Integrand size = 28, antiderivative size = 225 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {3 a^3 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{512 \sqrt {2} c^{11/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}} \] Output:
-3/1024*a^3*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^(11/2)/f+1/5*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(15/2)-1/8 *a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(11/2)+1/16*a^3*cos(f*x+e)/c^2/f/(c-c *sin(f*x+e))^(7/2)-1/128*a^3*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(5/2)-3/512 *a^3*cos(f*x+e)/c^4/f/(c-c*sin(f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 9.63 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.93 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2048 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-2688 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+992 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-20 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7-15 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9+(15+15 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{10}+4096 \sin \left (\frac {1}{2} (e+f x)\right )-5376 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+1984 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )-40 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \sin \left (\frac {1}{2} (e+f x)\right )-30 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3}{2560 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{11/2}} \] Input:
Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]
Output:
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(2048*(Cos[(e + f*x)/2] - Sin[( e + f*x)/2]) - 2688*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 992*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - 20*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) ^7 - 15*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9 + (15 + 15*I)*(-1)^(1/4)*A rcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - S in[(e + f*x)/2])^10 + 4096*Sin[(e + f*x)/2] - 5376*(Cos[(e + f*x)/2] - Sin [(e + f*x)/2])^2*Sin[(e + f*x)/2] + 1984*(Cos[(e + f*x)/2] - Sin[(e + f*x) /2])^4*Sin[(e + f*x)/2] - 40*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[( e + f*x)/2] - 30*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*Sin[(e + f*x)/2]) *(1 + Sin[e + f*x])^3)/(2560*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(11/2))
Time = 1.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 3215, 3042, 3159, 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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^{11/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^{11/2}}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{17/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{17/2}}dx\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}}dx}{2 c^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{13/2}}dx}{2 c^2}\right )\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3159 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{5 c f (c-c \sin (e+f x))^{15/2}}-\frac {\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}}{2 c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]
Output:
a^3*c^3*(Cos[e + f*x]^5/(5*c*f*(c - c*Sin[e + f*x])^(15/2)) - (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*S in[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*Sin[e + f*x]])]/(2*Sqr t[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))/(2*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 = 2.18 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.57
method | result | size |
default | \(\frac {a^{3} \left (15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{7} \sin \left (f x +e \right )^{5}-30 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {9}{2}} c^{\frac {5}{2}}-75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{4} c^{7}+280 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} c^{\frac {7}{2}}+150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{3} c^{7}+1024 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {9}{2}}-150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{7}-1120 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {11}{2}}+75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{7}+480 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {13}{2}}-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{7}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{5120 c^{\frac {25}{2}} \left (-1+\sin \left (f x +e \right )\right )^{4} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(353\) |
parts | \(\text {Expression too large to display}\) | \(1440\) |
Input:
int((a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^(11/2),x,method=_RETURNVERBOSE)
Output:
1/5120/c^(25/2)*a^3*(15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/ 2)/c^(1/2))*c^7*sin(f*x+e)^5-30*(c*(1+sin(f*x+e)))^(9/2)*c^(5/2)-75*2^(1/2 )*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^4*c^7+2 80*(c*(1+sin(f*x+e)))^(7/2)*c^(7/2)+150*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^7+1024*(c*(1+sin(f*x+e)))^(5/2) *c^(9/2)-150*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^7-1120*(c*(1+sin(f*x+e)))^(3/2)*c^(11/2)+75*2^(1/2)*arctan h(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^7+480*(c*(1+s in(f*x+e)))^(1/2)*c^(13/2)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2) *2^(1/2)/c^(1/2))*c^7)*(c*(1+sin(f*x+e)))^(1/2)/(-1+sin(f*x+e))^4/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. 600 vs. \(2 (194) = 388\).
Time = 0.11 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.67 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx =\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas" )
Output:
1/10240*(15*sqrt(2)*(a^3*cos(f*x + e)^6 - 5*a^3*cos(f*x + e)^5 - 18*a^3*co s(f*x + e)^4 + 20*a^3*cos(f*x + e)^3 + 48*a^3*cos(f*x + e)^2 - 16*a^3*cos( f*x + e) - 32*a^3 + (a^3*cos(f*x + e)^5 + 6*a^3*cos(f*x + e)^4 - 12*a^3*co s(f*x + e)^3 - 32*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 32*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)*s in(f*x + e) - cos(f*x + e) - 2)) + 4*(15*a^3*cos(f*x + e)^5 - 65*a^3*cos(f *x + e)^4 + 812*a^3*cos(f*x + e)^3 + 1796*a^3*cos(f*x + e)^2 - 1144*a^3*co s(f*x + e) - 2048*a^3 + (15*a^3*cos(f*x + e)^4 + 80*a^3*cos(f*x + e)^3 + 8 92*a^3*cos(f*x + e)^2 - 904*a^3*cos(f*x + e) - 2048*a^3)*sin(f*x + e))*sqr t(-c*sin(f*x + e) + c))/(c^6*f*cos(f*x + e)^6 - 5*c^6*f*cos(f*x + e)^5 - 1 8*c^6*f*cos(f*x + e)^4 + 20*c^6*f*cos(f*x + e)^3 + 48*c^6*f*cos(f*x + e)^2 - 16*c^6*f*cos(f*x + e) - 32*c^6*f + (c^6*f*cos(f*x + e)^5 + 6*c^6*f*cos( f*x + e)^4 - 12*c^6*f*cos(f*x + e)^3 - 32*c^6*f*cos(f*x + e)^2 + 16*c^6*f* cos(f*x + e) + 32*c^6*f)*sin(f*x + e))
Timed out. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(11/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima" )
Output:
integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(11/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),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. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{11/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(11/2),x)
Output:
int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(11/2), x)
\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c}\, a^{3} \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x +\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x \right )+3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{6}-6 \sin \left (f x +e \right )^{5}+15 \sin \left (f x +e \right )^{4}-20 \sin \left (f x +e \right )^{3}+15 \sin \left (f x +e \right )^{2}-6 \sin \left (f x +e \right )+1}d x \right )\right )}{c^{6}} \] Input:
int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),x)
Output:
(sqrt(c)*a**3*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**6 - 6*sin(e + f*x)**5 + 15*sin(e + f*x)**4 - 20*sin(e + f*x)**3 + 15*sin(e + f*x)**2 - 6 *sin(e + f*x) + 1),x) + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(s in(e + f*x)**6 - 6*sin(e + f*x)**5 + 15*sin(e + f*x)**4 - 20*sin(e + f*x)* *3 + 15*sin(e + f*x)**2 - 6*sin(e + f*x) + 1),x) + 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**6 - 6*sin(e + f*x)**5 + 15*sin(e + f*x)**4 - 20*sin(e + f*x)**3 + 15*sin(e + f*x)**2 - 6*sin(e + f*x) + 1) ,x) + 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**6 - 6* sin(e + f*x)**5 + 15*sin(e + f*x)**4 - 20*sin(e + f*x)**3 + 15*sin(e + f*x )**2 - 6*sin(e + f*x) + 1),x)))/c**6