3.224 \(\int \sin (a+\frac{b}{(c+d x)^{2/3}}) \, dx\)

Optimal. Leaf size=141 \[ \frac{2 \sqrt{2 \pi } b^{3/2} \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{d}+\frac{2 \sqrt{2 \pi } b^{3/2} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d}+\frac{2 b \sqrt [3]{c+d x} \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d} \]

[Out]

(2*b*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)])/d + (2*b^(3/2)*Sqrt[2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]
)/(c + d*x)^(1/3)])/d + (2*b^(3/2)*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/d + ((c +
d*x)*Sin[a + b/(c + d*x)^(2/3)])/d

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Rubi [A]  time = 0.11186, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3363, 3409, 3387, 3388, 3353, 3352, 3351} \[ \frac{2 \sqrt{2 \pi } b^{3/2} \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )}{d}+\frac{2 \sqrt{2 \pi } b^{3/2} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{d}+\frac{(c+d x) \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d}+\frac{2 b \sqrt [3]{c+d x} \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)],x]

[Out]

(2*b*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)])/d + (2*b^(3/2)*Sqrt[2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]
)/(c + d*x)^(1/3)])/d + (2*b^(3/2)*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/d + ((c +
d*x)*Sin[a + b/(c + d*x)^(2/3)])/d

Rule 3363

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

Rule 3409

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

Rule 3387

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

Rule 3388

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

Rule 3353

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

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

Mathematica [A]  time = 0.151286, size = 146, normalized size = 1.04 \[ \frac{2 \sqrt{2 \pi } b^{3/2} \sin (a) \text{FresnelC}\left (\frac{\sqrt{\frac{2}{\pi }} \sqrt{b}}{\sqrt [3]{c+d x}}\right )+2 \sqrt{2 \pi } b^{3/2} \cos (a) S\left (\frac{\sqrt{b} \sqrt{\frac{2}{\pi }}}{\sqrt [3]{c+d x}}\right )+c \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )+d x \sin \left (a+\frac{b}{(c+d x)^{2/3}}\right )+2 b \sqrt [3]{c+d x} \cos \left (a+\frac{b}{(c+d x)^{2/3}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)],x]

[Out]

(2*b*(c + d*x)^(1/3)*Cos[a + b/(c + d*x)^(2/3)] + 2*b^(3/2)*Sqrt[2*Pi]*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c
+ d*x)^(1/3)] + 2*b^(3/2)*Sqrt[2*Pi]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] + c*Sin[a + b/(c +
d*x)^(2/3)] + d*x*Sin[a + b/(c + d*x)^(2/3)])/d

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Maple [A]  time = 0.013, size = 105, normalized size = 0.7 \begin{align*} 3\,{\frac{1}{d} \left ( 1/3\, \left ( dx+c \right ) \sin \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2/3}}} \right ) -2/3\,b \left ( -\sqrt [3]{dx+c}\cos \left ( a+{\frac{b}{ \left ( dx+c \right ) ^{2/3}}} \right ) -\sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{b}\sqrt{2}}{\sqrt{\pi }\sqrt [3]{dx+c}}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{b}\sqrt{2}}{\sqrt{\pi }\sqrt [3]{dx+c}}} \right ) \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/(d*x+c)^(2/3)),x)

[Out]

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

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Maxima [C]  time = 1.44145, size = 716, normalized size = 5.08 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="maxima")

[Out]

1/2*(4*(d*x + c)^(2/3)*b*sqrt(abs(b)/(d*x + c)^(2/3))*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + (((I*sqrt
(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - I*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(1/4*pi + 1/2
*arctan2(0, b)) + (I*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - I*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3)
)) - 1))*cos(-1/4*pi + 1/2*arctan2(0, b)) + (sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + sqrt(pi)*(erf(sqr
t(-I*b/(d*x + c)^(2/3))) - 1))*sin(1/4*pi + 1/2*arctan2(0, b)) - (sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1
) + sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(a) + ((sqrt(pi)*(erf
(sqrt(I*b/(d*x + c)^(2/3))) - 1) + sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(1/4*pi + 1/2*arctan2(0,
b)) + (sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(-1
/4*pi + 1/2*arctan2(0, b)) + (-I*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + I*sqrt(pi)*(erf(sqrt(-I*b/(d*
x + c)^(2/3))) - 1))*sin(1/4*pi + 1/2*arctan2(0, b)) + (I*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - I*sq
rt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*sin(-1/4*pi + 1/2*arctan2(0, b)))*sin(a))*b^2 + 2*(d*x + c)^(4/3
)*sqrt(abs(b)/(d*x + c)^(2/3))*sin(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)))/((d*x + c)^(1/3)*d*sqrt(abs(b)/(d
*x + c)^(2/3)))

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Fricas [A]  time = 2.0195, size = 406, normalized size = 2.88 \begin{align*} \frac{2 \, \sqrt{2} \pi b \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) + 2 \, \sqrt{2} \pi b \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\frac{\sqrt{2} \sqrt{\frac{b}{\pi }}}{{\left (d x + c\right )}^{\frac{1}{3}}}\right ) \sin \left (a\right ) + 2 \,{\left (d x + c\right )}^{\frac{1}{3}} b \cos \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{1}{3}} b}{d x + c}\right ) +{\left (d x + c\right )} \sin \left (\frac{a d x + a c +{\left (d x + c\right )}^{\frac{1}{3}} b}{d x + c}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="fricas")

[Out]

(2*sqrt(2)*pi*b*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*sqrt(b/pi)/(d*x + c)^(1/3)) + 2*sqrt(2)*pi*b*sqrt(b/pi)*
fresnel_cos(sqrt(2)*sqrt(b/pi)/(d*x + c)^(1/3))*sin(a) + 2*(d*x + c)^(1/3)*b*cos((a*d*x + a*c + (d*x + c)^(1/3
)*b)/(d*x + c)) + (d*x + c)*sin((a*d*x + a*c + (d*x + c)^(1/3)*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}{\left (c + d x\right )^{\frac{2}{3}}} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)**(2/3)),x)

[Out]

Integral(sin(a + b/(c + d*x)**(2/3)), 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{2}{3}}}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3)),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3)), x)