Integrand size = 27, antiderivative size = 127 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {(c-d) (c+7 d) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)}}\right )}{6 \sqrt {6} f}+\frac {(c-5 d) d \cos (e+f x)}{6 f \sqrt {3+3 \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (3+3 \sin (e+f x))^{3/2}} \]
-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)* (c+7*d)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^( 3/2)/f*2^(1/2)+1/2*(c-5*d)*d*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 1.25 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-(c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} \left (c^2+6 c d-7 d^2\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-4 d^2 \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+4 d^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2\right )}{6 \sqrt {3} f (1+\sin (e+f x))^{3/2}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)^2*Sin[(e + f*x)/2] - (c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (1 + I)*(-1)^(3/4)*(c^2 + 6 *c*d - 7*d^2)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos [(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 4*d^2*Cos[(e + f*x)/2]*(Cos[(e + f*x )/2] + Sin[(e + f*x)/2])^2 + 4*d^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Si n[(e + f*x)/2])^2))/(6*Sqrt[3]*f*(1 + Sin[e + f*x])^(3/2))
Time = 0.53 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3239, 27, 3042, 3230, 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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3239 |
| \(\displaystyle -\frac {\int -\frac {a \left (c^2+5 d c-2 d^2\right )-a (c-5 d) d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a \left (c^2+5 d c-2 d^2\right )-a (c-5 d) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a \left (c^2+5 d c-2 d^2\right )-a (c-5 d) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {a (c-d) (c+7 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (c-d) (c+7 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {2 a (c-d) (c+7 d) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {2 a d (c-5 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {\sqrt {2} \sqrt {a} (c-d) (c+7 d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{2 f (a \sin (e+f x)+a)^{3/2}}\) |
-1/2*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(f*(a + a*Sin[e + f*x])^( 3/2)) + (-((Sqrt[2]*Sqrt[a]*(c - d)*(c + 7*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x ])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f) + (2*a*(c - 5*d)*d*Cos[e + f*x] )/(f*Sqrt[a + a*Sin[e + f*x]]))/(4*a^2)
3.6.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f* x])^m*((c + d*Sin[e + f*x])/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*c*d*(m - 1) + b*(d^2 + c^2*(m + 1) ) + d*(a*d*(m - 1) + b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(315\) vs. \(2(119)=238\).
Time = 2.70 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.49
| method | result | size |
| default | \(-\frac {\left (\sin \left (f x +e \right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,c^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c d -7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,d^{2}+8 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, d^{2}\right )+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,c^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a c d -7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \,d^{2}+2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c^{2}-4 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, c d +10 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, d^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(316\) |
| parts | \(-\frac {c^{2} \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (f x +e \right )+2 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}}+\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {d^{2} \left (7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (f x +e \right )+7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a -8 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, \sin \left (f x +e \right )-10 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {c d \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a \sin \left (f x +e \right )-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a +2 \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{2 a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(402\) |
-1/4/a^(5/2)*(sin(f*x+e)*(2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/ 2)/a^(1/2))*a*c^2+6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^( 1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))* a*d^2+8*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d^2)+2^(1/2)*arctanh(1/2*(a-a*sin(f *x+e))^(1/2)*2^(1/2)/a^(1/2))*a*c^2+6*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e)) ^(1/2)*2^(1/2)/a^(1/2))*a*c*d-7*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2) *2^(1/2)/a^(1/2))*a*d^2+2*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*c^2-4*(a-a*sin(f* x+e))^(1/2)*a^(1/2)*c*d+10*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*d^2)*(-a*(sin(f* x+e)-1))^(1/2)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (119) = 238\).
Time = 0.32 (sec) , antiderivative size = 380, normalized size of antiderivative = 2.99 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c^{2} - 12 \, c d + 14 \, d^{2} - {\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (2 \, c^{2} + 12 \, c d - 14 \, d^{2} + {\left (c^{2} + 6 \, c d - 7 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (4 \, d^{2} \cos \left (f x + e\right )^{2} + c^{2} - 2 \, c d + d^{2} + {\left (c^{2} - 2 \, c d + 5 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left (4 \, d^{2} \cos \left (f x + e\right ) - c^{2} + 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{8 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]
-1/8*(sqrt(2)*((c^2 + 6*c*d - 7*d^2)*cos(f*x + e)^2 - 2*c^2 - 12*c*d + 14* d^2 - (c^2 + 6*c*d - 7*d^2)*cos(f*x + e) - (2*c^2 + 12*c*d - 14*d^2 + (c^2 + 6*c*d - 7*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e) ^2 + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(c os(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*( 4*d^2*cos(f*x + e)^2 + c^2 - 2*c*d + d^2 + (c^2 - 2*c*d + 5*d^2)*cos(f*x + e) + (4*d^2*cos(f*x + e) - c^2 + 2*c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f* x + e) + a))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f *cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
\[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{2}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (119) = 238\).
Time = 0.35 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.09 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=\frac {\frac {16 \, \sqrt {2} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (\sqrt {a} c^{2} + 6 \, \sqrt {a} c d - 7 \, \sqrt {a} d^{2}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (\sqrt {a} c^{2} + 6 \, \sqrt {a} c d - 7 \, \sqrt {a} d^{2}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, \sqrt {2} {\left (\sqrt {a} c^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, \sqrt {a} c d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + \sqrt {a} d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{8 \, f} \]
1/8*(16*sqrt(2)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)/(a^(3/2)*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e))) + sqrt(2)*(sqrt(a)*c^2 + 6*sqrt(a)*c*d - 7*sqrt(a)* d^2)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*f* x + 1/2*e))) - sqrt(2)*(sqrt(a)*c^2 + 6*sqrt(a)*c*d - 7*sqrt(a)*d^2)*log(- sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e ))) - 2*sqrt(2)*(sqrt(a)*c^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 2*sqrt(a)*c* d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + sqrt(a)*d^2*sin(-1/4*pi + 1/2*f*x + 1/2 *e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f
Timed out. \[ \int \frac {(c+d \sin (e+f x))^2}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]