\(\int x^5 (a+b \sin (c+d x^3))^2 \, dx\) [69]

Optimal result

Integrand size = 18, antiderivative size = 107 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {a^2 x^6}{6}+\frac {b^2 x^6}{12}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{3 d}+\frac {2 a b \sin \left (c+d x^3\right )}{3 d^2}-\frac {b^2 x^3 \cos \left (c+d x^3\right ) \sin \left (c+d x^3\right )}{6 d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{12 d^2} \] Output:

1/6*a^2*x^6+1/12*b^2*x^6-2/3*a*b*x^3*cos(d*x^3+c)/d+2/3*a*b*sin(d*x^3+c)/d 
^2-1/6*b^2*x^3*cos(d*x^3+c)*sin(d*x^3+c)/d+1/12*b^2*sin(d*x^3+c)^2/d^2
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {4 a^2 d^2 x^6+2 b^2 d^2 x^6-16 a b d x^3 \cos \left (c+d x^3\right )-b^2 \cos \left (2 \left (c+d x^3\right )\right )+16 a b \sin \left (c+d x^3\right )-2 b^2 d x^3 \sin \left (2 \left (c+d x^3\right )\right )}{24 d^2} \] Input:

Integrate[x^5*(a + b*Sin[c + d*x^3])^2,x]
 

Output:

(4*a^2*d^2*x^6 + 2*b^2*d^2*x^6 - 16*a*b*d*x^3*Cos[c + d*x^3] - b^2*Cos[2*( 
c + d*x^3)] + 16*a*b*Sin[c + d*x^3] - 2*b^2*d*x^3*Sin[2*(c + d*x^3)])/(24* 
d^2)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, 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.

\(\displaystyle \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx\)

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{3} \int x^3 \left (a+b \sin \left (d x^3+c\right )\right )^2dx^3\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int x^3 \left (a+b \sin \left (d x^3+c\right )\right )^2dx^3\)

\(\Big \downarrow \) 3798

\(\displaystyle \frac {1}{3} \int \left (a^2 x^3+b^2 \sin ^2\left (d x^3+c\right ) x^3+2 a b \sin \left (d x^3+c\right ) x^3\right )dx^3\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} \left (\frac {a^2 x^6}{2}+\frac {2 a b \sin \left (c+d x^3\right )}{d^2}-\frac {2 a b x^3 \cos \left (c+d x^3\right )}{d}+\frac {b^2 \sin ^2\left (c+d x^3\right )}{4 d^2}-\frac {b^2 x^3 \sin \left (c+d x^3\right ) \cos \left (c+d x^3\right )}{2 d}+\frac {b^2 x^6}{4}\right )\)

Input:

Int[x^5*(a + b*Sin[c + d*x^3])^2,x]
 

Output:

((a^2*x^6)/2 + (b^2*x^6)/4 - (2*a*b*x^3*Cos[c + d*x^3])/d + (2*a*b*Sin[c + 
 d*x^3])/d^2 - (b^2*x^3*Cos[c + d*x^3]*Sin[c + d*x^3])/(2*d) + (b^2*Sin[c 
+ d*x^3]^2)/(4*d^2))/3
 

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 = 2.72 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {x^{6} a^{2}}{6}+\frac {b^{2} x^{6}}{12}-\frac {2 a b \,x^{3} \cos \left (d \,x^{3}+c \right )}{3 d}+\frac {2 a b \sin \left (d \,x^{3}+c \right )}{3 d^{2}}-\frac {b^{2} \cos \left (2 d \,x^{3}+2 c \right )}{24 d^{2}}-\frac {b^{2} x^{3} \sin \left (2 d \,x^{3}+2 c \right )}{12 d}\)	\(92\)
parallelrisch	\(\frac {4 a^{2} d^{2} x^{6}+2 b^{2} d^{2} x^{6}-16 a b \,x^{3} \cos \left (d \,x^{3}+c \right ) d -2 b^{2} x^{3} \sin \left (2 d \,x^{3}+2 c \right ) d +16 \sin \left (d \,x^{3}+c \right ) a b -b^{2} \cos \left (2 d \,x^{3}+2 c \right )+b^{2}}{24 d^{2}}\)	\(96\)
parts	\(\frac {x^{6} a^{2}}{6}+\frac {\frac {b^{2} x^{6}}{12}+\frac {x^{6} b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{6}+\frac {x^{6} b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{4}}{12}+\frac {b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3 d^{2}}-\frac {b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d}+\frac {b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{3}}{3 d}}{{\left (1+\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}\right )}^{2}}+\frac {\frac {4 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}-\frac {2 a b \,x^{3}}{3 d}+\frac {2 a b \,x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3 d}}{1+\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}\)	\(216\)
default	\(\frac {b^{2} x^{6}}{6}+\frac {x^{6} a^{2}}{6}+\frac {-\frac {b^{2} x^{6}}{6}-\frac {x^{6} b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3}-\frac {x^{6} b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{4}}{6}+\frac {2 b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3 d^{2}}-\frac {2 b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d}+\frac {2 b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{3}}{3 d}}{2 {\left (1+\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}\right )}^{2}}+\frac {\frac {8 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}-\frac {4 a b \,x^{3}}{3 d}+\frac {4 a b \,x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3 d}}{2+2 \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}\)	\(226\)
norman	\(\frac {\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6}+\left (\frac {a^{2}}{3}+\frac {b^{2}}{6}\right ) x^{6} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}+\left (\frac {a^{2}}{6}+\frac {b^{2}}{12}\right ) x^{6} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{4}+\frac {b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}}{3 d^{2}}-\frac {b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d}+\frac {b^{2} x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {4 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{3 d^{2}}+\frac {4 a b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{3}}{3 d^{2}}-\frac {2 a b \,x^{3}}{3 d}+\frac {2 a b \,x^{3} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{4}}{3 d}}{{\left (1+\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2}\right )}^{2}}\)	\(229\)
orering	\(\frac {\left (81 d^{4} x^{12}+990 x^{6} d^{2}+2380\right ) {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}}{486 x^{6} d^{4}}-\frac {\left (135 x^{6} d^{2}+616\right ) \left (5 x^{4} {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}+6 x^{7} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b d \cos \left (d \,x^{3}+c \right )\right )}{324 d^{4} x^{10}}+\frac {\left (45 x^{6} d^{2}+556\right ) \left (20 x^{3} {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}+72 x^{6} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b d \cos \left (d \,x^{3}+c \right )+18 x^{9} b^{2} d^{2} \cos \left (d \,x^{3}+c \right )^{2}-18 x^{9} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{3}+c \right )\right )}{1944 d^{4} x^{9}}-\frac {19 \left (60 x^{2} {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}+552 x^{5} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b d \cos \left (d \,x^{3}+c \right )+378 x^{8} b^{2} d^{2} \cos \left (d \,x^{3}+c \right )^{2}-378 x^{8} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{3}+c \right )-162 \sin \left (d \,x^{3}+c \right ) \cos \left (d \,x^{3}+c \right ) b^{2} d^{3} x^{11}-54 x^{11} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{3} \cos \left (d \,x^{3}+c \right )\right )}{972 d^{4} x^{8}}+\frac {120 x {\left (a +b \sin \left (d \,x^{3}+c \right )\right )}^{2}+3120 x^{4} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b d \cos \left (d \,x^{3}+c \right )+4680 x^{7} b^{2} d^{2} \cos \left (d \,x^{3}+c \right )^{2}-4680 x^{7} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{2} \sin \left (d \,x^{3}+c \right )-5184 x^{10} b^{2} d^{3} \cos \left (d \,x^{3}+c \right ) \sin \left (d \,x^{3}+c \right )-1728 x^{10} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{3} \cos \left (d \,x^{3}+c \right )-648 d^{4} x^{13} \cos \left (d \,x^{3}+c \right )^{2} b^{2}+486 \sin \left (d \,x^{3}+c \right )^{2} d^{4} x^{13} b^{2}+162 x^{13} \left (a +b \sin \left (d \,x^{3}+c \right )\right ) b \,d^{4} \sin \left (d \,x^{3}+c \right )}{1944 d^{4} x^{7}}\)	\(613\)

Input:

int(x^5*(a+b*sin(d*x^3+c))^2,x,method=_RETURNVERBOSE)
 

Output:

1/6*x^6*a^2+1/12*b^2*x^6-2/3*a*b*x^3*cos(d*x^3+c)/d+2/3*a*b*sin(d*x^3+c)/d 
^2-1/24*b^2/d^2*cos(2*d*x^3+2*c)-1/12*b^2*x^3/d*sin(2*d*x^3+2*c)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {{\left (2 \, a^{2} + b^{2}\right )} d^{2} x^{6} - 8 \, a b d x^{3} \cos \left (d x^{3} + c\right ) - b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, {\left (b^{2} d x^{3} \cos \left (d x^{3} + c\right ) - 4 \, a b\right )} \sin \left (d x^{3} + c\right )}{12 \, d^{2}} \] Input:

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="fricas")
 

Output:

1/12*((2*a^2 + b^2)*d^2*x^6 - 8*a*b*d*x^3*cos(d*x^3 + c) - b^2*cos(d*x^3 + 
 c)^2 - 2*(b^2*d*x^3*cos(d*x^3 + c) - 4*a*b)*sin(d*x^3 + c))/d^2
 

Sympy [A] (verification not implemented)

Time = 0.53 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.34 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{6}}{6} - \frac {2 a b x^{3} \cos {\left (c + d x^{3} \right )}}{3 d} + \frac {2 a b \sin {\left (c + d x^{3} \right )}}{3 d^{2}} + \frac {b^{2} x^{6} \sin ^{2}{\left (c + d x^{3} \right )}}{12} + \frac {b^{2} x^{6} \cos ^{2}{\left (c + d x^{3} \right )}}{12} - \frac {b^{2} x^{3} \sin {\left (c + d x^{3} \right )} \cos {\left (c + d x^{3} \right )}}{6 d} - \frac {b^{2} \cos ^{2}{\left (c + d x^{3} \right )}}{12 d^{2}} & \text {for}\: d \neq 0 \\\frac {x^{6} \left (a + b \sin {\left (c \right )}\right )^{2}}{6} & \text {otherwise} \end {cases} \] Input:

integrate(x**5*(a+b*sin(d*x**3+c))**2,x)
 

Output:

Piecewise((a**2*x**6/6 - 2*a*b*x**3*cos(c + d*x**3)/(3*d) + 2*a*b*sin(c + 
d*x**3)/(3*d**2) + b**2*x**6*sin(c + d*x**3)**2/12 + b**2*x**6*cos(c + d*x 
**3)**2/12 - b**2*x**3*sin(c + d*x**3)*cos(c + d*x**3)/(6*d) - b**2*cos(c 
+ d*x**3)**2/(12*d**2), Ne(d, 0)), (x**6*(a + b*sin(c))**2/6, True))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.81 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {1}{6} \, a^{2} x^{6} - \frac {2 \, {\left (d x^{3} \cos \left (d x^{3} + c\right ) - \sin \left (d x^{3} + c\right )\right )} a b}{3 \, d^{2}} + \frac {{\left (2 \, d^{2} x^{6} - 2 \, d x^{3} \sin \left (2 \, d x^{3} + 2 \, c\right ) - \cos \left (2 \, d x^{3} + 2 \, c\right )\right )} b^{2}}{24 \, d^{2}} \] Input:

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="maxima")
 

Output:

1/6*a^2*x^6 - 2/3*(d*x^3*cos(d*x^3 + c) - sin(d*x^3 + c))*a*b/d^2 + 1/24*( 
2*d^2*x^6 - 2*d*x^3*sin(2*d*x^3 + 2*c) - cos(2*d*x^3 + 2*c))*b^2/d^2
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.54 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {4 \, {\left (d x^{3} + c\right )}^{2} a^{2} + 2 \, {\left (d x^{3} + c\right )}^{2} b^{2} - 16 \, {\left (d x^{3} + c\right )} a b \cos \left (d x^{3} + c\right ) - 2 \, {\left (d x^{3} + c\right )} b^{2} \sin \left (2 \, d x^{3} + 2 \, c\right ) - b^{2} \cos \left (2 \, d x^{3} + 2 \, c\right ) + 16 \, a b \sin \left (d x^{3} + c\right )}{24 \, d^{2}} - \frac {4 \, {\left (d x^{3} + c\right )} a^{2} c + {\left (2 \, d x^{3} + 2 \, c - \sin \left (2 \, d x^{3} + 2 \, c\right )\right )} b^{2} c - 8 \, a b c \cos \left (d x^{3} + c\right )}{12 \, d^{2}} \] Input:

integrate(x^5*(a+b*sin(d*x^3+c))^2,x, algorithm="giac")
 

Output:

1/24*(4*(d*x^3 + c)^2*a^2 + 2*(d*x^3 + c)^2*b^2 - 16*(d*x^3 + c)*a*b*cos(d 
*x^3 + c) - 2*(d*x^3 + c)*b^2*sin(2*d*x^3 + 2*c) - b^2*cos(2*d*x^3 + 2*c) 
+ 16*a*b*sin(d*x^3 + c))/d^2 - 1/12*(4*(d*x^3 + c)*a^2*c + (2*d*x^3 + 2*c 
- sin(2*d*x^3 + 2*c))*b^2*c - 8*a*b*c*cos(d*x^3 + c))/d^2
 

Mupad [B] (verification not implemented)

Time = 39.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=-\frac {b^2\,{\cos \left (d\,x^3+c\right )}^2-2\,a^2\,d^2\,x^6-b^2\,d^2\,x^6-8\,a\,b\,\sin \left (d\,x^3+c\right )+8\,a\,b\,d\,x^3\,\cos \left (d\,x^3+c\right )+2\,b^2\,d\,x^3\,\cos \left (d\,x^3+c\right )\,\sin \left (d\,x^3+c\right )}{12\,d^2} \] Input:

int(x^5*(a + b*sin(c + d*x^3))^2,x)
 

Output:

-(b^2*cos(c + d*x^3)^2 - 2*a^2*d^2*x^6 - b^2*d^2*x^6 - 8*a*b*sin(c + d*x^3 
) + 8*a*b*d*x^3*cos(c + d*x^3) + 2*b^2*d*x^3*cos(c + d*x^3)*sin(c + d*x^3) 
)/(12*d^2)
 

Reduce [F]

\[ \int x^5 \left (a+b \sin \left (c+d x^3\right )\right )^2 \, dx=\frac {-4 \cos \left (d \,x^{3}+c \right ) a b d \,x^{3}+6 \left (\int \sin \left (d \,x^{3}+c \right )^{2} x^{5}d x \right ) b^{2} d^{2}+4 \sin \left (d \,x^{3}+c \right ) a b +a^{2} d^{2} x^{6}}{6 d^{2}} \] Input:

int(x^5*(a+b*sin(d*x^3+c))^2,x)
 

Output:

( - 4*cos(c + d*x**3)*a*b*d*x**3 + 6*int(sin(c + d*x**3)**2*x**5,x)*b**2*d 
**2 + 4*sin(c + d*x**3)*a*b + a**2*d**2*x**6)/(6*d**2)