3.22 \(\int x \cos ^7(a+b x^2) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0537244, antiderivative size = 67, normalized size of antiderivative = 1., 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 = {3380, 2633} \[ -\frac{\sin ^7\left (a+b x^2\right )}{14 b}+\frac{3 \sin ^5\left (a+b x^2\right )}{10 b}-\frac{\sin ^3\left (a+b x^2\right )}{2 b}+\frac{\sin \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x^2]^7,x]

[Out]

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

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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]))

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]

Rubi steps

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

Mathematica [A]  time = 0.0878097, size = 54, normalized size = 0.81 \[ \frac{-5 \sin ^7\left (a+b x^2\right )+21 \sin ^5\left (a+b x^2\right )-35 \sin ^3\left (a+b x^2\right )+35 \sin \left (a+b x^2\right )}{70 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x^2]^7,x]

[Out]

(35*Sin[a + b*x^2] - 35*Sin[a + b*x^2]^3 + 21*Sin[a + b*x^2]^5 - 5*Sin[a + b*x^2]^7)/(70*b)

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Maple [A]  time = 0.028, size = 50, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( b{x}^{2}+a \right ) }{14\,b} \left ({\frac{16}{5}}+ \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17826, size = 74, normalized size = 1.1 \begin{align*} \frac{5 \, \sin \left (7 \, b x^{2} + 7 \, a\right ) + 49 \, \sin \left (5 \, b x^{2} + 5 \, a\right ) + 245 \, \sin \left (3 \, b x^{2} + 3 \, a\right ) + 1225 \, \sin \left (b x^{2} + a\right )}{4480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.70206, size = 123, normalized size = 1.84 \begin{align*} \frac{{\left (5 \, \cos \left (b x^{2} + a\right )^{6} + 6 \, \cos \left (b x^{2} + a\right )^{4} + 8 \, \cos \left (b x^{2} + a\right )^{2} + 16\right )} \sin \left (b x^{2} + a\right )}{70 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/70*(5*cos(b*x^2 + a)^6 + 6*cos(b*x^2 + a)^4 + 8*cos(b*x^2 + a)^2 + 16)*sin(b*x^2 + a)/b

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Sympy [A]  time = 17.4554, size = 94, normalized size = 1.4 \begin{align*} \begin{cases} \frac{8 \sin ^{7}{\left (a + b x^{2} \right )}}{35 b} + \frac{4 \sin ^{5}{\left (a + b x^{2} \right )} \cos ^{2}{\left (a + b x^{2} \right )}}{5 b} + \frac{\sin ^{3}{\left (a + b x^{2} \right )} \cos ^{4}{\left (a + b x^{2} \right )}}{b} + \frac{\sin{\left (a + b x^{2} \right )} \cos ^{6}{\left (a + b x^{2} \right )}}{2 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \cos ^{7}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27403, size = 70, normalized size = 1.04 \begin{align*} -\frac{5 \, \sin \left (b x^{2} + a\right )^{7} - 21 \, \sin \left (b x^{2} + a\right )^{5} + 35 \, \sin \left (b x^{2} + a\right )^{3} - 35 \, \sin \left (b x^{2} + a\right )}{70 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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