3.1.22 \(\int x \cos ^7(a+b x^2) \, dx\) [22]

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

[Out]

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

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Rubi [A]
time = 0.03, 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 = {3461, 2713} \begin {gather*} -\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} \end {gather*}

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 2713

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 3461

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]))

Rubi steps

\begin {align*} \int x \cos ^7\left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int \cos ^7(a+b x) \, dx,x,x^2\right )\\ &=-\frac {\text {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*}

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

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x^2]^7,x]

[Out]

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

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Maple [A]
time = 0.08, size = 50, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {\left (\frac {16}{5}+\cos ^{6}\left (b \,x^{2}+a \right )+\frac {6 \left (\cos ^{4}\left (b \,x^{2}+a \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (b \,x^{2}+a \right )\right )}{5}\right ) \sin \left (b \,x^{2}+a \right )}{14 b}\)	\(50\)
default	\(\frac {\left (\frac {16}{5}+\cos ^{6}\left (b \,x^{2}+a \right )+\frac {6 \left (\cos ^{4}\left (b \,x^{2}+a \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (b \,x^{2}+a \right )\right )}{5}\right ) \sin \left (b \,x^{2}+a \right )}{14 b}\)	\(50\)
risch	\(\frac {35 \sin \left (b \,x^{2}+a \right )}{128 b}+\frac {\sin \left (7 b \,x^{2}+7 a \right )}{896 b}+\frac {7 \sin \left (5 b \,x^{2}+5 a \right )}{640 b}+\frac {7 \sin \left (3 b \,x^{2}+3 a \right )}{128 b}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.30, size = 55, normalized size = 0.82 \begin {gather*} \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 {gather*}

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 = 0.38, size = 51, normalized size = 0.76 \begin {gather*} \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 {gather*}

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 = 0.90, size = 94, normalized size = 1.40 \begin {gather*} \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 {gather*}

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 = 0.43, size = 52, normalized size = 0.78 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.75, size = 55, normalized size = 0.82 \begin {gather*} \frac {245\,\sin \left (3\,b\,x^2+3\,a\right )+49\,\sin \left (5\,b\,x^2+5\,a\right )+5\,\sin \left (7\,b\,x^2+7\,a\right )+1225\,\sin \left (b\,x^2+a\right )}{4480\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x^2)^7,x)

[Out]

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

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