3.33 \(\int x \sin ^7(a+b x^2) \, dx\)

Optimal. Leaf size=67 \[ \frac {\cos ^7\left (a+b x^2\right )}{14 b}-\frac {3 \cos ^5\left (a+b x^2\right )}{10 b}+\frac {\cos ^3\left (a+b x^2\right )}{2 b}-\frac {\cos \left (a+b x^2\right )}{2 b} \]

[Out]

-1/2*cos(b*x^2+a)/b+1/2*cos(b*x^2+a)^3/b-3/10*cos(b*x^2+a)^5/b+1/14*cos(b*x^2+a)^7/b

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3379, 2633} \[ \frac {\cos ^7\left (a+b x^2\right )}{14 b}-\frac {3 \cos ^5\left (a+b x^2\right )}{10 b}+\frac {\cos ^3\left (a+b x^2\right )}{2 b}-\frac {\cos \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*x^2]^7,x]

[Out]

-Cos[a + b*x^2]/(2*b) + Cos[a + b*x^2]^3/(2*b) - (3*Cos[a + b*x^2]^5)/(10*b) + Cos[a + b*x^2]^7/(14*b)

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 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 \sin ^7\left (a+b x^2\right ) \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \sin ^7(a+b x) \, dx,x,x^2\right )\\ &=-\frac {\operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cos \left (a+b x^2\right )\right )}{2 b}\\ &=-\frac {\cos \left (a+b x^2\right )}{2 b}+\frac {\cos ^3\left (a+b x^2\right )}{2 b}-\frac {3 \cos ^5\left (a+b x^2\right )}{10 b}+\frac {\cos ^7\left (a+b x^2\right )}{14 b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 67, normalized size = 1.00 \[ -\frac {35 \cos \left (a+b x^2\right )}{128 b}+\frac {7 \cos \left (3 \left (a+b x^2\right )\right )}{128 b}-\frac {7 \cos \left (5 \left (a+b x^2\right )\right )}{640 b}+\frac {\cos \left (7 \left (a+b x^2\right )\right )}{896 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*x^2]^7,x]

[Out]

(-35*Cos[a + b*x^2])/(128*b) + (7*Cos[3*(a + b*x^2)])/(128*b) - (7*Cos[5*(a + b*x^2)])/(640*b) + Cos[7*(a + b*
x^2)]/(896*b)

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fricas [A]  time = 0.57, size = 52, normalized size = 0.78 \[ \frac {5 \, \cos \left (b x^{2} + a\right )^{7} - 21 \, \cos \left (b x^{2} + a\right )^{5} + 35 \, \cos \left (b x^{2} + a\right )^{3} - 35 \, \cos \left (b x^{2} + a\right )}{70 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x^2+a)^7,x, algorithm="fricas")

[Out]

1/70*(5*cos(b*x^2 + a)^7 - 21*cos(b*x^2 + a)^5 + 35*cos(b*x^2 + a)^3 - 35*cos(b*x^2 + a))/b

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giac [A]  time = 0.51, size = 52, normalized size = 0.78 \[ \frac {5 \, \cos \left (b x^{2} + a\right )^{7} - 21 \, \cos \left (b x^{2} + a\right )^{5} + 35 \, \cos \left (b x^{2} + a\right )^{3} - 35 \, \cos \left (b x^{2} + a\right )}{70 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x^2+a)^7,x, algorithm="giac")

[Out]

1/70*(5*cos(b*x^2 + a)^7 - 21*cos(b*x^2 + a)^5 + 35*cos(b*x^2 + a)^3 - 35*cos(b*x^2 + a))/b

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maple [A]  time = 0.08, size = 50, normalized size = 0.75 \[ -\frac {\left (\frac {16}{5}+\sin ^{6}\left (b \,x^{2}+a \right )+\frac {6 \left (\sin ^{4}\left (b \,x^{2}+a \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (b \,x^{2}+a \right )\right )}{5}\right ) \cos \left (b \,x^{2}+a \right )}{14 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(b*x^2+a)^7,x)

[Out]

-1/14/b*(16/5+sin(b*x^2+a)^6+6/5*sin(b*x^2+a)^4+8/5*sin(b*x^2+a)^2)*cos(b*x^2+a)

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maxima [A]  time = 0.32, size = 55, normalized size = 0.82 \[ \frac {5 \, \cos \left (7 \, b x^{2} + 7 \, a\right ) - 49 \, \cos \left (5 \, b x^{2} + 5 \, a\right ) + 245 \, \cos \left (3 \, b x^{2} + 3 \, a\right ) - 1225 \, \cos \left (b x^{2} + a\right )}{4480 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x^2+a)^7,x, algorithm="maxima")

[Out]

1/4480*(5*cos(7*b*x^2 + 7*a) - 49*cos(5*b*x^2 + 5*a) + 245*cos(3*b*x^2 + 3*a) - 1225*cos(b*x^2 + a))/b

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mupad [B]  time = 4.99, size = 55, normalized size = 0.82 \[ \frac {245\,\cos \left (3\,b\,x^2+3\,a\right )-49\,\cos \left (5\,b\,x^2+5\,a\right )+5\,\cos \left (7\,b\,x^2+7\,a\right )-1225\,\cos \left (b\,x^2+a\right )}{4480\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*x^2)^7,x)

[Out]

(245*cos(3*a + 3*b*x^2) - 49*cos(5*a + 5*b*x^2) + 5*cos(7*a + 7*b*x^2) - 1225*cos(a + b*x^2))/(4480*b)

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sympy [A]  time = 9.99, size = 95, normalized size = 1.42 \[ \begin {cases} - \frac {\sin ^{6}{\left (a + b x^{2} \right )} \cos {\left (a + b x^{2} \right )}}{2 b} - \frac {\sin ^{4}{\left (a + b x^{2} \right )} \cos ^{3}{\left (a + b x^{2} \right )}}{b} - \frac {4 \sin ^{2}{\left (a + b x^{2} \right )} \cos ^{5}{\left (a + b x^{2} \right )}}{5 b} - \frac {8 \cos ^{7}{\left (a + b x^{2} \right )}}{35 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \sin ^{7}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(b*x**2+a)**7,x)

[Out]

Piecewise((-sin(a + b*x**2)**6*cos(a + b*x**2)/(2*b) - sin(a + b*x**2)**4*cos(a + b*x**2)**3/b - 4*sin(a + b*x
**2)**2*cos(a + b*x**2)**5/(5*b) - 8*cos(a + b*x**2)**7/(35*b), Ne(b, 0)), (x**2*sin(a)**7/2, True))

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