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

Optimal. Leaf size=136 \[ \frac{b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^3 \sin (a) \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}+\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.161785, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {3361, 3297, 3303, 3299, 3302} \[ \frac{b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^3 \sin (a) \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}+\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(1/3)],x]

[Out]

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

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]

Rule 3297

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*Sin[e + f*x])/(d*(
m + 1)), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right ) \, dx &=-\frac{3 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^4} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^3} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^2} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d}\\ &=\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}+\frac{b^3 \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d}\\ &=\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}+\frac{\left (b^3 \cos (a)\right ) \operatorname{Subst}\left (\int \frac{\cos (b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{\left (b^3 \sin (a)\right ) \operatorname{Subst}\left (\int \frac{\sin (b x)}{x} \, dx,x,\frac{1}{\sqrt [3]{c+d x}}\right )}{2 d}\\ &=\frac{b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{b^3 \cos (a) \text{Ci}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}-\frac{b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}+\frac{(c+d x) \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{d}-\frac{b^3 \sin (a) \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.10153, size = 133, normalized size = 0.98 \[ \frac{b^3 \cos (a) \text{CosIntegral}\left (\frac{b}{\sqrt [3]{c+d x}}\right )-b^3 \sin (a) \text{Si}\left (\frac{b}{\sqrt [3]{c+d x}}\right )-b^2 \sqrt [3]{c+d x} \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+2 c \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+2 d x \sin \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )+b (c+d x)^{2/3} \cos \left (a+\frac{b}{\sqrt [3]{c+d x}}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 108, normalized size = 0.8 \begin{align*} -3\,{\frac{{b}^{3}}{d} \left ( -1/3\,{\frac{dx+c}{{b}^{3}}\sin \left ( a+{\frac{b}{\sqrt [3]{dx+c}}} \right ) }-1/6\,{\frac{ \left ( dx+c \right ) ^{2/3}}{{b}^{2}}\cos \left ( a+{\frac{b}{\sqrt [3]{dx+c}}} \right ) }+1/6\,{\frac{\sqrt [3]{dx+c}}{b}\sin \left ( a+{\frac{b}{\sqrt [3]{dx+c}}} \right ) }+1/6\,{\it Si} \left ({\frac{b}{\sqrt [3]{dx+c}}} \right ) \sin \left ( a \right ) -1/6\,{\it Ci} \left ({\frac{b}{\sqrt [3]{dx+c}}} \right ) \cos \left ( a \right ) \right ) } \end{align*}

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

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Maxima [C]  time = 1.2256, size = 186, normalized size = 1.37 \begin{align*} \frac{{\left ({\left ({\rm Ei}\left (\frac{i \, b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) +{\rm Ei}\left (-\frac{i \, b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right )\right )} \cos \left (a\right ) +{\left (i \,{\rm Ei}\left (\frac{i \, b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) - i \,{\rm Ei}\left (-\frac{i \, b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right )\right )} \sin \left (a\right )\right )} b^{3} + 2 \,{\left (d x + c\right )}^{\frac{2}{3}} b \cos \left (\frac{{\left (d x + c\right )}^{\frac{1}{3}} a + b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) - 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b^{2} - 2 \, d x - 2 \, c\right )} \sin \left (\frac{{\left (d x + c\right )}^{\frac{1}{3}} a + b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((Ei(I*b/(d*x + c)^(1/3)) + Ei(-I*b/(d*x + c)^(1/3)))*cos(a) + (I*Ei(I*b/(d*x + c)^(1/3)) - I*Ei(-I*b/(d*
x + c)^(1/3)))*sin(a))*b^3 + 2*(d*x + c)^(2/3)*b*cos(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) - 2*((d*x + c)^(
1/3)*b^2 - 2*d*x - 2*c)*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)))/d

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Fricas [A]  time = 1.82137, size = 412, normalized size = 3.03 \begin{align*} \frac{b^{3} \cos \left (a\right ) \operatorname{Ci}\left (\frac{b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) + b^{3} \cos \left (a\right ) \operatorname{Ci}\left (-\frac{b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) - 2 \, b^{3} \sin \left (a\right ) \operatorname{Si}\left (\frac{b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) + 2 \,{\left (d x + c\right )}^{\frac{2}{3}} b \cos \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right ) - 2 \,{\left ({\left (d x + c\right )}^{\frac{1}{3}} b^{2} - 2 \, d x - 2 \, c\right )} \sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{2}{3}} b}{d x + c}\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^3*cos(a)*cos_integral(b/(d*x + c)^(1/3)) + b^3*cos(a)*cos_integral(-b/(d*x + c)^(1/3)) - 2*b^3*sin(a)*s
in_integral(b/(d*x + c)^(1/3)) + 2*(d*x + c)^(2/3)*b*cos((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + c)) - 2*((d*
x + c)^(1/3)*b^2 - 2*d*x - 2*c)*sin((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + c)))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (a + \frac{b}{\sqrt [3]{c + d x}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (a + \frac{b}{{\left (d x + c\right )}^{\frac{1}{3}}}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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