\(\int (a+b \sin (d+e x)) (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x))^{3/2} \, dx\) [447]

Optimal result

Integrand size = 41, antiderivative size = 331 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=-\frac {b \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{4 e}-\frac {\left (4 a^4+28 a^2 b^2+3 b^4\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{6 e (b+a \sin (d+e x))^3}-\frac {\left (4 a^2+3 b^2\right ) \cos (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{12 e (b+a \sin (d+e x))}+\frac {5 a^4 b \left (3 a^2+4 b^2\right ) x \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{8 \left (a b+a^2 \sin (d+e x)\right )^3}-\frac {a^4 b \left (29 a^2+6 b^2\right ) \cos (d+e x) \sin (d+e x) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}{24 e \left (a b+a^2 \sin (d+e x)\right )^3} \] Output:

-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
 

Mathematica [A] (verified)

Time = 1.97 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.42 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {\sqrt {(b+a \sin (d+e x))^2} \left (-24 \left (3 a^4+21 a^2 b^2+4 b^4\right ) \cos (d+e x)+8 a \left (a^3+3 a b^2\right ) \cos (3 (d+e x))+3 a b \left (20 \left (3 a^2+4 b^2\right ) (d+e x)-8 \left (4 a^2+3 b^2\right ) \sin (2 (d+e x))+a^2 \sin (4 (d+e x))\right )\right )}{96 e (b+a \sin (d+e x))} \] Input:

Integrate[(a + b*Sin[d + e*x])*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x 
]^2)^(3/2),x]
 

Output:

(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]))
 

Rubi [A] (verified)

Time = 0.78 (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

\(\displaystyle \frac {\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2} \left (\frac {1}{4} \left (\frac {1}{3} \left (-\frac {a^4 b \left (29 a^2+6 b^2\right ) \sin (d+e x) \cos (d+e x)}{2 e}+\frac {15}{2} a^4 b x \left (3 a^2+4 b^2\right )-\frac {2 a^3 \left (4 a^4+28 a^2 b^2+3 b^4\right ) \cos (d+e x)}{e}\right )-\frac {a \left (4 a^2+3 b^2\right ) \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^2}{3 e}\right )-\frac {b \cos (d+e x) \left (a^2 \sin (d+e x)+a b\right )^3}{4 e}\right )}{\left (a^2 \sin (d+e x)+a b\right )^3}\)

Input:

Int[(a + b*Sin[d + e*x])*(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^( 
3/2),x]
 

Output:

((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
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3213

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]
 

rule 3232

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]
 

rule 3771

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.32 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {\left (45 \left (e x +d \right ) a^{3} b +60 \left (e x +d \right ) a \,b^{3}+\left (8 \cos \left (e x +d \right )^{3}-24 \cos \left (e x +d \right )-16\right ) a^{4}+\sin \left (e x +d \right ) \cos \left (e x +d \right ) \left (6 \cos \left (e x +d \right )^{2}-51\right ) a^{3} b +\left (24 \cos \left (e x +d \right )^{3}-144 \cos \left (e x +d \right )-120\right ) b^{2} a^{2}-36 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,b^{3}+\left (-24 \cos \left (e x +d \right )-24\right ) b^{4}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{24 e}\)	\(152\)
parts	\(\frac {a \left (9 a^{2} b \left (e x +d \right )+6 \left (e x +d \right ) b^{3}+\left (2 \cos \left (e x +d \right )^{3}-6 \cos \left (e x +d \right )-4\right ) a^{3}-9 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{2} b +\left (-18 \cos \left (e x +d \right )-18\right ) a \,b^{2}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{6 e}+\frac {b \left (3 \left (e x +d \right ) a^{3}+12 a \,b^{2} \left (e x +d \right )+\sin \left (e x +d \right ) \cos \left (e x +d \right ) \left (2 \cos \left (e x +d \right )^{2}-5\right ) a^{3}+\left (8 \cos \left (e x +d \right )^{3}-24 \cos \left (e x +d \right )-16\right ) b \,a^{2}-12 \sin \left (e x +d \right ) \cos \left (e x +d \right ) a \,b^{2}+\left (-8 \cos \left (e x +d \right )-8\right ) b^{3}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{8 e}\)	\(222\)

Input:

int((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,metho 
d=_RETURNVERBOSE)
 

Output:

1/24/e*(45*(e*x+d)*a^3*b+60*(e*x+d)*a*b^3+(8*cos(e*x+d)^3-24*cos(e*x+d)-16 
)*a^4+sin(e*x+d)*cos(e*x+d)*(6*cos(e*x+d)^2-51)*a^3*b+(24*cos(e*x+d)^3-144 
*cos(e*x+d)-120)*b^2*a^2-36*cos(e*x+d)*sin(e*x+d)*a*b^3+(-24*cos(e*x+d)-24 
)*b^4)*csgn(b+a*sin(e*x+d))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.34 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {8 \, {\left (a^{4} + 3 \, a^{2} b^{2}\right )} \cos \left (e x + d\right )^{3} + 15 \, {\left (3 \, a^{3} b + 4 \, a b^{3}\right )} e x - 24 \, {\left (a^{4} + 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (e x + d\right ) + 3 \, {\left (2 \, a^{3} b \cos \left (e x + d\right )^{3} - {\left (17 \, a^{3} b + 12 \, a b^{3}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{24 \, e} \] Input:

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")
 

Output:

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
 

Sympy [F(-1)]

Timed out. \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\text {Timed out} \] Input:

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)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 556, normalized size of antiderivative = 1.68 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx =\text {Too large to display} \] Input:

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")
 

Output:

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
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.69 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) \sin \left (4 \, e x + 4 \, d\right )}{32 \, e} + \frac {5}{8} \, {\left (3 \, a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 4 \, a b^{3} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} x + \frac {{\left (a^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 3 \, a^{2} b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \cos \left (3 \, e x + 3 \, d\right )}{12 \, e} - \frac {{\left (3 \, a^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 21 \, a^{2} b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 4 \, b^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \cos \left (e x + d\right )}{4 \, e} - \frac {{\left (4 \, a^{3} b \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) + 3 \, a b^{3} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \] Input:

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")
 

Output:

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
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\int \left (a+b\,\sin \left (d+e\,x\right )\right )\,{\left (a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2\right )}^{3/2} \,d x \] Input:

int((a + b*sin(d + e*x))*(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^( 
3/2),x)
 

Output:

int((a + b*sin(d + e*x))*(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^( 
3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.52 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2} \, dx=\frac {-6 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{3} a^{3} b -8 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2} a^{4}-24 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2} a^{2} b^{2}-45 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3} b -36 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,b^{3}-16 \cos \left (e x +d \right ) a^{4}-120 \cos \left (e x +d \right ) a^{2} b^{2}-24 \cos \left (e x +d \right ) b^{4}+16 a^{4}+45 a^{3} b e x +120 a^{2} b^{2}+60 a \,b^{3} e x +24 b^{4}}{24 e} \] Input:

int((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)
 

Output:

( - 6*cos(d + e*x)*sin(d + e*x)**3*a**3*b - 8*cos(d + e*x)*sin(d + e*x)**2 
*a**4 - 24*cos(d + e*x)*sin(d + e*x)**2*a**2*b**2 - 45*cos(d + e*x)*sin(d 
+ e*x)*a**3*b - 36*cos(d + e*x)*sin(d + e*x)*a*b**3 - 16*cos(d + e*x)*a**4 
 - 120*cos(d + e*x)*a**2*b**2 - 24*cos(d + e*x)*b**4 + 16*a**4 + 45*a**3*b 
*e*x + 120*a**2*b**2 + 60*a*b**3*e*x + 24*b**4)/(24*e)