Integrand size = 28, antiderivative size = 190 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}} \] Output:
3/512*a^2*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+1/4*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^2* cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(7/2)+1/64*a^2*cos(f*x+e)/c^2/f/(c-c*sin(f *x+e))^(5/2)+3/256*a^2*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 5.27 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.95 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (128 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-96 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+4 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7-(3+3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+256 \sin \left (\frac {1}{2} (e+f x)\right )-192 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+8 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )+6 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{256 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^{9/2}} \] Input:
Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(128*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 96*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 4*(Cos[(e + f*x )/2] - Sin[(e + f*x)/2])^5 + 3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7 - ( 3 + 3*I)*(-1)^(1/4)*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 + 256*Sin[(e + f*x)/2] - 192*(Cos[ (e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 8*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] + 6*(Cos[(e + f*x)/2] - Sin[(e + f* x)/2])^6*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(256*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2))
Time = 0.88 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 3042, 3129, 3042, 3129, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{13/2}}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]
Output:
a^2*c^2*(Cos[e + f*x]^3/(4*c*f*(c - c*Sin[e + f*x])^(11/2)) - (3*(Cos[e + f*x]/(3*c*f*(c - c*Sin[e + f*x])^(7/2)) - (Cos[e + f*x]/(4*f*(c - c*Sin[e + f*x])^(5/2)) + (3*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Si n[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[e + f*x]/(2*f*(c - c*Sin[e + f*x ])^(3/2))))/(8*c))/(6*c^2)))/(8*c^2))
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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 1.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(-\frac {a^{2} \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{6} \sin \left (f x +e \right )^{4}-6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} c^{\frac {5}{2}}-12 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{3} c^{6}+44 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {7}{2}}+18 \sin \left (f x +e \right )^{2} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{6}+88 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {9}{2}}-12 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{6}-48 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {11}{2}}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{512 c^{\frac {21}{2}} \left (-1+\sin \left (f x +e \right )\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(299\) |
| parts | \(\text {Expression too large to display}\) | \(918\) |
Input:
int((a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
Output:
-1/512/c^(21/2)*a^2*(3*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2 ))*2^(1/2)*c^6*sin(f*x+e)^4-6*(c*(1+sin(f*x+e)))^(7/2)*c^(5/2)-12*2^(1/2)* arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^6+44* (c*(1+sin(f*x+e)))^(5/2)*c^(7/2)+18*sin(f*x+e)^2*arctanh(1/2*(c*(1+sin(f*x +e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^6+88*(c*(1+sin(f*x+e)))^(3/2)*c^(9/ 2)-12*sin(f*x+e)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^( 1/2)*c^6-48*(c*(1+sin(f*x+e)))^(1/2)*c^(11/2)+3*2^(1/2)*arctanh(1/2*(c*(1+ sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^6)*(c*(1+sin(f*x+e)))^(1/2)/(-1+sin( f*x+e))^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. 523 vs. \(2 (163) = 326\).
Time = 0.11 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.75 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{5} + 5 \, a^{2} \cos \left (f x + e\right )^{4} - 8 \, a^{2} \cos \left (f x + e\right )^{3} - 20 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{4} - 4 \, a^{2} \cos \left (f x + e\right )^{3} - 12 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, 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 (3 \, a^{2} \cos \left (f x + e\right )^{4} + 13 \, a^{2} \cos \left (f x + e\right )^{3} + 86 \, a^{2} \cos \left (f x + e\right )^{2} - 52 \, a^{2} \cos \left (f x + e\right ) - 128 \, a^{2} - {\left (3 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 76 \, a^{2} \cos \left (f x + e\right ) + 128 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1024 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f - {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")
Output:
1/1024*(3*sqrt(2)*(a^2*cos(f*x + e)^5 + 5*a^2*cos(f*x + e)^4 - 8*a^2*cos(f *x + e)^3 - 20*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + 16*a^2 - (a^2*cos (f*x + e)^4 - 4*a^2*cos(f*x + e)^3 - 12*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + 16*a^2)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*s qrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*c os(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*a^2*cos(f*x + e)^4 + 13*a^2*cos(f*x + e)^3 + 86*a^2*cos(f*x + e)^2 - 52*a^2*cos(f*x + e ) - 128*a^2 - (3*a^2*cos(f*x + e)^3 - 10*a^2*cos(f*x + e)^2 + 76*a^2*cos(f *x + e) + 128*a^2)*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. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(9/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(9/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="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. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(9/2),x)
Output:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(9/2), x)
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {c}\, a^{2} \left (-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )-2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{5}} \] Input:
int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x)
Output:
(sqrt(c)*a**2*( - 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) - 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) - 2*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)))/c**5