3.97 \(\int \cos (a+b \sqrt [3]{c+d x}) \, dx\)

Optimal. Leaf size=85 \[ -\frac{6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac{6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

[Out]

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

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Rubi [A]  time = 0.0551264, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3362, 3296, 2637} \[ -\frac{6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac{6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*(c + d*x)^(1/3)],x]

[Out]

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

Rule 3362

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Cos[c + d*x])^p, x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && In
tegerQ[1/n]

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos \left (a+b \sqrt [3]{c+d x}\right ) \, dx &=\frac{3 \operatorname{Subst}\left (\int x^2 \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac{3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac{6 \operatorname{Subst}\left (\int x \sin (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b d}\\ &=\frac{6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac{3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d}-\frac{6 \operatorname{Subst}\left (\int \cos (a+b x) \, dx,x,\sqrt [3]{c+d x}\right )}{b^2 d}\\ &=\frac{6 \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}-\frac{6 \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}+\frac{3 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b d}\\ \end{align*}

Mathematica [A]  time = 0.111096, size = 65, normalized size = 0.76 \[ \frac{3 \left (b^2 (c+d x)^{2/3}-2\right ) \sin \left (a+b \sqrt [3]{c+d x}\right )+6 b \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*(c + d*x)^(1/3)],x]

[Out]

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

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Maple [A]  time = 0.034, size = 131, normalized size = 1.5 \begin{align*} 3\,{\frac{ \left ( a+b\sqrt [3]{dx+c} \right ) ^{2}\sin \left ( a+b\sqrt [3]{dx+c} \right ) -2\,\sin \left ( a+b\sqrt [3]{dx+c} \right ) +2\, \left ( a+b\sqrt [3]{dx+c} \right ) \cos \left ( a+b\sqrt [3]{dx+c} \right ) -2\,a \left ( \cos \left ( a+b\sqrt [3]{dx+c} \right ) + \left ( a+b\sqrt [3]{dx+c} \right ) \sin \left ( a+b\sqrt [3]{dx+c} \right ) \right ) +{a}^{2}\sin \left ( a+b\sqrt [3]{dx+c} \right ) }{d{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a+b*(d*x+c)^(1/3)),x)

[Out]

3/d/b^3*((a+b*(d*x+c)^(1/3))^2*sin(a+b*(d*x+c)^(1/3))-2*sin(a+b*(d*x+c)^(1/3))+2*(a+b*(d*x+c)^(1/3))*cos(a+b*(
d*x+c)^(1/3))-2*a*(cos(a+b*(d*x+c)^(1/3))+(a+b*(d*x+c)^(1/3))*sin(a+b*(d*x+c)^(1/3)))+a^2*sin(a+b*(d*x+c)^(1/3
)))

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Maxima [A]  time = 1.11794, size = 159, normalized size = 1.87 \begin{align*} \frac{3 \,{\left (a^{2} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) - 2 \,{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) + \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )} a + 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )} \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) +{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )}}{b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3)),x, algorithm="maxima")

[Out]

3*(a^2*sin((d*x + c)^(1/3)*b + a) - 2*(((d*x + c)^(1/3)*b + a)*sin((d*x + c)^(1/3)*b + a) + cos((d*x + c)^(1/3
)*b + a))*a + 2*((d*x + c)^(1/3)*b + a)*cos((d*x + c)^(1/3)*b + a) + (((d*x + c)^(1/3)*b + a)^2 - 2)*sin((d*x
+ c)^(1/3)*b + a))/(b^3*d)

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Fricas [A]  time = 1.62542, size = 155, normalized size = 1.82 \begin{align*} \frac{3 \,{\left (2 \,{\left (d x + c\right )}^{\frac{1}{3}} b \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right ) +{\left ({\left (d x + c\right )}^{\frac{2}{3}} b^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )\right )}}{b^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3)),x, algorithm="fricas")

[Out]

3*(2*(d*x + c)^(1/3)*b*cos((d*x + c)^(1/3)*b + a) + ((d*x + c)^(2/3)*b^2 - 2)*sin((d*x + c)^(1/3)*b + a))/(b^3
*d)

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Sympy [A]  time = 1.67818, size = 94, normalized size = 1.11 \begin{align*} \begin{cases} x \cos{\left (a \right )} & \text{for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \cos{\left (a + b \sqrt [3]{c} \right )} & \text{for}\: d = 0 \\\frac{3 \left (c + d x\right )^{\frac{2}{3}} \sin{\left (a + b \sqrt [3]{c + d x} \right )}}{b d} + \frac{6 \sqrt [3]{c + d x} \cos{\left (a + b \sqrt [3]{c + d x} \right )}}{b^{2} d} - \frac{6 \sin{\left (a + b \sqrt [3]{c + d x} \right )}}{b^{3} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)**(1/3)),x)

[Out]

Piecewise((x*cos(a), Eq(b, 0) & (Eq(b, 0) | Eq(d, 0))), (x*cos(a + b*c**(1/3)), Eq(d, 0)), (3*(c + d*x)**(2/3)
*sin(a + b*(c + d*x)**(1/3))/(b*d) + 6*(c + d*x)**(1/3)*cos(a + b*(c + d*x)**(1/3))/(b**2*d) - 6*sin(a + b*(c
+ d*x)**(1/3))/(b**3*d), True))

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Giac [A]  time = 1.26277, size = 109, normalized size = 1.28 \begin{align*} \frac{3 \,{\left (\frac{2 \,{\left (d x + c\right )}^{\frac{1}{3}} \cos \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{b} + \frac{{\left ({\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}^{2} - 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )} a + a^{2} - 2\right )} \sin \left ({\left (d x + c\right )}^{\frac{1}{3}} b + a\right )}{b^{2}}\right )}}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*(d*x+c)^(1/3)),x, algorithm="giac")

[Out]

3*(2*(d*x + c)^(1/3)*cos((d*x + c)^(1/3)*b + a)/b + (((d*x + c)^(1/3)*b + a)^2 - 2*((d*x + c)^(1/3)*b + a)*a +
a^2 - 2)*sin((d*x + c)^(1/3)*b + a)/b^2)/(b*d)