3.23 \(\int x^5 \sin ^3(a+b x^2) \, dx\)

Optimal. Leaf size=117 \[ -\frac {\cos ^3\left (a+b x^2\right )}{27 b^3}+\frac {7 \cos \left (a+b x^2\right )}{9 b^3}+\frac {x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}+\frac {2 x^2 \sin \left (a+b x^2\right )}{3 b^2}-\frac {x^4 \cos \left (a+b x^2\right )}{3 b}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3379, 3311, 3296, 2638, 2633} \[ \frac {x^2 \sin ^3\left (a+b x^2\right )}{9 b^2}+\frac {2 x^2 \sin \left (a+b x^2\right )}{3 b^2}-\frac {\cos ^3\left (a+b x^2\right )}{27 b^3}+\frac {7 \cos \left (a+b x^2\right )}{9 b^3}-\frac {x^4 \cos \left (a+b x^2\right )}{3 b}-\frac {x^4 \sin ^2\left (a+b x^2\right ) \cos \left (a+b x^2\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sin[a + b*x^2]^3,x]

[Out]

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

Rule 2633

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

Rule 2638

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

Rule 3296

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

Rule 3311

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

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rubi steps

\begin {align*} \int x^5 \sin ^3\left (a+b x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^2 \sin ^3(a+b x) \, dx,x,x^2\right )\\ &=-\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}+\frac {1}{3} \operatorname {Subst}\left (\int x^2 \sin (a+b x) \, dx,x,x^2\right )-\frac {\operatorname {Subst}\left (\int \sin ^3(a+b x) \, dx,x,x^2\right )}{9 b^2}\\ &=-\frac {x^4 \cos \left (a+b x^2\right )}{3 b}-\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}+\frac {\operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos \left (a+b x^2\right )\right )}{9 b^3}+\frac {2 \operatorname {Subst}\left (\int x \cos (a+b x) \, dx,x,x^2\right )}{3 b}\\ &=\frac {\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}-\frac {2 \operatorname {Subst}\left (\int \sin (a+b x) \, dx,x,x^2\right )}{3 b^2}\\ &=\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}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 75, normalized size = 0.64 \[ \frac {-81 \left (b^2 x^4-2\right ) \cos \left (a+b x^2\right )+\left (9 b^2 x^4-2\right ) \cos \left (3 \left (a+b x^2\right )\right )-6 b x^2 \left (\sin \left (3 \left (a+b x^2\right )\right )-27 \sin \left (a+b x^2\right )\right )}{216 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sin[a + b*x^2]^3,x]

[Out]

(-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)

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fricas [A]  time = 0.60, size = 79, normalized size = 0.68 \[ \frac {{\left (9 \, b^{2} x^{4} - 2\right )} \cos \left (b x^{2} + a\right )^{3} - 3 \, {\left (9 \, b^{2} x^{4} - 14\right )} \cos \left (b x^{2} + a\right ) - 6 \, {\left (b x^{2} \cos \left (b x^{2} + a\right )^{2} - 7 \, b x^{2}\right )} \sin \left (b x^{2} + a\right )}{54 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 122, normalized size = 1.04 \[ -\frac {\frac {6 \, x^{2} \sin \left (3 \, b x^{2} + 3 \, a\right )}{b} - \frac {162 \, x^{2} \sin \left (b x^{2} + a\right )}{b} - \frac {{\left (9 \, {\left (b x^{2} + a\right )}^{2} - 18 \, {\left (b x^{2} + a\right )} a + 9 \, a^{2} - 2\right )} \cos \left (3 \, b x^{2} + 3 \, a\right )}{b^{2}} + \frac {81 \, {\left ({\left (b x^{2} + a\right )}^{2} - 2 \, {\left (b x^{2} + a\right )} a + a^{2} - 2\right )} \cos \left (b x^{2} + a\right )}{b^{2}}}{216 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/216*(6*x^2*sin(3*b*x^2 + 3*a)/b - 162*x^2*sin(b*x^2 + a)/b - (9*(b*x^2 + a)^2 - 18*(b*x^2 + a)*a + 9*a^2 -
2)*cos(3*b*x^2 + 3*a)/b^2 + 81*((b*x^2 + a)^2 - 2*(b*x^2 + a)*a + a^2 - 2)*cos(b*x^2 + a)/b^2)/b

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maple [A]  time = 0.03, size = 113, normalized size = 0.97 \[ -\frac {3 x^{4} \cos \left (b \,x^{2}+a \right )}{8 b}+\frac {\frac {3 x^{2} \sin \left (b \,x^{2}+a \right )}{4 b}+\frac {3 \cos \left (b \,x^{2}+a \right )}{4 b^{2}}}{b}+\frac {x^{4} \cos \left (3 b \,x^{2}+3 a \right )}{24 b}-\frac {\frac {x^{2} \sin \left (3 b \,x^{2}+3 a \right )}{6 b}+\frac {\cos \left (3 b \,x^{2}+3 a \right )}{18 b^{2}}}{6 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*x^4*cos(b*x^2+a)/b+3/2/b*(1/2/b*x^2*sin(b*x^2+a)+1/2/b^2*cos(b*x^2+a))+1/24/b*x^4*cos(3*b*x^2+3*a)-1/6/b*
(1/6/b*x^2*sin(3*b*x^2+3*a)+1/18/b^2*cos(3*b*x^2+3*a))

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maxima [A]  time = 0.35, size = 79, normalized size = 0.68 \[ -\frac {6 \, b x^{2} \sin \left (3 \, b x^{2} + 3 \, a\right ) - 162 \, b x^{2} \sin \left (b x^{2} + a\right ) - {\left (9 \, b^{2} x^{4} - 2\right )} \cos \left (3 \, b x^{2} + 3 \, a\right ) + 81 \, {\left (b^{2} x^{4} - 2\right )} \cos \left (b x^{2} + a\right )}{216 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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mupad [B]  time = 4.97, size = 94, normalized size = 0.80 \[ \frac {\frac {3\,\cos \left (b\,x^2+a\right )}{4}-\frac {\cos \left (3\,b\,x^2+3\,a\right )}{108}+b\,\left (\frac {3\,x^2\,\sin \left (b\,x^2+a\right )}{4}-\frac {x^2\,\sin \left (3\,b\,x^2+3\,a\right )}{36}\right )+b^2\,\left (\frac {x^4\,\cos \left (3\,b\,x^2+3\,a\right )}{24}-\frac {3\,x^4\,\cos \left (b\,x^2+a\right )}{8}\right )}{b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*sin(a + b*x^2)^3,x)

[Out]

((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

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sympy [A]  time = 10.33, size = 143, normalized size = 1.22 \[ \begin {cases} - \frac {x^{4} \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{2 b} - \frac {x^{4} \cos ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {7 x^{2} \sin ^{3}{\left (a + b x^{2} \right )}}{9 b^{2}} + \frac {2 x^{2} \sin {\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{3 b^{2}} + \frac {7 \sin ^{2}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{9 b^{3}} + \frac {20 \cos ^{3}{\left (a + b x^{2} \right )}}{27 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{6} \sin ^{3}{\relax (a )}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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, True))

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