3.3.19 \(\int \sin (a+\frac {b}{\sqrt [3]{c+d x}}) \, dx\) [219]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 = {3442, 3378, 3384, 3380, 3383} \begin {gather*} \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} \end {gather*}

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 3378

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 3380

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 3383

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]

Rule 3384

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 3442

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+\frac {b}{\sqrt [3]{c+d x}}\right ) \, dx &=-\frac {3 \text {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 \text {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 \text {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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.08, size = 133, normalized size = 0.98 \begin {gather*} \frac {b (c+d x)^{2/3} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^3 \cos (a) \text {Ci}\left (\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^2 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-b^3 \sin (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{2 d} \end {gather*}

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 = 0.02, size = 108, normalized size = 0.79

method	result	size
derivativedivides	\(-\frac {3 b^{3} \left (-\frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )}{3 b^{3}}-\frac {\cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )^{\frac {2}{3}}}{6 b^{2}}+\frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )^{\frac {1}{3}}}{6 b}+\frac {\sinIntegral \left (\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \sin \left (a \right )}{6}-\frac {\cosineIntegral \left (\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \cos \left (a \right )}{6}\right )}{d}\)	\(108\)
default	\(-\frac {3 b^{3} \left (-\frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )}{3 b^{3}}-\frac {\cos \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )^{\frac {2}{3}}}{6 b^{2}}+\frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \left (d x +c \right )^{\frac {1}{3}}}{6 b}+\frac {\sinIntegral \left (\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \sin \left (a \right )}{6}-\frac {\cosineIntegral \left (\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right ) \cos \left (a \right )}{6}\right )}{d}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/d*b^3*(-1/3*sin(a+b/(d*x+c)^(1/3))/b^3*(d*x+c)-1/6*cos(a+b/(d*x+c)^(1/3))/b^2*(d*x+c)^(2/3)+1/6*sin(a+b/(d*
x+c)^(1/3))/b*(d*x+c)^(1/3)+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] Result contains complex when optimal does not.
time = 0.38, size = 138, normalized size = 1.01 \begin {gather*} \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 {gather*}

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 = 0.38, size = 139, normalized size = 1.02 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin {\left (a + \frac {b}{\sqrt [3]{c + d x}} \right )}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 663 vs. \(2 (114) = 228\).
time = 5.08, size = 663, normalized size = 4.88 \begin {gather*} \frac {a^{3} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + a^{3} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )} a^{2} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {1}{3}}} - \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )} a^{2} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {1}{3}}} + \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{2} a b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {2}{3}}} + \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{2} a b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {2}{3}}} - \frac {{\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{3} b^{4} \cos \left (a\right ) \operatorname {Ci}\left (-a + \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{d x + c} - \frac {{\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{3} b^{4} \sin \left (a\right ) \operatorname {Si}\left (a - \frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{d x + c} + a^{2} b^{4} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - \frac {2 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )} a b^{4} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {1}{3}}} + \frac {{\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{2} b^{4} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {2}{3}}} + a b^{4} \cos \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - \frac {{\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )} b^{4} \cos \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{{\left (d x + c\right )}^{\frac {1}{3}}} - 2 \, b^{4} \sin \left (\frac {{\left (d x + c\right )}^{\frac {1}{3}} a + b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )}{2 \, {\left (a^{3} - \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )} a^{2}}{{\left (d x + c\right )}^{\frac {1}{3}}} + \frac {3 \, {\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{2} a}{{\left (d x + c\right )}^{\frac {2}{3}}} - \frac {{\left ({\left (d x + c\right )}^{\frac {1}{3}} a + b\right )}^{3}}{d x + c}\right )} b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3*b^4*cos(a)*cos_integral(-a + ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) + a^3*b^4*sin(a)*sin_integral(a
- ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) - 3*((d*x + c)^(1/3)*a + b)*a^2*b^4*cos(a)*cos_integral(-a + ((d*x
+ c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(1/3) - 3*((d*x + c)^(1/3)*a + b)*a^2*b^4*sin(a)*sin_integral(a
- ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(1/3) + 3*((d*x + c)^(1/3)*a + b)^2*a*b^4*cos(a)*cos_inte
gral(-a + ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(2/3) + 3*((d*x + c)^(1/3)*a + b)^2*a*b^4*sin(a)*
sin_integral(a - ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(2/3) - ((d*x + c)^(1/3)*a + b)^3*b^4*cos(
a)*cos_integral(-a + ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c) - ((d*x + c)^(1/3)*a + b)^3*b^4*sin(a)
*sin_integral(a - ((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c) + a^2*b^4*sin(((d*x + c)^(1/3)*a + b)/(d*
x + c)^(1/3)) - 2*((d*x + c)^(1/3)*a + b)*a*b^4*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(1/3) +
((d*x + c)^(1/3)*a + b)^2*b^4*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/(d*x + c)^(2/3) + a*b^4*cos(((d*x
+ c)^(1/3)*a + b)/(d*x + c)^(1/3)) - ((d*x + c)^(1/3)*a + b)*b^4*cos(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))/
(d*x + c)^(1/3) - 2*b^4*sin(((d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)))/((a^3 - 3*((d*x + c)^(1/3)*a + b)*a^2/(d
*x + c)^(1/3) + 3*((d*x + c)^(1/3)*a + b)^2*a/(d*x + c)^(2/3) - ((d*x + c)^(1/3)*a + b)^3/(d*x + c))*b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b/(c + d*x)^(1/3)),x)

[Out]

int(sin(a + b/(c + d*x)^(1/3)), x)

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