\(\int x^3 (a+b \sin (c+d x^2))^2 \, dx\) [13]

Optimal result

Integrand size = 18, antiderivative size = 102 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {a^2 x^4}{4}+\frac {b^2 x^4}{8}-\frac {a b x^2 \cos \left (c+d x^2\right )}{d}+\frac {a b \sin \left (c+d x^2\right )}{d^2}-\frac {b^2 x^2 \cos \left (c+d x^2\right ) \sin \left (c+d x^2\right )}{4 d}+\frac {b^2 \sin ^2\left (c+d x^2\right )}{8 d^2} \] Output:

1/4*a^2*x^4+1/8*b^2*x^4-a*b*x^2*cos(d*x^2+c)/d+a*b*sin(d*x^2+c)/d^2-1/4*b^ 
2*x^2*cos(d*x^2+c)*sin(d*x^2+c)/d+1/8*b^2*sin(d*x^2+c)^2/d^2
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.90 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {4 a^2 d^2 x^4+2 b^2 d^2 x^4-16 a b d x^2 \cos \left (c+d x^2\right )-b^2 \cos \left (2 \left (c+d x^2\right )\right )+16 a b \sin \left (c+d x^2\right )-2 b^2 d x^2 \sin \left (2 \left (c+d x^2\right )\right )}{16 d^2} \] Input:

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

Output:

(4*a^2*d^2*x^4 + 2*b^2*d^2*x^4 - 16*a*b*d*x^2*Cos[c + d*x^2] - b^2*Cos[2*( 
c + d*x^2)] + 16*a*b*Sin[c + d*x^2] - 2*b^2*d*x^2*Sin[2*(c + d*x^2)])/(16* 
d^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3860, 3042, 3798, 2009}

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.

\(\displaystyle \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{2} \int x^2 \left (a+b \sin \left (d x^2+c\right )\right )^2dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int x^2 \left (a+b \sin \left (d x^2+c\right )\right )^2dx^2\)

\(\Big \downarrow \) 3798

\(\displaystyle \frac {1}{2} \int \left (a^2 x^2+b^2 \sin ^2\left (d x^2+c\right ) x^2+2 a b \sin \left (d x^2+c\right ) x^2\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {a^2 x^4}{2}+\frac {2 a b \sin \left (c+d x^2\right )}{d^2}-\frac {2 a b x^2 \cos \left (c+d x^2\right )}{d}+\frac {b^2 \sin ^2\left (c+d x^2\right )}{4 d^2}-\frac {b^2 x^2 \sin \left (c+d x^2\right ) \cos \left (c+d x^2\right )}{2 d}+\frac {b^2 x^4}{4}\right )\)

Input:

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

Output:

((a^2*x^4)/2 + (b^2*x^4)/4 - (2*a*b*x^2*Cos[c + d*x^2])/d + (2*a*b*Sin[c + 
 d*x^2])/d^2 - (b^2*x^2*Cos[c + d*x^2]*Sin[c + d*x^2])/(2*d) + (b^2*Sin[c 
+ d*x^2]^2)/(4*d^2))/2
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ 
m, 0] || NeQ[a^2 - b^2, 0])
 

rule 3860

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ 
(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ 
(m + 1)/n], 0]))
 

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.87

method	result	size
parts	\(\frac {a^{2} x^{4}}{4}+b^{2} \left (\frac {x^{4}}{8}-\frac {x^{2} \sin \left (2 d \,x^{2}+2 c \right )}{8 d}-\frac {\cos \left (2 d \,x^{2}+2 c \right )}{16 d^{2}}\right )+2 a b \left (-\frac {x^{2} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\sin \left (d \,x^{2}+c \right )}{2 d^{2}}\right )\)	\(89\)
default	\(\frac {\left (a^{2}+\frac {b^{2}}{2}\right ) x^{4}}{4}-\frac {b^{2} \left (\frac {x^{2} \sin \left (2 d \,x^{2}+2 c \right )}{4 d}+\frac {\cos \left (2 d \,x^{2}+2 c \right )}{8 d^{2}}\right )}{2}+2 a b \left (-\frac {x^{2} \cos \left (d \,x^{2}+c \right )}{2 d}+\frac {\sin \left (d \,x^{2}+c \right )}{2 d^{2}}\right )\)	\(91\)
risch	\(\frac {a^{2} x^{4}}{4}+\frac {b^{2} x^{4}}{8}-\frac {a b \,x^{2} \cos \left (d \,x^{2}+c \right )}{d}+\frac {a b \sin \left (d \,x^{2}+c \right )}{d^{2}}-\frac {b^{2} \cos \left (2 d \,x^{2}+2 c \right )}{16 d^{2}}-\frac {b^{2} x^{2} \sin \left (2 d \,x^{2}+2 c \right )}{8 d}\)	\(91\)
parallelrisch	\(\frac {4 a^{2} d^{2} x^{4}+2 b^{2} d^{2} x^{4}-2 b^{2} x^{2} \sin \left (2 d \,x^{2}+2 c \right ) d -16 a b \,x^{2} \cos \left (d \,x^{2}+c \right ) d +16 \sin \left (d \,x^{2}+c \right ) a b -b^{2} \cos \left (2 d \,x^{2}+2 c \right )+b^{2}}{16 d^{2}}\)	\(96\)
norman	\(\frac {\left (\frac {a^{2}}{4}+\frac {b^{2}}{8}\right ) x^{4}+\left (\frac {a^{2}}{2}+\frac {b^{2}}{4}\right ) x^{4} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a^{2}}{4}+\frac {b^{2}}{8}\right ) x^{4} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{4}+\frac {a b \,x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {b^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}}{2 d^{2}}-\frac {b^{2} x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{2 d}+\frac {b^{2} x^{2} \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{3}}{2 d}+\frac {2 a b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{d^{2}}+\frac {2 a b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{3}}{d^{2}}-\frac {a b \,x^{2}}{d}}{{\left (1+\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2}\right )}^{2}}\)	\(228\)
orering	\(\frac {\left (64 x^{8} d^{4}+700 x^{4} d^{2}+1485\right ) {\left (a +b \sin \left (d \,x^{2}+c \right )\right )}^{2}}{256 x^{4} d^{4}}-\frac {5 \left (44 x^{4} d^{2}+169\right ) \left (3 x^{2} {\left (a +b \sin \left (d \,x^{2}+c \right )\right )}^{2}+4 x^{4} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b d \cos \left (d \,x^{2}+c \right )\right )}{256 d^{4} x^{6}}+\frac {\left (20 x^{4} d^{2}+197\right ) \left (6 x {\left (a +b \sin \left (d \,x^{2}+c \right )\right )}^{2}+28 x^{3} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b d \cos \left (d \,x^{2}+c \right )+8 x^{5} b^{2} d^{2} \cos \left (d \,x^{2}+c \right )^{2}-8 x^{5} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{2}+c \right )\right )}{256 d^{4} x^{5}}-\frac {11 \left (6 {\left (a +b \sin \left (d \,x^{2}+c \right )\right )}^{2}+108 x^{2} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b d \cos \left (d \,x^{2}+c \right )+96 x^{4} b^{2} d^{2} \cos \left (d \,x^{2}+c \right )^{2}-96 x^{4} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{2}+c \right )-48 \sin \left (d \,x^{2}+c \right ) \cos \left (d \,x^{2}+c \right ) b^{2} d^{3} x^{6}-16 x^{6} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{3} \cos \left (d \,x^{2}+c \right )\right )}{128 x^{4} d^{4}}+\frac {240 \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b d x \cos \left (d \,x^{2}+c \right )+600 x^{3} b^{2} d^{2} \cos \left (d \,x^{2}+c \right )^{2}-600 x^{3} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{2}+c \right )-864 x^{5} b^{2} d^{3} \cos \left (d \,x^{2}+c \right ) \sin \left (d \,x^{2}+c \right )-288 x^{5} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{3} \cos \left (d \,x^{2}+c \right )-128 d^{4} x^{7} \cos \left (d \,x^{2}+c \right )^{2} b^{2}+96 \sin \left (d \,x^{2}+c \right )^{2} d^{4} x^{7} b^{2}+32 x^{7} \left (a +b \sin \left (d \,x^{2}+c \right )\right ) b \,d^{4} \sin \left (d \,x^{2}+c \right )}{256 d^{4} x^{3}}\)	\(589\)

Input:

int(x^3*(a+b*sin(d*x^2+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*a^2*x^4+b^2*(1/8*x^4-1/8/d*x^2*sin(2*d*x^2+2*c)-1/16/d^2*cos(2*d*x^2+2 
*c))+2*a*b*(-1/2/d*x^2*cos(d*x^2+c)+1/2/d^2*sin(d*x^2+c))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{4} - 8 \, a b d x^{2} \cos \left (d x^{2} + c\right ) - b^{2} \cos \left (d x^{2} + c\right )^{2} - 2 \, {\left (b^{2} d x^{2} \cos \left (d x^{2} + c\right ) - 4 \, a b\right )} \sin \left (d x^{2} + c\right )}{8 \, d^{2}} \] Input:

integrate(x^3*(a+b*sin(d*x^2+c))^2,x, algorithm="fricas")
 

Output:

1/8*((2*a^2 + b^2)*d^2*x^4 - 8*a*b*d*x^2*cos(d*x^2 + c) - b^2*cos(d*x^2 + 
c)^2 - 2*(b^2*d*x^2*cos(d*x^2 + c) - 4*a*b)*sin(d*x^2 + c))/d^2
 

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.33 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} - \frac {a b x^{2} \cos {\left (c + d x^{2} \right )}}{d} + \frac {a b \sin {\left (c + d x^{2} \right )}}{d^{2}} + \frac {b^{2} x^{4} \sin ^{2}{\left (c + d x^{2} \right )}}{8} + \frac {b^{2} x^{4} \cos ^{2}{\left (c + d x^{2} \right )}}{8} - \frac {b^{2} x^{2} \sin {\left (c + d x^{2} \right )} \cos {\left (c + d x^{2} \right )}}{4 d} + \frac {b^{2} \sin ^{2}{\left (c + d x^{2} \right )}}{8 d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{4} \left (a + b \sin {\left (c \right )}\right )^{2}}{4} & \text {otherwise} \end {cases} \] Input:

integrate(x**3*(a+b*sin(d*x**2+c))**2,x)
 

Output:

Piecewise((a**2*x**4/4 - a*b*x**2*cos(c + d*x**2)/d + a*b*sin(c + d*x**2)/ 
d**2 + b**2*x**4*sin(c + d*x**2)**2/8 + b**2*x**4*cos(c + d*x**2)**2/8 - b 
**2*x**2*sin(c + d*x**2)*cos(c + d*x**2)/(4*d) + b**2*sin(c + d*x**2)**2/( 
8*d**2), Ne(d, 0)), (x**4*(a + b*sin(c))**2/4, True))
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.85 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {1}{4} \, a^{2} x^{4} - \frac {{\left (d x^{2} \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right )} a b}{d^{2}} + \frac {{\left (2 \, d^{2} x^{4} - 2 \, d x^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) - \cos \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2}}{16 \, d^{2}} \] Input:

integrate(x^3*(a+b*sin(d*x^2+c))^2,x, algorithm="maxima")
 

Output:

1/4*a^2*x^4 - (d*x^2*cos(d*x^2 + c) - sin(d*x^2 + c))*a*b/d^2 + 1/16*(2*d^ 
2*x^4 - 2*d*x^2*sin(2*d*x^2 + 2*c) - cos(2*d*x^2 + 2*c))*b^2/d^2
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.62 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {4 \, {\left (d x^{2} + c\right )}^{2} a^{2} + 2 \, {\left (d x^{2} + c\right )}^{2} b^{2} - 16 \, {\left (d x^{2} + c\right )} a b \cos \left (d x^{2} + c\right ) - 2 \, {\left (d x^{2} + c\right )} b^{2} \sin \left (2 \, d x^{2} + 2 \, c\right ) - b^{2} \cos \left (2 \, d x^{2} + 2 \, c\right ) + 16 \, a b \sin \left (d x^{2} + c\right )}{16 \, d^{2}} - \frac {4 \, {\left (d x^{2} + c\right )} a^{2} c + {\left (2 \, d x^{2} + 2 \, c - \sin \left (2 \, d x^{2} + 2 \, c\right )\right )} b^{2} c - 8 \, a b c \cos \left (d x^{2} + c\right )}{8 \, d^{2}} \] Input:

integrate(x^3*(a+b*sin(d*x^2+c))^2,x, algorithm="giac")
 

Output:

1/16*(4*(d*x^2 + c)^2*a^2 + 2*(d*x^2 + c)^2*b^2 - 16*(d*x^2 + c)*a*b*cos(d 
*x^2 + c) - 2*(d*x^2 + c)*b^2*sin(2*d*x^2 + 2*c) - b^2*cos(2*d*x^2 + 2*c) 
+ 16*a*b*sin(d*x^2 + c))/d^2 - 1/8*(4*(d*x^2 + c)*a^2*c + (2*d*x^2 + 2*c - 
 sin(2*d*x^2 + 2*c))*b^2*c - 8*a*b*c*cos(d*x^2 + c))/d^2
 

Mupad [B] (verification not implemented)

Time = 38.70 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=-\frac {b^2\,{\cos \left (d\,x^2+c\right )}^2-2\,a^2\,d^2\,x^4-b^2\,d^2\,x^4-8\,a\,b\,\sin \left (d\,x^2+c\right )+8\,a\,b\,d\,x^2\,\cos \left (d\,x^2+c\right )+2\,b^2\,d\,x^2\,\cos \left (d\,x^2+c\right )\,\sin \left (d\,x^2+c\right )}{8\,d^2} \] Input:

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

Output:

-(b^2*cos(c + d*x^2)^2 - 2*a^2*d^2*x^4 - b^2*d^2*x^4 - 8*a*b*sin(c + d*x^2 
) + 8*a*b*d*x^2*cos(c + d*x^2) + 2*b^2*d*x^2*cos(c + d*x^2)*sin(c + d*x^2) 
)/(8*d^2)
 

Reduce [F]

\[ \int x^3 \left (a+b \sin \left (c+d x^2\right )\right )^2 \, dx=\frac {-4 \cos \left (d \,x^{2}+c \right ) a b d \,x^{2}+4 \left (\int \sin \left (d \,x^{2}+c \right )^{2} x^{3}d x \right ) b^{2} d^{2}+4 \sin \left (d \,x^{2}+c \right ) a b +a^{2} d^{2} x^{4}}{4 d^{2}} \] Input:

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

Output:

( - 4*cos(c + d*x**2)*a*b*d*x**2 + 4*int(sin(c + d*x**2)**2*x**3,x)*b**2*d 
**2 + 4*sin(c + d*x**2)*a*b + a**2*d**2*x**4)/(4*d**2)