Integrand size = 23, antiderivative size = 145 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=-\frac {11 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cos (c+d x) \sin ^2(c+d x)}{2 d (a+a \sin (c+d x))^{3/2}}+\frac {13 \cos (c+d x)}{3 a d \sqrt {a+a \sin (c+d x)}}-\frac {7 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{6 a^2 d} \]
1/2*cos(d*x+c)*sin(d*x+c)^2/d/(a+a*sin(d*x+c))^(3/2)-11/4*arctanh(1/2*cos( d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+13/3*cos( d*x+c)/a/d/(a+a*sin(d*x+c))^(1/2)-7/6*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^ 2/d
Result contains complex when optimal does not.
Time = 0.47 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (11 \cos \left (\frac {1}{2} (c+d x)\right )+7 \cos \left (\frac {3}{2} (c+d x)\right )+\cos \left (\frac {5}{2} (c+d x)\right )-11 \sin \left (\frac {1}{2} (c+d x)\right )+(33+33 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) (1+\sin (c+d x))+7 \sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {5}{2} (c+d x)\right )\right )}{6 d (a (1+\sin (c+d x)))^{3/2}} \]
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(11*Cos[(c + d*x)/2] + 7*Cos[(3*(c + d*x))/2] + Cos[(5*(c + d*x))/2] - 11*Sin[(c + d*x)/2] + (33 + 33*I)*(-1) ^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(1 + Sin[c + d*x]) + 7*Sin[(3*(c + d*x))/2] - Sin[(5*(c + d*x))/2]))/(6*d*(a*(1 + Sin [c + d*x]))^(3/2))
Time = 0.74 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{(a \sin (c+d x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3244 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {\sin (c+d x) (4 a-7 a \sin (c+d x))}{2 \sqrt {\sin (c+d x) a+a}}dx}{2 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {\sin (c+d x) (4 a-7 a \sin (c+d x))}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {\sin (c+d x) (4 a-7 a \sin (c+d x))}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {4 a \sin (c+d x)-7 a \sin ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\int \frac {4 a \sin (c+d x)-7 a \sin (c+d x)^2}{\sqrt {\sin (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {2 \int -\frac {7 a^2-26 a^2 \sin (c+d x)}{2 \sqrt {\sin (c+d x) a+a}}dx}{3 a}+\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{4 a^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {7 a^2-26 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\int \frac {7 a^2-26 a^2 \sin (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{3 a}}{4 a^2}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {33 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {52 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {33 a^2 \int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx+\frac {52 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{3 a}}{4 a^2}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {52 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {66 a^2 \int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{3 a}}{4 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sin ^2(c+d x) \cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}-\frac {\frac {14 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {\frac {52 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {33 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}\) |
(Cos[c + d*x]*Sin[c + d*x]^2)/(2*d*(a + a*Sin[c + d*x])^(3/2)) - ((14*Cos[ c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - ((-33*Sqrt[2]*a^(3/2)*ArcTanh[( Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/d + (52*a^2*Cos [c + d*x])/(d*Sqrt[a + a*Sin[c + d*x]]))/(3*a))/(4*a^2)
3.1.69.3.1 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[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.66 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.33
| method | result | size |
| default | \(\frac {\left (8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )-33 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+8 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}+24 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )-33 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{12 a^{\frac {7}{2}} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(193\) |
1/12/a^(7/2)*(8*(a-a*sin(d*x+c))^(3/2)*a^(1/2)*sin(d*x+c)-33*2^(1/2)*arcta nh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*sin(d*x+c)+8*(a-a*sin(d *x+c))^(3/2)*a^(1/2)+24*(a-a*sin(d*x+c))^(1/2)*a^(3/2)*sin(d*x+c)-33*2^(1/ 2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+30*(a-a*sin(d*x +c))^(1/2)*a^(3/2))*(-a*(sin(d*x+c)-1))^(1/2)/cos(d*x+c)/(a+a*sin(d*x+c))^ (1/2)/d
Leaf count of result is larger than twice the leaf count of optimal. 295 vs. \(2 (122) = 244\).
Time = 0.28 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.03 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {33 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (4 \, \cos \left (d x + c\right )^{3} + 16 \, \cos \left (d x + c\right )^{2} - {\left (4 \, \cos \left (d x + c\right )^{2} - 12 \, \cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) + 15 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d \cos \left (d x + c\right ) - 2 \, a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + 2 \, a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]
1/24*(33*sqrt(2)*(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d *x + c) - 2)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(d*x + c ) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*c os(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2 )*sin(d*x + c) - cos(d*x + c) - 2)) - 4*(4*cos(d*x + c)^3 + 16*cos(d*x + c )^2 - (4*cos(d*x + c)^2 - 12*cos(d*x + c) + 3)*sin(d*x + c) + 15*cos(d*x + c) + 3)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^2 - a^2*d*cos(d*x + c) - 2*a^2*d - (a^2*d*cos(d*x + c) + 2*a^2*d)*sin(d*x + c))
Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{3}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.79 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.36 \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\frac {33 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {33 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {6 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2} {\left (2 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{\frac {9}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{6} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{24 \, d} \]
1/24*(33*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos( -1/4*pi + 1/2*d*x + 1/2*c))) - 33*sqrt(2)*log(-sin(-1/4*pi + 1/2*d*x + 1/2 *c) + 1)/(a^(3/2)*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) + 6*sqrt(2)*sin(-1/ 4*pi + 1/2*d*x + 1/2*c)/((sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^(3/2)*sg n(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 16*sqrt(2)*(2*a^(9/2)*sin(-1/4*pi + 1 /2*d*x + 1/2*c)^3 + 3*a^(9/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^6*sgn(cos (-1/4*pi + 1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\sin ^3(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^3}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \]