2*A + 9*B)*(Co s[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(3*(e + f*x))/2] - 3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(5*(e + f*x))/2]))/(30*f*(Cos[(e + f*x)/ 2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(3/2))
Time = 1.16 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.97, 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, 3158, 3042, 3158, 3042, 3158, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{5/2}}dx}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^3 c^3 \left (\frac {(A+B) \cos ^7(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {(5 A+9 B) \left (\frac {2 \left (\frac {2 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )}{c}-\frac {2 \cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{5/2}}\right )}{4 c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(3/ 2),x]
Output:
a^3*c^3*(((A + B)*Cos[e + f*x]^7)/(2*f*(c - c*Sin[e + f*x])^(9/2)) - ((5*A + 9*B)*((-2*Cos[e + f*x]^5)/(5*c*f*(c - c*Sin[e + f*x])^(5/2)) + (2*((-2* Cos[e + f*x]^3)/(3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (2*((2*Sqrt[2]*ArcTan h[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) - (2*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/c))/c))/(4*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[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[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]))
Time = 1.88 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.62
| method | result | size |
| default | \(-\frac {2 a^{3} \left (\sin \left (f x +e \right ) \left (60 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}+5 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+120 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}+15 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+3 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c}-75 A \,\operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3}-135 B \,\operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3}\right )-90 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}-5 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-150 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {5}{2}}-15 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}-3 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} \sqrt {c}+75 A \,\operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3}+135 B \,\operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{15 c^{\frac {9}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(354\) |
| parts | \(\text {Expression too large to display}\) | \(787\) |
Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x,method=_R ETURNVERBOSE)
Output:
-2/15/c^(9/2)*a^3*(sin(f*x+e)*(60*A*(c+c*sin(f*x+e))^(1/2)*c^(5/2)+5*A*(c+ c*sin(f*x+e))^(3/2)*c^(3/2)+120*B*(c+c*sin(f*x+e))^(1/2)*c^(5/2)+15*B*(c+c *sin(f*x+e))^(3/2)*c^(3/2)+3*B*(c+c*sin(f*x+e))^(5/2)*c^(1/2)-75*A*arctanh (1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3-135*B*arctanh(1/2 *(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3)-90*A*(c+c*sin(f*x+e) )^(1/2)*c^(5/2)-5*A*(c+c*sin(f*x+e))^(3/2)*c^(3/2)-150*B*(c+c*sin(f*x+e))^ (1/2)*c^(5/2)-15*B*(c+c*sin(f*x+e))^(3/2)*c^(3/2)-3*B*(c+c*sin(f*x+e))^(5/ 2)*c^(1/2)+75*A*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2 )*c^3+135*B*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^ 3)*(c*(1+sin(f*x+e)))^(1/2)/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. 430 vs. \(2 (195) = 390\).
Time = 0.10 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.97 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left ({\left (5 \, A + 9 \, B\right )} a^{3} c \cos \left (f x + e\right )^{2} - {\left (5 \, A + 9 \, B\right )} a^{3} c \cos \left (f x + e\right ) - 2 \, {\left (5 \, A + 9 \, B\right )} a^{3} c + {\left ({\left (5 \, A + 9 \, B\right )} a^{3} c \cos \left (f x + e\right ) + 2 \, {\left (5 \, A + 9 \, B\right )} a^{3} c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + 2 \, {\left (3 \, B a^{3} \cos \left (f x + e\right )^{4} - {\left (5 \, A + 18 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - {\left (65 \, A + 141 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 30 \, {\left (3 \, A + 5 \, B\right )} a^{3} \cos \left (f x + e\right ) - 30 \, {\left (A + B\right )} a^{3} - {\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} + {\left (5 \, A + 21 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 60 \, {\left (A + 2 \, B\right )} a^{3} \cos \left (f x + e\right ) + 30 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, al gorithm="fricas")
Output:
1/15*(15*sqrt(2)*((5*A + 9*B)*a^3*c*cos(f*x + e)^2 - (5*A + 9*B)*a^3*c*cos (f*x + e) - 2*(5*A + 9*B)*a^3*c + ((5*A + 9*B)*a^3*c*cos(f*x + e) + 2*(5*A + 9*B)*a^3*c)*sin(f*x + e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin (f*x + e) - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2) *sin(f*x + e) - cos(f*x + e) - 2))/sqrt(c) + 2*(3*B*a^3*cos(f*x + e)^4 - ( 5*A + 18*B)*a^3*cos(f*x + e)^3 - (65*A + 141*B)*a^3*cos(f*x + e)^2 - 30*(3 *A + 5*B)*a^3*cos(f*x + e) - 30*(A + B)*a^3 - (3*B*a^3*cos(f*x + e)^3 + (5 *A + 21*B)*a^3*cos(f*x + e)^2 - 60*(A + 2*B)*a^3*cos(f*x + e) + 30*(A + B) *a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2 *f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),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 )^(3/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),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. \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(3/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^(3/ 2), x)
\[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, a^{3} \left (\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) b +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) a +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) b +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) a +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) b +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right ) b \right )}{c^{2}} \] Input:
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x)
Output:
(sqrt(c)*a**3*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**4)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*b + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*a + 3*int((sqrt( - si n(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x) *b + 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*a + 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)** 2)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*b + 3*int((sqrt( - sin(e + f* x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*a + int((s qrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x)*b))/c**2