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

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

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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 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 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 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
]

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*}

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Mathematica [A]  time = 0.16, size = 146, normalized size = 1.04 \[ \frac {2 \sqrt {2 \pi } b^{3/2} \sin (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+2 \sqrt {2 \pi } b^{3/2} \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+c \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+d x \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+2 b \sqrt [3]{c+d x} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{d} \]

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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fricas [A]  time = 0.61, size = 143, normalized size = 1.01 \[ \frac {2 \, \sqrt {2} \pi b \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {S}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + 2 \, \sqrt {2} \pi b \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} \sqrt {\frac {b}{\pi }}}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) \sin \relax (a) + 2 \, {\left (d x + c\right )}^{\frac {1}{3}} b \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right ) + {\left (d x + c\right )} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{d} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\,{d x} \]

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)

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maple [A]  time = 0.03, size = 105, normalized size = 0.74 \[ \frac {\left (d x +c \right ) \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-2 b \left (-\left (d x +c \right )^{\frac {1}{3}} \cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \mathrm {S}\left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )+\sin \relax (a ) \FresnelC \left (\frac {\sqrt {b}\, \sqrt {2}}{\sqrt {\pi }\, \left (d x +c \right )^{\frac {1}{3}}}\right )\right )\right )}{d} \]

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 = 0.49, size = 219, normalized size = 1.55 \[ \frac {\sqrt {2} {\left (2 \, \sqrt {2} {\left (d x + c\right )}^{\frac {2}{3}} \sqrt {\frac {1}{{\left (d x + c\right )}^{\frac {4}{3}}}} b^{2} \cos \left (\frac {{\left (d x + c\right )}^{\frac {2}{3}} a + b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \sqrt {2} {\left (d x + c\right )}^{\frac {4}{3}} \sqrt {\frac {1}{{\left (d x + c\right )}^{\frac {4}{3}}}} b \sin \left (\frac {{\left (d x + c\right )}^{\frac {2}{3}} a + b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + {\left ({\left (\left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )} - \left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )}\right )} \cos \relax (a) + {\left (-\left (i - 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )} + \left (i + 1\right ) \, \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 1\right )}\right )} \sin \relax (a)\right )} b^{2} \left (\frac {b^{2}}{{\left (d x + c\right )}^{\frac {4}{3}}}\right )^{\frac {1}{4}}\right )} \sqrt {{\left (d x + c\right )}^{\frac {4}{3}}}}{2 \, {\left (d x + c\right )}^{\frac {1}{3}} b d} \]

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*sqrt(2)*(2*sqrt(2)*(d*x + c)^(2/3)*sqrt((d*x + c)^(-4/3))*b^2*cos(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3))
+ sqrt(2)*(d*x + c)^(4/3)*sqrt((d*x + c)^(-4/3))*b*sin(((d*x + c)^(2/3)*a + b)/(d*x + c)^(2/3)) + (((I + 1)*s
qrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) - (I - 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3))) - 1))*cos(a)
+ (-(I - 1)*sqrt(pi)*(erf(sqrt(I*b/(d*x + c)^(2/3))) - 1) + (I + 1)*sqrt(pi)*(erf(sqrt(-I*b/(d*x + c)^(2/3)))
- 1))*sin(a))*b^2*(b^2/(d*x + c)^(4/3))^(1/4))*sqrt((d*x + c)^(4/3))/((d*x + c)^(1/3)*b*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/(c + d*x)^(2/3)),x)

[Out]

int(sin(a + b/(c + d*x)^(2/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin {\left (a + \frac {b}{\left (c + d x\right )^{\frac {2}{3}}} \right )}\, dx \]

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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