Integrand size = 27, antiderivative size = 180 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=-\frac {(c-d) \left (c^2+6 c d+25 d^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} \cos (e+f x)}{\sqrt {3+3 \sin (e+f x)}}\right )}{48 \sqrt {6} f}-\frac {(c-d)^2 (3 c+13 d) \cos (e+f x)}{48 f (3+3 \sin (e+f x))^{3/2}}+\frac {(c-9 d) d^2 \cos (e+f x)}{36 f \sqrt {3+3 \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (3+3 \sin (e+f x))^{5/2}} \]
-1/16*(c-d)^2*(3*c+13*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)*c os(f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^(5/2)-3/32*(c-d)*(c^2+6*c* d+25*d^2)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a ^(5/2)/f*2^(1/2)+1/4*(c-9*d)*d^2*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 1.93 (sec) , antiderivative size = 403, normalized size of antiderivative = 2.24 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-11 c^3 \cos \left (\frac {1}{2} (e+f x)\right )+9 c^2 d \cos \left (\frac {1}{2} (e+f x)\right )+15 c d^2 \cos \left (\frac {1}{2} (e+f x)\right )-45 d^3 \cos \left (\frac {1}{2} (e+f x)\right )-3 c^3 \cos \left (\frac {3}{2} (e+f x)\right )-15 c^2 d \cos \left (\frac {3}{2} (e+f x)\right )+39 c d^2 \cos \left (\frac {3}{2} (e+f x)\right )-69 d^3 \cos \left (\frac {3}{2} (e+f x)\right )+16 d^3 \cos \left (\frac {5}{2} (e+f x)\right )+11 c^3 \sin \left (\frac {1}{2} (e+f x)\right )-9 c^2 d \sin \left (\frac {1}{2} (e+f x)\right )-15 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+45 d^3 \sin \left (\frac {1}{2} (e+f x)\right )+(6+6 i) (-1)^{3/4} \left (c^3+5 c^2 d+19 c d^2-25 d^3\right ) \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-3 c^3 \sin \left (\frac {3}{2} (e+f x)\right )-15 c^2 d \sin \left (\frac {3}{2} (e+f x)\right )+39 c d^2 \sin \left (\frac {3}{2} (e+f x)\right )-69 d^3 \sin \left (\frac {3}{2} (e+f x)\right )-16 d^3 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{288 \sqrt {3} f (1+\sin (e+f x))^{5/2}} \]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-11*c^3*Cos[(e + f*x)/2] + 9*c^2*d *Cos[(e + f*x)/2] + 15*c*d^2*Cos[(e + f*x)/2] - 45*d^3*Cos[(e + f*x)/2] - 3*c^3*Cos[(3*(e + f*x))/2] - 15*c^2*d*Cos[(3*(e + f*x))/2] + 39*c*d^2*Cos[ (3*(e + f*x))/2] - 69*d^3*Cos[(3*(e + f*x))/2] + 16*d^3*Cos[(5*(e + f*x))/ 2] + 11*c^3*Sin[(e + f*x)/2] - 9*c^2*d*Sin[(e + f*x)/2] - 15*c*d^2*Sin[(e + f*x)/2] + 45*d^3*Sin[(e + f*x)/2] + (6 + 6*I)*(-1)^(3/4)*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4] )]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 3*c^3*Sin[(3*(e + f*x))/2] - 15*c^2*d*Sin[(3*(e + f*x))/2] + 39*c*d^2*Sin[(3*(e + f*x))/2] - 69*d^3*Sin [(3*(e + f*x))/2] - 16*d^3*Sin[(5*(e + f*x))/2]))/(288*Sqrt[3]*f*(1 + Sin[ e + f*x])^(5/2))
Time = 1.03 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3498, 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.
\(\displaystyle \int \frac {(c+d \sin (e+f x))^3}{(a \sin (e+f x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d \sin (e+f x))^3}{(a \sin (e+f x)+a)^{5/2}}dx\) |
\(\Big \downarrow \) 3244 |
\(\displaystyle -\frac {\int -\frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{2 (\sin (e+f x) a+a)^{3/2}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (a \left (3 c^2+9 d c-4 d^2\right )-a (c-9 d) d \sin (e+f x)\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\int \frac {-a (c-9 d) d^2 \sin ^2(e+f x)+\left (a d \left (3 c^2+9 d c-4 d^2\right )-a c (c-9 d) d\right ) \sin (e+f x)+a c \left (3 c^2+9 d c-4 d^2\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {-a (c-9 d) d^2 \sin (e+f x)^2+\left (a d \left (3 c^2+9 d c-4 d^2\right )-a c (c-9 d) d\right ) \sin (e+f x)+a c \left (3 c^2+9 d c-4 d^2\right )}{(\sin (e+f x) a+a)^{3/2}}dx}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3498 |
\(\displaystyle \frac {-\frac {\int -\frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a}}dx}{2 a^2}-\frac {a (3 c+13 d) (c-d)^2 \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {a^2 \left (3 c^3+15 d c^2+53 d^2 c-39 d^3\right )-4 a^2 (c-9 d) d^2 \sin (e+f x)}{\sqrt {\sin (e+f x) a+a}}dx}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {\frac {3 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx+\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {6 a^2 (c-d) \left (c^2+6 c d+25 d^2\right ) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {8 a^2 d^2 (c-9 d) \cos (e+f x)}{f \sqrt {a \sin (e+f x)+a}}-\frac {3 \sqrt {2} a^{3/2} (c-d) \left (c^2+6 c d+25 d^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f}}{4 a^2}-\frac {a (c-d)^2 (3 c+13 d) \cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2}}}{8 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}}\) |
-1/4*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]) ^(5/2)) + (-1/2*(a*(c - d)^2*(3*c + 13*d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^(3/2)) + ((-3*Sqrt[2]*a^(3/2)*(c - d)*(c^2 + 6*c*d + 25*d^2)*ArcTanh [(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/f + (8*a^2*(c - 9*d)*d^2*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]))/(4*a^2))/(8*a^2)
3.6.57.3.1 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_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(687\) vs. \(2(171)=342\).
Time = 4.15 (sec) , antiderivative size = 688, normalized size of antiderivative = 3.82
method | result | size |
default | \(\frac {\left (\left (64 a^{\frac {3}{2}} d^{3} \sqrt {a -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} c^{3}+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +57 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-2 \sin \left (f x +e \right ) \left (64 a^{\frac {3}{2}} d^{3} \sqrt {a -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} c^{3}+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +57 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-75 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right )+6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c^{3}+30 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c^{2} d -78 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c \,d^{2}+42 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} d^{3} \sqrt {a}-20 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{3}-36 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{2} d +132 a^{\frac {3}{2}} c \,d^{2} \sqrt {a -a \sin \left (f x +e \right )}-204 a^{\frac {3}{2}} d^{3} \sqrt {a -a \sin \left (f x +e \right )}-6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}-30 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d -114 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}+150 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {9}{2}} \left (\sin \left (f x +e \right )+1\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(688\) |
parts | \(\text {Expression too large to display}\) | \(840\) |
1/32/a^(9/2)*((64*a^(3/2)*d^3*(a-a*sin(f*x+e))^(1/2)+3*2^(1/2)*arctanh(1/2 *(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+15*2^(1/2)*arctanh(1/2*(a -a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d+57*2^(1/2)*arctanh(1/2*(a- a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d^2-75*2^(1/2)*arctanh(1/2*(a-a *sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3)*cos(f*x+e)^2-2*sin(f*x+e)*(64 *a^(3/2)*d^3*(a-a*sin(f*x+e))^(1/2)+3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e)) ^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3+15*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1 /2)*2^(1/2)/a^(1/2))*a^2*c^2*d+57*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/ 2)*2^(1/2)/a^(1/2))*a^2*c*d^2-75*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2 )*2^(1/2)/a^(1/2))*a^2*d^3)+6*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^3+30*(a-a*s in(f*x+e))^(3/2)*a^(1/2)*c^2*d-78*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d^2+42* (a-a*sin(f*x+e))^(3/2)*d^3*a^(1/2)-20*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^3-3 6*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2*d+132*a^(3/2)*c*d^2*(a-a*sin(f*x+e))^ (1/2)-204*a^(3/2)*d^3*(a-a*sin(f*x+e))^(1/2)-6*2^(1/2)*arctanh(1/2*(a-a*si n(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3-30*2^(1/2)*arctanh(1/2*(a-a*sin(f *x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d-114*2^(1/2)*arctanh(1/2*(a-a*sin(f *x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d^2+150*2^(1/2)*arctanh(1/2*(a-a*sin(f *x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3)*(-a*(sin(f*x+e)-1))^(1/2)/(sin(f*x+ e)+1)/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. 608 vs. \(2 (171) = 342\).
Time = 0.30 (sec) , antiderivative size = 608, normalized size of antiderivative = 3.38 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=-\frac {3 \, \sqrt {2} {\left ({\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 4 \, c^{3} - 20 \, c^{2} d - 76 \, c d^{2} + 100 \, d^{3} + 3 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{3} + 20 \, c^{2} d + 76 \, c d^{2} - 100 \, d^{3} - {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\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 (32 \, d^{3} \cos \left (f x + e\right )^{3} - 4 \, c^{3} + 12 \, c^{2} d - 12 \, c d^{2} + 4 \, d^{3} - {\left (3 \, c^{3} + 15 \, c^{2} d - 39 \, c d^{2} + 53 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (7 \, c^{3} + 3 \, c^{2} d - 27 \, c d^{2} + 81 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (32 \, d^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} + 12 \, c^{2} d - 12 \, c d^{2} + 4 \, d^{3} + {\left (3 \, c^{3} + 15 \, c^{2} d - 39 \, c d^{2} + 85 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
-1/64*(3*sqrt(2)*((c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e)^3 - 4*c ^3 - 20*c^2*d - 76*c*d^2 + 100*d^3 + 3*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3) *cos(f*x + e)^2 - 2*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e) - (4* c^3 + 20*c^2*d + 76*c*d^2 - 100*d^3 - (c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)* cos(f*x + e)^2 + 2*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*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)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*si n(f*x + e) - cos(f*x + e) - 2)) + 4*(32*d^3*cos(f*x + e)^3 - 4*c^3 + 12*c^ 2*d - 12*c*d^2 + 4*d^3 - (3*c^3 + 15*c^2*d - 39*c*d^2 + 53*d^3)*cos(f*x + e)^2 - (7*c^3 + 3*c^2*d - 27*c*d^2 + 81*d^3)*cos(f*x + e) - (32*d^3*cos(f* x + e)^2 - 4*c^3 + 12*c^2*d - 12*c*d^2 + 4*d^3 + (3*c^3 + 15*c^2*d - 39*c* d^2 + 85*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f *cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
Timed out. \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (171) = 342\).
Time = 0.47 (sec) , antiderivative size = 404, normalized size of antiderivative = 2.24 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {\frac {128 \, \sqrt {2} d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {3 \, \sqrt {2} {\left (\sqrt {a} c^{3} + 5 \, \sqrt {a} c^{2} d + 19 \, \sqrt {a} c d^{2} - 25 \, \sqrt {a} d^{3}\right )} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3 \, \sqrt {2} {\left (\sqrt {a} c^{3} + 5 \, \sqrt {a} c^{2} d + 19 \, \sqrt {a} c d^{2} - 25 \, \sqrt {a} d^{3}\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{3} \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 (3 \, \sqrt {a} c^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, \sqrt {a} c^{2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 39 \, \sqrt {a} c d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 21 \, \sqrt {a} d^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, \sqrt {a} c^{3} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, \sqrt {a} c^{2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 33 \, \sqrt {a} c d^{2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 19 \, \sqrt {a} d^{3} \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 )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{64 \, f} \]
1/64*(128*sqrt(2)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)/(a^(5/2)*sgn(cos(-1/4 *pi + 1/2*f*x + 1/2*e))) + 3*sqrt(2)*(sqrt(a)*c^3 + 5*sqrt(a)*c^2*d + 19*s qrt(a)*c*d^2 - 25*sqrt(a)*d^3)*log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^ 3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(sqrt(a)*c^3 + 5*sqrt(a )*c^2*d + 19*sqrt(a)*c*d^2 - 25*sqrt(a)*d^3)*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 2*sqrt(2)*(3*sqrt( a)*c^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 15*sqrt(a)*c^2*d*sin(-1/4*pi + 1 /2*f*x + 1/2*e)^3 - 39*sqrt(a)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 21 *sqrt(a)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 - 5*sqrt(a)*c^3*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 9*sqrt(a)*c^2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 33*s qrt(a)*c*d^2*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 19*sqrt(a)*d^3*sin(-1/4*pi + 1/2*f*x + 1/2*e))/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(- 1/4*pi + 1/2*f*x + 1/2*e))))/f
Timed out. \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]