Integrand size = 41, antiderivative size = 331 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=-\frac {b \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{4 e}-\frac {\left (4 a^4+28 a^2 b^2+3 b^4\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{6 e (b+a \sin (d+e x))^3}-\frac {\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{12 e (b+a \sin (d+e x))}+\frac {5 a^4 b \left (3 a^2+4 b^2\right ) x \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{8 \left (a b+a^2 \sin (d+e x)\right )^3}-\frac {a^4 b \left (29 a^2+6 b^2\right ) \cos (d+e x) \sin (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{24 e \left (a b+a^2 \sin (d+e x)\right )^3} \]
-1/4*b*cos(e*x+d)*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2)/e-1/6*(4*a ^4+28*a^2*b^2+3*b^4)*cos(e*x+d)*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3 /2)/e/(b+a*sin(e*x+d))^3-1/12*(4*a^2+3*b^2)*cos(e*x+d)*(b^2+2*a*b*sin(e*x+ d)+a^2*sin(e*x+d)^2)^(3/2)/e/(b+a*sin(e*x+d))+5/8*a^4*b*(3*a^2+4*b^2)*x*(b ^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2)/(a*b+a^2*sin(e*x+d))^3-1/24*a^ 4*b*(29*a^2+6*b^2)*cos(e*x+d)*sin(e*x+d)*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x +d)^2)^(3/2)/e/(a*b+a^2*sin(e*x+d))^3
Time = 2.55 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.42 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {\sqrt {(b+a \sin (d+e x))^2} \left (-24 \left (3 a^4+21 a^2 b^2+4 b^4\right ) \cos (d+e x)+8 a \left (a^3+3 a b^2\right ) \cos (3 (d+e x))+3 a b \left (20 \left (3 a^2+4 b^2\right ) (d+e x)-8 \left (4 a^2+3 b^2\right ) \sin (2 (d+e x))+a^2 \sin (4 (d+e x))\right )\right )}{96 e (b+a \sin (d+e x))} \]
(Sqrt[(b + a*Sin[d + e*x])^2]*(-24*(3*a^4 + 21*a^2*b^2 + 4*b^4)*Cos[d + e* x] + 8*a*(a^3 + 3*a*b^2)*Cos[3*(d + e*x)] + 3*a*b*(20*(3*a^2 + 4*b^2)*(d + e*x) - 8*(4*a^2 + 3*b^2)*Sin[2*(d + e*x)] + a^2*Sin[4*(d + e*x)])))/(96*e *(b + a*Sin[d + e*x]))
Time = 0.75 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.65, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {3042, 3771, 27, 3042, 3232, 3042, 3232, 3042, 3213}
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.
| \(\displaystyle \int (a+b \sin (d+e x)) \left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+b \sin (d+e x)) \left (a^2 \sin (d+e x)^2+2 a b \sin (d+e x)+b^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 3771 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \int 8 \left (\sin (d+e x) a^2+b a\right )^3 (a+b \sin (d+e x))dx}{8 \left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \int \left (\sin (d+e x) a^2+b a\right )^3 (a+b \sin (d+e x))dx}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \int \left (\sin (d+e x) a^2+b a\right )^3 (a+b \sin (d+e x))dx}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \int \left (\sin (d+e x) a^2+b a\right )^2 \left (7 b a^2+\left (4 a^2+3 b^2\right ) \sin (d+e x) a\right )dx-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \int \left (\sin (d+e x) a^2+b a\right )^2 \left (7 b a^2+\left (4 a^2+3 b^2\right ) \sin (d+e x) a\right )dx-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (\sin (d+e x) a^2+b a\right ) \left (\left (8 a^2+27 b^2\right ) a^3+b \left (29 a^2+6 b^2\right ) \sin (d+e x) a^2\right )dx-\frac {a \left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^2}{3 e}\right )-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (\sin (d+e x) a^2+b a\right ) \left (\left (8 a^2+27 b^2\right ) a^3+b \left (29 a^2+6 b^2\right ) \sin (d+e x) a^2\right )dx-\frac {a \left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^2}{3 e}\right )-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {a^4 b \left (29 a^2+6 b^2\right ) \sin (d+e x) \cos (d+e x)}{2 e}+\frac {15}{2} a^4 b x \left (3 a^2+4 b^2\right )-\frac {2 a^3 \left (4 a^4+28 a^2 b^2+3 b^4\right ) \cos (d+e x)}{e}\right )-\frac {a \left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^2}{3 e}\right )-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\) |
((b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^(3/2)*(-1/4*(b*Cos[d + e* x]*(a*b + a^2*Sin[d + e*x])^3)/e + (-1/3*(a*(4*a^2 + 3*b^2)*Cos[d + e*x]*( a*b + a^2*Sin[d + e*x])^2)/e + ((15*a^4*b*(3*a^2 + 4*b^2)*x)/2 - (2*a^3*(4 *a^4 + 28*a^2*b^2 + 3*b^4)*Cos[d + e*x])/e - (a^4*b*(29*a^2 + 6*b^2)*Cos[d + e*x]*Sin[d + e*x])/(2*e))/3)/4))/(a*b + a^2*Sin[d + e*x])^3
3.6.4.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)* (x_)] + (c_.)*sin[(d_.) + (e_.)*(x_)]^2)^(n_), x_Symbol] :> Simp[(a + b*Sin [d + e*x] + c*Sin[d + e*x]^2)^n/(b + 2*c*Sin[d + e*x])^(2*n) Int[(A + B*S in[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.70 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.58
| method | result | size |
| default | \(\frac {\operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right ) \left (a \,b^{3} \left (e x +d \right )-3 \cos \left (e x +d \right ) a^{2} b^{2}+3 a^{3} b \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-\frac {a^{4} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}-\cos \left (e x +d \right ) b^{4}+3 a \,b^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-a^{2} b^{2} \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )+a^{3} b \left (-\frac {\left (\sin \left (e x +d \right )^{3}+\frac {3 \sin \left (e x +d \right )}{2}\right ) \cos \left (e x +d \right )}{4}+\frac {3 e x}{8}+\frac {3 d}{8}\right )\right )}{e}\) | \(191\) |
| parts | \(\frac {a \left (2 \cos \left (e x +d \right )^{3} a^{3}-9 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{2} b -6 a^{3} \cos \left (e x +d \right )-18 \cos \left (e x +d \right ) a \,b^{2}+9 a^{2} b \left (e x +d \right )+6 \left (e x +d \right ) b^{3}-4 a^{3}-18 a \,b^{2}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{6 e}+\frac {b \left (2 \cos \left (e x +d \right )^{3} \sin \left (e x +d \right ) a^{3}+8 a^{2} b \cos \left (e x +d \right )^{3}-5 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3}-12 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,b^{2}-24 \cos \left (e x +d \right ) a^{2} b -8 b^{3} \cos \left (e x +d \right )+3 \left (e x +d \right ) a^{3}+12 a \,b^{2} \left (e x +d \right )-16 a^{2} b -8 b^{3}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{8 e}\) | \(247\) |
csgn(b+a*sin(e*x+d))/e*(a*b^3*(e*x+d)-3*cos(e*x+d)*a^2*b^2+3*a^3*b*(-1/2*c os(e*x+d)*sin(e*x+d)+1/2*e*x+1/2*d)-1/3*a^4*(2+sin(e*x+d)^2)*cos(e*x+d)-co s(e*x+d)*b^4+3*a*b^3*(-1/2*cos(e*x+d)*sin(e*x+d)+1/2*e*x+1/2*d)-a^2*b^2*(2 +sin(e*x+d)^2)*cos(e*x+d)+a^3*b*(-1/4*(sin(e*x+d)^3+3/2*sin(e*x+d))*cos(e* x+d)+3/8*e*x+3/8*d))
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.34 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {8 \, {\left (a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (e x + d\right )^{3} + 15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} e x - 24 \, {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (e x + d\right ) + 3 \, {\left (2 \, a^{3} b \cos \left (e x + d\right )^{3} - {\left (17 \, a^{3} b + 12 \, a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{24 \, e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="fricas")
1/24*(8*(a^4 + 3*a^2*b^2)*cos(e*x + d)^3 + 15*(3*a^3*b + 4*a*b^3)*e*x - 24 *(a^4 + 6*a^2*b^2 + b^4)*cos(e*x + d) + 3*(2*a^3*b*cos(e*x + d)^3 - (17*a^ 3*b + 12*a*b^3)*cos(e*x + d))*sin(e*x + d))/e
Timed out. \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.35 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.68 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {4 \, {\left (3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {4 \, a^{3} + 18 \, a b^{2} + \frac {9 \, a^{2} b \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {18 \, a b^{2} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} - \frac {9 \, a^{2} b \sin \left (e x + d\right )^{5}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{5}} + \frac {12 \, {\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}}}{\frac {3 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + \frac {\sin \left (e x + d\right )^{6}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{6}} + 1}\right )} a + 3 \, {\left (3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \arctan \left (\frac {\sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1}\right ) - \frac {16 \, a^{2} b + 8 \, b^{3} + \frac {8 \, b^{3} \sin \left (e x + d\right )^{6}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{6}} + \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \sin \left (e x + d\right )}{\cos \left (e x + d\right ) + 1} + \frac {8 \, {\left (8 \, a^{2} b + 3 \, b^{3}\right )} \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {{\left (11 \, a^{3} + 12 \, a b^{2}\right )} \sin \left (e x + d\right )^{3}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{3}} + \frac {24 \, {\left (2 \, a^{2} b + b^{3}\right )} \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} - \frac {{\left (11 \, a^{3} + 12 \, a b^{2}\right )} \sin \left (e x + d\right )^{5}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{5}} - \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \sin \left (e x + d\right )^{7}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{7}}}{\frac {4 \, \sin \left (e x + d\right )^{2}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{2}} + \frac {6 \, \sin \left (e x + d\right )^{4}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{4}} + \frac {4 \, \sin \left (e x + d\right )^{6}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{6}} + \frac {\sin \left (e x + d\right )^{8}}{{\left (\cos \left (e x + d\right ) + 1\right )}^{8}} + 1}\right )} b}{12 \, e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="maxima")
1/12*(4*(3*(3*a^2*b + 2*b^3)*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - (4* a^3 + 18*a*b^2 + 9*a^2*b*sin(e*x + d)/(cos(e*x + d) + 1) + 18*a*b^2*sin(e* x + d)^4/(cos(e*x + d) + 1)^4 - 9*a^2*b*sin(e*x + d)^5/(cos(e*x + d) + 1)^ 5 + 12*(a^3 + 3*a*b^2)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2)/(3*sin(e*x + d )^2/(cos(e*x + d) + 1)^2 + 3*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 + sin(e*x + d)^6/(cos(e*x + d) + 1)^6 + 1))*a + 3*(3*(a^3 + 4*a*b^2)*arctan(sin(e*x + d)/(cos(e*x + d) + 1)) - (16*a^2*b + 8*b^3 + 8*b^3*sin(e*x + d)^6/(cos( e*x + d) + 1)^6 + 3*(a^3 + 4*a*b^2)*sin(e*x + d)/(cos(e*x + d) + 1) + 8*(8 *a^2*b + 3*b^3)*sin(e*x + d)^2/(cos(e*x + d) + 1)^2 + (11*a^3 + 12*a*b^2)* sin(e*x + d)^3/(cos(e*x + d) + 1)^3 + 24*(2*a^2*b + b^3)*sin(e*x + d)^4/(c os(e*x + d) + 1)^4 - (11*a^3 + 12*a*b^2)*sin(e*x + d)^5/(cos(e*x + d) + 1) ^5 - 3*(a^3 + 4*a*b^2)*sin(e*x + d)^7/(cos(e*x + d) + 1)^7)/(4*sin(e*x + d )^2/(cos(e*x + d) + 1)^2 + 6*sin(e*x + d)^4/(cos(e*x + d) + 1)^4 + 4*sin(e *x + d)^6/(cos(e*x + d) + 1)^6 + sin(e*x + d)^8/(cos(e*x + d) + 1)^8 + 1)) *b)/e
Time = 0.30 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.69 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) \sin \left (4 \, e x + 4 \, d\right )}{32 \, e} + \frac {5}{8} \, {\left (3 \, a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 4 \, a b^{3} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} x + \frac {{\left (a^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 3 \, a^{2} b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \cos \left (3 \, e x + 3 \, d\right )}{12 \, e} - \frac {{\left (3 \, a^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 21 \, a^{2} b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 4 \, b^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \cos \left (e x + d\right )}{4 \, e} - \frac {{\left (4 \, a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 3 \, a b^{3} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \]
integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="giac")
1/32*a^3*b*sgn(a*sin(e*x + d) + b)*sin(4*e*x + 4*d)/e + 5/8*(3*a^3*b*sgn(a *sin(e*x + d) + b) + 4*a*b^3*sgn(a*sin(e*x + d) + b))*x + 1/12*(a^4*sgn(a* sin(e*x + d) + b) + 3*a^2*b^2*sgn(a*sin(e*x + d) + b))*cos(3*e*x + 3*d)/e - 1/4*(3*a^4*sgn(a*sin(e*x + d) + b) + 21*a^2*b^2*sgn(a*sin(e*x + d) + b) + 4*b^4*sgn(a*sin(e*x + d) + b))*cos(e*x + d)/e - 1/4*(4*a^3*b*sgn(a*sin(e *x + d) + b) + 3*a*b^3*sgn(a*sin(e*x + d) + b))*sin(2*e*x + 2*d)/e
Timed out. \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\int \left (a+b\,\sin \left (d+e\,x\right )\right )\,{\left (a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2\right )}^{3/2} \,d x \]