3.3.42 \(\int \sqrt [3]{c e+d e x} \sin (a+\frac {b}{\sqrt [3]{c+d x}}) \, dx\) [242]

Optimal. Leaf size=247 \[ -\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^4 \sqrt [3]{e (c+d x)} \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right ) \sin (a)}{8 d \sqrt [3]{c+d x}}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^4 \sqrt [3]{e (c+d x)} \cos (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}} \]

[Out]

-1/8*b^3*(e*(d*x+c))^(1/3)*cos(a+b/(d*x+c)^(1/3))/d+1/4*b*(d*x+c)^(2/3)*(e*(d*x+c))^(1/3)*cos(a+b/(d*x+c)^(1/3
))/d-1/8*b^4*(e*(d*x+c))^(1/3)*cos(a)*Si(b/(d*x+c)^(1/3))/d/(d*x+c)^(1/3)-1/8*b^4*(e*(d*x+c))^(1/3)*Ci(b/(d*x+
c)^(1/3))*sin(a)/d/(d*x+c)^(1/3)-1/8*b^2*(d*x+c)^(1/3)*(e*(d*x+c))^(1/3)*sin(a+b/(d*x+c)^(1/3))/d+3/4*(d*x+c)*
(e*(d*x+c))^(1/3)*sin(a+b/(d*x+c)^(1/3))/d

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Rubi [A]
time = 0.16, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3512, 15, 3378, 3384, 3380, 3383} \begin {gather*} -\frac {b^4 \sin (a) \sqrt [3]{e (c+d x)} \text {CosIntegral}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}-\frac {b^4 \cos (a) \sqrt [3]{e (c+d x)} \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}-\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(b^3*(e*(c + d*x))^(1/3)*Cos[a + b/(c + d*x)^(1/3)])/d + (b*(c + d*x)^(2/3)*(e*(c + d*x))^(1/3)*Cos[a + b
/(c + d*x)^(1/3)])/(4*d) - (b^4*(e*(c + d*x))^(1/3)*CosIntegral[b/(c + d*x)^(1/3)]*Sin[a])/(8*d*(c + d*x)^(1/3
)) - (b^2*(c + d*x)^(1/3)*(e*(c + d*x))^(1/3)*Sin[a + b/(c + d*x)^(1/3)])/(8*d) + (3*(c + d*x)*(e*(c + d*x))^(
1/3)*Sin[a + b/(c + d*x)^(1/3)])/(4*d) - (b^4*(e*(c + d*x))^(1/3)*Cos[a]*SinIntegral[b/(c + d*x)^(1/3)])/(8*d*
(c + d*x)^(1/3))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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 3512

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)])^(p_.), x_Symbol] :
> Dist[1/(n*f), Subst[Int[ExpandIntegrand[(a + b*Sin[c + d*x])^p, x^(1/n - 1)*(g - e*(h/f) + h*(x^(1/n)/f))^m,
x], x], x, (e + f*x)^n], x] /; FreeQ[{a, b, c, d, e, f, g, h, m}, x] && IGtQ[p, 0] && IntegerQ[1/n]

Rubi steps

\begin {align*} \int \sqrt [3]{c e+d e x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right ) \, dx &=-\frac {3 \text {Subst}\left (\int \frac {\sqrt [3]{\frac {e}{x^3}} \sin (a+b x)}{x^4} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=-\frac {\left (3 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x^5} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d \sqrt [3]{c+d x}}\\ &=\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {\left (3 b \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \frac {\cos (a+b x)}{x^4} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 d \sqrt [3]{c+d x}}\\ &=\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}+\frac {\left (b^2 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x^3} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 d \sqrt [3]{c+d x}}\\ &=\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}+\frac {\left (b^3 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \frac {\cos (a+b x)}{x^2} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}\\ &=-\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {\left (b^4 \sqrt [3]{e (c+d x)}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}\\ &=-\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {\left (b^4 \sqrt [3]{e (c+d x)} \cos (a)\right ) \text {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}-\frac {\left (b^4 \sqrt [3]{e (c+d x)} \sin (a)\right ) \text {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}\\ &=-\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^4 \sqrt [3]{e (c+d x)} \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right ) \sin (a)}{8 d \sqrt [3]{c+d x}}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}-\frac {b^4 \sqrt [3]{e (c+d x)} \cos (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 208, normalized size = 0.84 \begin {gather*} -\frac {\sqrt [3]{e (c+d x)} \left (-2 b c \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-2 b d x \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^3 \sqrt [3]{c+d x} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^4 \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right ) \sin (a)-6 c \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-6 d x \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^2 (c+d x)^{2/3} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^4 \cos (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )\right )}{8 d \sqrt [3]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*((e*(c + d*x))^(1/3)*(-2*b*c*Cos[a + b/(c + d*x)^(1/3)] - 2*b*d*x*Cos[a + b/(c + d*x)^(1/3)] + b^3*(c + d
*x)^(1/3)*Cos[a + b/(c + d*x)^(1/3)] + b^4*CosIntegral[b/(c + d*x)^(1/3)]*Sin[a] - 6*c*(c + d*x)^(1/3)*Sin[a +
b/(c + d*x)^(1/3)] - 6*d*x*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(1/3)] + b^2*(c + d*x)^(2/3)*Sin[a + b/(c + d*
x)^(1/3)] + b^4*Cos[a]*SinIntegral[b/(c + d*x)^(1/3)]))/(d*(c + d*x)^(1/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {1}{3}} \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.46, size = 128, normalized size = 0.52 \begin {gather*} -\frac {3 \, {\left ({\left (-i \, \Gamma \left (-4, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + i \, \Gamma \left (-4, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - i \, \Gamma \left (-4, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (-4, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (-4, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (-4, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (-4, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-4, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \sin \left (a\right )\right )} b^{4} e^{\frac {1}{3}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*((-I*gamma(-4, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(-4, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma
(-4, I*b/(d*x + c)^(1/3)) + I*gamma(-4, -I*b/(d*x + c)^(1/3)))*cos(a) - (gamma(-4, I*b*conjugate((d*x + c)^(-1
/3))) + gamma(-4, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(-4, I*b/(d*x + c)^(1/3)) + gamma(-4, -I*b/(d*x + c
)^(1/3)))*sin(a))*b^4*e^(1/3)/d

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)^(1/3)*e^(1/3)*sin((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + c)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{e \left (c + d x\right )} \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((d*e*x+c*e)**(1/3)*sin(a+b/(d*x+c)**(1/3)),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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