Integrand size = 38, antiderivative size = 175 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 a^2 (A+9 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {a^2 (A+9 B) \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^2 (A+9 B) \cos (e+f x)}{8 c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:
3/8*a^2*(A+9*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^(5/2)/f+1/4*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^( 9/2)-1/8*a^2*(A+9*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(5/2)-3/8*a^2*(A+9*B) *cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 11.92 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.97 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-(5 A+13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(3+3 i) \sqrt [4]{-1} (A+9 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-8 B \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+8 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-2 (5 A+13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )-8 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^{5/2}} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x ])^(5/2),x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (5*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (3 + 3*I)*(-1)^(1/4)*(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])^4 - 8*B*Cos[(e + f*x)/2] *(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 8*(A + B)*Sin[(e + f*x)/2] - 2* (5*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] - 8* B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f *x])^2)/(4*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^ (5/2))
Time = 0.97 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.97, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3338 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{8 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^2 c^2 \left (\frac {(A+B) \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{9/2}}-\frac {(A+9 B) \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{8 c}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/ 2),x]
Output:
a^2*c^2*(((A + B)*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(9/2)) - ((A + 9*B)*(Cos[e + f*x]^3/(c*f*(c - c*Sin[e + f*x])^(5/2)) - (3*((2*Sqrt[2]*Ar cTanh[(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]])))/(2*c^2)))/(8*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_), 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. \(385\) vs. \(2(152)=304\).
Time = 0.77 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.21
| method | result | size |
| default | \(-\frac {a^{2} \left (3 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}+27 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}-6 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c^{2}-54 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c^{2}-16 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )^{2}+3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2}+10 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+27 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2}+26 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+32 B \,c^{\frac {3}{2}} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sin \left (f x +e \right )-12 A \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}-60 B \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{8 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(386\) |
| parts | \(\text {Expression too large to display}\) | \(828\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,method=_R ETURNVERBOSE)
Output:
-1/8/c^(9/2)*a^2*(3*A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2) /c^(1/2))*sin(f*x+e)^2*c^2+27*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/ 2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-6*A*arctanh(1/2*(c*(1+sin(f*x+e)))^(1 /2)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c^2-54*B*arctanh(1/2*(c*(1+sin(f*x +e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c^2-16*B*(c*(1+sin(f*x+e)) )^(1/2)*c^(3/2)*sin(f*x+e)^2+3*A*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1 /2)/c^(1/2))*2^(1/2)*c^2+10*A*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+27*B*arctan h(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^2+26*B*(c*(1+sin (f*x+e)))^(3/2)*c^(1/2)+32*B*c^(3/2)*(c*(1+sin(f*x+e)))^(1/2)*sin(f*x+e)-1 2*A*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)-60*B*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2) )*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/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. 449 vs. \(2 (152) = 304\).
Time = 0.16 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.57 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (A + 9 \, B\right )} a^{2} - {\left ({\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (A + 9 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (A + 9 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (8 \, B a^{2} \cos \left (f x + e\right )^{3} - {\left (5 \, A + 21 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - {\left (A + 25 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \, {\left (A + B\right )} a^{2} + {\left (8 \, B a^{2} \cos \left (f x + e\right )^{2} + {\left (5 \, A + 29 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, al gorithm="fricas")
Output:
1/16*(3*sqrt(2)*((A + 9*B)*a^2*cos(f*x + e)^3 + 3*(A + 9*B)*a^2*cos(f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e) - 4*(A + 9*B)*a^2 - ((A + 9*B)*a^2*cos (f*x + e)^2 - 2*(A + 9*B)*a^2*cos(f*x + e) - 4*(A + 9*B)*a^2)*sin(f*x + e) )*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqr t(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*(8*B*a^2*cos(f*x + e)^3 - (5*A + 21*B)*a^2*c os(f*x + e)^2 - (A + 25*B)*a^2*cos(f*x + e) + 4*(A + B)*a^2 + (8*B*a^2*cos (f*x + e)^2 + (5*A + 29*B)*a^2*cos(f*x + e) + 4*(A + B)*a^2)*sin(f*x + e)) *sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f* x + e) - 4*c^3*f)*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, al gorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c )^(5/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),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. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(5/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(5/ 2), x)
\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, a^{2} \left (-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a -2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b -2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b \right )}{c^{3}} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x)
Output:
(sqrt(c)*a**2*( - int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*a - int((sqrt( - sin(e + f*x) + 1)*sin (e + f*x)**3)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x )*b - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 - 3 *sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*a - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x ) - 1),x)*b - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) **3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*a - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f *x) - 1),x)*b))/c**3