\(\int x^5 \sin ^3(a+b x^2) \, dx\) [23]

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\frac {7 \cos \left (a+b x^2\right )}{9 b^3}-\frac {x^4 \cos \left (a+b x^2\right )}{3 b}-\frac {\cos ^3\left (a+b x^2\right )}{27 b^3}+\frac {2 x^2 \sin \left (a+b x^2\right )}{3 b^2}-\frac {x^4 \cos \left (a+b x^2\right ) \sin ^2\left (a+b x^2\right )}{6 b}+\frac {x^2 \sin ^3\left (a+b x^2\right )}{9 b^2} \] Output:

7/9*cos(b*x^2+a)/b^3-1/3*x^4*cos(b*x^2+a)/b-1/27*cos(b*x^2+a)^3/b^3+2/3*x^ 
2*sin(b*x^2+a)/b^2-1/6*x^4*cos(b*x^2+a)*sin(b*x^2+a)^2/b+1/9*x^2*sin(b*x^2 
+a)^3/b^2
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\frac {-81 \left (-2+b^2 x^4\right ) \cos \left (a+b x^2\right )+\left (-2+9 b^2 x^4\right ) \cos \left (3 \left (a+b x^2\right )\right )-6 b x^2 \left (-27 \sin \left (a+b x^2\right )+\sin \left (3 \left (a+b x^2\right )\right )\right )}{216 b^3} \] Input:

Integrate[x^5*Sin[a + b*x^2]^3,x]
 

Output:

(-81*(-2 + b^2*x^4)*Cos[a + b*x^2] + (-2 + 9*b^2*x^4)*Cos[3*(a + b*x^2)] - 
 6*b*x^2*(-27*Sin[a + b*x^2] + Sin[3*(a + b*x^2)]))/(216*b^3)
 

Rubi [A] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3860, 3042, 3792, 3042, 3113, 2009, 3777, 3042, 3777, 25, 3042, 3118}

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.

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

\(\Big \downarrow \) 3860

\(\displaystyle \frac {1}{2} \int x^4 \sin ^3\left (b x^2+a\right )dx^2\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int x^4 \sin \left (b x^2+a\right )^3dx^2\)

\(\Big \downarrow \) 3792

\(\displaystyle \frac {1}{2} \left (-\frac {2 \int \sin ^3\left (b x^2+a\right )dx^2}{9 b^2}+\frac {2}{3} \int x^4 \sin \left (b x^2+a\right )dx^2+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (-\frac {2 \int \sin \left (b x^2+a\right )^3dx^2}{9 b^2}+\frac {2}{3} \int x^4 \sin \left (b x^2+a\right )dx^2+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3113

\(\displaystyle \frac {1}{2} \left (\frac {2 \int \left (1-x^4\right )d\cos \left (b x^2+a\right )}{9 b^3}+\frac {2}{3} \int x^4 \sin \left (b x^2+a\right )dx^2+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \int x^4 \sin \left (b x^2+a\right )dx^2+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {2 \int x^2 \cos \left (b x^2+a\right )dx^2}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {2 \int x^2 \sin \left (b x^2+a+\frac {\pi }{2}\right )dx^2}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3777

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {2 \left (\frac {\int -\sin \left (b x^2+a\right )dx^2}{b}+\frac {x^2 \sin \left (a+b x^2\right )}{b}\right )}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {2 \left (\frac {x^2 \sin \left (a+b x^2\right )}{b}-\frac {\int \sin \left (b x^2+a\right )dx^2}{b}\right )}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {2}{3} \left (\frac {2 \left (\frac {x^2 \sin \left (a+b x^2\right )}{b}-\frac {\int \sin \left (b x^2+a\right )dx^2}{b}\right )}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )+\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

\(\Big \downarrow \) 3118

\(\displaystyle \frac {1}{2} \left (\frac {2 \left (\cos \left (a+b x^2\right )-\frac {x^6}{3}\right )}{9 b^3}+\frac {2 x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}+\frac {2}{3} \left (\frac {2 \left (\frac {\cos \left (a+b x^2\right )}{b^2}+\frac {x^2 \sin \left (a+b x^2\right )}{b}\right )}{b}-\frac {x^4 \cos \left (a+b x^2\right )}{b}\right )-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{3 b}\right )\)

Input:

Int[x^5*Sin[a + b*x^2]^3,x]
 

Output:

((2*(-1/3*x^6 + Cos[a + b*x^2]))/(9*b^3) - (x^4*Cos[a + b*x^2]*Sin[a + b*x 
^2]^2)/(3*b) + (2*x^2*Sin[a + b*x^2]^3)/(9*b^2) + (2*(-((x^4*Cos[a + b*x^2 
])/b) + (2*(Cos[a + b*x^2]/b^2 + (x^2*Sin[a + b*x^2])/b))/b))/3)/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3113

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] 
 && IGtQ[(n - 1)/2, 0]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3792

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbo 
l] :> Simp[d*m*(c + d*x)^(m - 1)*((b*Sin[e + f*x])^n/(f^2*n^2)), x] + (-Sim 
p[b*(c + d*x)^m*Cos[e + f*x]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x] + Simp[b^ 
2*((n - 1)/n)   Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[d^2 
*m*((m - 1)/(f^2*n^2))   Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x]) 
/; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]
 

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 = 1.38 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73

method	result	size
risch	\(-\frac {3 \left (b^{2} x^{4}-2\right ) \cos \left (b \,x^{2}+a \right )}{8 b^{3}}+\frac {3 x^{2} \sin \left (b \,x^{2}+a \right )}{4 b^{2}}+\frac {\left (9 b^{2} x^{4}-2\right ) \cos \left (3 b \,x^{2}+3 a \right )}{216 b^{3}}-\frac {x^{2} \sin \left (3 b \,x^{2}+3 a \right )}{36 b^{2}}\)	\(85\)
default	\(-\frac {3 x^{4} \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {\frac {3 x^{2} \sin \left (b \,x^{2}+a \right )}{4 b}+\frac {3 \cos \left (b \,x^{2}+a \right )}{4 b^{2}}}{b}+\frac {x^{4} \cos \left (3 b \,x^{2}+3 a \right )}{24 b}-\frac {\frac {x^{2} \sin \left (3 b \,x^{2}+3 a \right )}{6 b}+\frac {\cos \left (3 b \,x^{2}+3 a \right )}{18 b^{2}}}{6 b}\)	\(113\)
orering	\(\frac {5 \left (648 b^{4} x^{8}+91 b^{2} x^{4}-1260\right ) \sin \left (b \,x^{2}+a \right )^{3}}{1296 b^{6} x^{6}}-\frac {5 \left (72 b^{4} x^{8}+155 b^{2} x^{4}-492\right ) \left (5 x^{4} \sin \left (b \,x^{2}+a \right )^{3}+6 x^{6} \sin \left (b \,x^{2}+a \right )^{2} b \cos \left (b \,x^{2}+a \right )\right )}{1296 b^{6} x^{10}}+\frac {\left (11 b^{2} x^{4}-20\right ) \left (20 x^{3} \sin \left (b \,x^{2}+a \right )^{3}+66 x^{5} \sin \left (b \,x^{2}+a \right )^{2} b \cos \left (b \,x^{2}+a \right )+24 x^{7} \sin \left (b \,x^{2}+a \right ) b^{2} \cos \left (b \,x^{2}+a \right )^{2}-12 x^{7} \sin \left (b \,x^{2}+a \right )^{3} b^{2}\right )}{72 b^{6} x^{9}}-\frac {\left (9 b^{2} x^{4}-20\right ) \left (60 x^{2} \sin \left (b \,x^{2}+a \right )^{3}+450 x^{4} \sin \left (b \,x^{2}+a \right )^{2} b \cos \left (b \,x^{2}+a \right )+432 x^{6} \sin \left (b \,x^{2}+a \right ) b^{2} \cos \left (b \,x^{2}+a \right )^{2}-216 x^{6} \sin \left (b \,x^{2}+a \right )^{3} b^{2}+48 x^{8} b^{3} \cos \left (b \,x^{2}+a \right )^{3}-168 x^{8} \sin \left (b \,x^{2}+a \right )^{2} b^{3} \cos \left (b \,x^{2}+a \right )\right )}{1296 b^{6} x^{8}}\)	\(352\)

Input:

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

Output:

-3/8*(b^2*x^4-2)/b^3*cos(b*x^2+a)+3/4*x^2*sin(b*x^2+a)/b^2+1/216*(9*b^2*x^ 
4-2)/b^3*cos(3*b*x^2+3*a)-1/36*x^2/b^2*sin(3*b*x^2+3*a)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\frac {{\left (9 \, b^{2} x^{4} - 2\right )} \cos \left (b x^{2} + a\right )^{3} - 3 \, {\left (9 \, b^{2} x^{4} - 14\right )} \cos \left (b x^{2} + a\right ) - 6 \, {\left (b x^{2} \cos \left (b x^{2} + a\right )^{2} - 7 \, b x^{2}\right )} \sin \left (b x^{2} + a\right )}{54 \, b^{3}} \] Input:

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

Output:

1/54*((9*b^2*x^4 - 2)*cos(b*x^2 + a)^3 - 3*(9*b^2*x^4 - 14)*cos(b*x^2 + a) 
 - 6*(b*x^2*cos(b*x^2 + a)^2 - 7*b*x^2)*sin(b*x^2 + a))/b^3
 

Sympy [A] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.22 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\begin {cases} - \frac {x^{4} \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{2 b} - \frac {x^{4} \cos ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {7 x^{2} \sin ^{3}{\left (a + b x^{2} \right )}}{9 b^{2}} + \frac {2 x^{2} \sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{3 b^{2}} + \frac {7 \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{9 b^{3}} + \frac {20 \cos ^{3}{\left (a + b x^{2} \right )}}{27 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{6} \sin ^{3}{\left (a \right )}}{6} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-x**4*sin(a + b*x**2)**2*cos(a + b*x**2)/(2*b) - x**4*cos(a + b 
*x**2)**3/(3*b) + 7*x**2*sin(a + b*x**2)**3/(9*b**2) + 2*x**2*sin(a + b*x* 
*2)*cos(a + b*x**2)**2/(3*b**2) + 7*sin(a + b*x**2)**2*cos(a + b*x**2)/(9* 
b**3) + 20*cos(a + b*x**2)**3/(27*b**3), Ne(b, 0)), (x**6*sin(a)**3/6, Tru 
e))
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=-\frac {6 \, b x^{2} \sin \left (3 \, b x^{2} + 3 \, a\right ) - 162 \, b x^{2} \sin \left (b x^{2} + a\right ) - {\left (9 \, b^{2} x^{4} - 2\right )} \cos \left (3 \, b x^{2} + 3 \, a\right ) + 81 \, {\left (b^{2} x^{4} - 2\right )} \cos \left (b x^{2} + a\right )}{216 \, b^{3}} \] Input:

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

Output:

-1/216*(6*b*x^2*sin(3*b*x^2 + 3*a) - 162*b*x^2*sin(b*x^2 + a) - (9*b^2*x^4 
 - 2)*cos(3*b*x^2 + 3*a) + 81*(b^2*x^4 - 2)*cos(b*x^2 + a))/b^3
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.18 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=-\frac {x^{2} \sin \left (3 \, b x^{2} + 3 \, a\right )}{36 \, b^{2}} + \frac {3 \, x^{2} \sin \left (b x^{2} + a\right )}{4 \, b^{2}} + \frac {{\left (\cos \left (b x^{2} + a\right )^{3} - 3 \, \cos \left (b x^{2} + a\right )\right )} a^{2}}{6 \, b^{3}} + \frac {{\left (9 \, {\left (b x^{2} + a\right )}^{2} - 18 \, {\left (b x^{2} + a\right )} a - 2\right )} \cos \left (3 \, b x^{2} + 3 \, a\right )}{216 \, b^{3}} - \frac {3 \, {\left ({\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a - 2\right )} \cos \left (b x^{2} + a\right )}{8 \, b^{3}} \] Input:

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

Output:

-1/36*x^2*sin(3*b*x^2 + 3*a)/b^2 + 3/4*x^2*sin(b*x^2 + a)/b^2 + 1/6*(cos(b 
*x^2 + a)^3 - 3*cos(b*x^2 + a))*a^2/b^3 + 1/216*(9*(b*x^2 + a)^2 - 18*(b*x 
^2 + a)*a - 2)*cos(3*b*x^2 + 3*a)/b^3 - 3/8*((b*x^2 + a)^2 - 2*(b*x^2 + a) 
*a - 2)*cos(b*x^2 + a)/b^3
 

Mupad [B] (verification not implemented)

Time = 38.98 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.80 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\frac {\frac {3\,\cos \left (b\,x^2+a\right )}{4}-\frac {\cos \left (3\,b\,x^2+3\,a\right )}{108}+b\,\left (\frac {3\,x^2\,\sin \left (b\,x^2+a\right )}{4}-\frac {x^2\,\sin \left (3\,b\,x^2+3\,a\right )}{36}\right )+b^2\,\left (\frac {x^4\,\cos \left (3\,b\,x^2+3\,a\right )}{24}-\frac {3\,x^4\,\cos \left (b\,x^2+a\right )}{8}\right )}{b^3} \] Input:

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

Output:

((3*cos(a + b*x^2))/4 - cos(3*a + 3*b*x^2)/108 + b*((3*x^2*sin(a + b*x^2)) 
/4 - (x^2*sin(3*a + 3*b*x^2))/36) + b^2*((x^4*cos(3*a + 3*b*x^2))/24 - (3* 
x^4*cos(a + b*x^2))/8))/b^3
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int x^5 \sin ^3\left (a+b x^2\right ) \, dx=\frac {-9 \cos \left (b \,x^{2}+a \right ) \sin \left (b \,x^{2}+a \right )^{2} b^{2} x^{4}+2 \cos \left (b \,x^{2}+a \right ) \sin \left (b \,x^{2}+a \right )^{2}-18 \cos \left (b \,x^{2}+a \right ) b^{2} x^{4}+40 \cos \left (b \,x^{2}+a \right )+6 \sin \left (b \,x^{2}+a \right )^{3} b \,x^{2}+36 \sin \left (b \,x^{2}+a \right ) b \,x^{2}+16}{54 b^{3}} \] Input:

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

Output:

( - 9*cos(a + b*x**2)*sin(a + b*x**2)**2*b**2*x**4 + 2*cos(a + b*x**2)*sin 
(a + b*x**2)**2 - 18*cos(a + b*x**2)*b**2*x**4 + 40*cos(a + b*x**2) + 6*si 
n(a + b*x**2)**3*b*x**2 + 36*sin(a + b*x**2)*b*x**2 + 16)/(54*b**3)