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

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

[Out]

6*cos(a+b*(d*x+c)^(1/3))/b^3/d-3*(d*x+c)^(2/3)*cos(a+b*(d*x+c)^(1/3))/b/d+6*(d*x+c)^(1/3)*sin(a+b*(d*x+c)^(1/3
))/b^2/d

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Rubi [A]  time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, 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 = {3361, 3296, 2638} \[ \frac {6 \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^2 d}+\frac {6 \cos \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d}-\frac {3 (c+d x)^{2/3} \cos \left (a+b \sqrt [3]{c+d x}\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*(c + d*x)^(1/3)],x]

[Out]

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

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 3361

Int[((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :> Dist[1/(n*f), Subst[Int[x
^(1/n - 1)*(a + b*Sin[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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 65, normalized size = 0.76 \[ \frac {\left (6-3 b^2 (c+d x)^{2/3}\right ) \cos \left (a+b \sqrt [3]{c+d x}\right )+6 b \sqrt [3]{c+d x} \sin \left (a+b \sqrt [3]{c+d x}\right )}{b^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 58, normalized size = 0.68 \[ \frac {3 \, {\left (2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{2} - 2\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 82, normalized size = 0.96 \[ \frac {3 \, {\left (\frac {2 \, {\left (d x + c\right )}^{\frac {1}{3}} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b} - \frac {{\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} a + a^{2} - 2\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}{b^{2}}\right )}}{b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 134, normalized size = 1.58 \[ \frac {-3 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )^{2} \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )+6 \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-6 a \left (\sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )-\left (a +b \left (d x +c \right )^{\frac {1}{3}}\right ) \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )\right )-3 a^{2} \cos \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{d \,b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 120, normalized size = 1.41 \[ -\frac {3 \, {\left (a^{2} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - 2 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )} a + {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )}^{2} - 2\right )} \cos \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right ) - 2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )} \sin \left ({\left (d x + c\right )}^{\frac {1}{3}} b + a\right )\right )}}{b^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.61, size = 69, normalized size = 0.81 \[ \frac {3\,\left (2\,\cos \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )+2\,b\,\sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{1/3}-b^2\,\cos \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )\,{\left (c+d\,x\right )}^{2/3}\right )}{b^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*(c + d*x)^(1/3)),x)

[Out]

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

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sympy [A]  time = 1.42, size = 94, normalized size = 1.11 \[ \begin {cases} x \sin {\relax (a )} & \text {for}\: b = 0 \wedge \left (b = 0 \vee d = 0\right ) \\x \sin {\left (a + b \sqrt [3]{c} \right )} & \text {for}\: d = 0 \\- \frac {3 \left (c + d x\right )^{\frac {2}{3}} \cos {\left (a + b \sqrt [3]{c + d x} \right )}}{b d} + \frac {6 \sqrt [3]{c + d x} \sin {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{2} d} + \frac {6 \cos {\left (a + b \sqrt [3]{c + d x} \right )}}{b^{3} d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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