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

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

[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 = 0.24, 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 = {3431, 15, 3297, 3303, 3299, 3302} \[ -\frac {b^4 \sin (a) \sqrt [3]{e (c+d x)} \text {CosIntegral}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}-\frac {b^4 \cos (a) \sqrt [3]{e (c+d x)} \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d \sqrt [3]{c+d x}}-\frac {b^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}-\frac {b^3 \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{8 d}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d}+\frac {b (c+d x)^{2/3} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^3*(e*(c + d*x))^(1/3)*Cos[a + b/(c + d*x)^(1/3)])/(8*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 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 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]

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 3431

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 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.40, size = 208, normalized size = 0.84 \[ -\frac {\sqrt [3]{e (c+d x)} \left (b^4 \sin (a) \text {Ci}\left (\frac {b}{\sqrt [3]{c+d x}}\right )+b^4 \cos (a) \text {Si}\left (\frac {b}{\sqrt [3]{c+d x}}\right )+b^3 \sqrt [3]{c+d x} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+b^2 (c+d x)^{2/3} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-6 c \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-6 d x \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-2 b c \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )-2 b d x \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )\right )}{8 d \sqrt [3]{c+d x}} \]

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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fricas [F]  time = 1.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d e x + c e\right )}^{\frac {1}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ), x\right ) \]

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

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

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

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maple [F]  time = 0.07, 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]  time = 1.78, size = 129, normalized size = 0.52 \[ \frac {{\left ({\left (3 i \, \Gamma \left (-4, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) - 3 i \, \Gamma \left (-4, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + 3 i \, \Gamma \left (-4, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - 3 i \, \Gamma \left (-4, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \cos \relax (a) + 3 \, {\left (\Gamma \left (-4, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (-4, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {1}{3}}}}\right ) + \Gamma \left (-4, \frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-4, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {1}{3}}}\right )\right )} \sin \relax (a)\right )} b^{4} e^{\frac {1}{3}}}{4 \, d} \]

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]

1/4*((3*I*gamma(-4, I*b*conjugate((d*x + c)^(-1/3))) - 3*I*gamma(-4, -I*b*conjugate((d*x + c)^(-1/3))) + 3*I*g
amma(-4, I*b/(d*x + c)^(1/3)) - 3*I*gamma(-4, -I*b/(d*x + c)^(1/3)))*cos(a) + 3*(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{1/3}}\right )\,{\left (c\,e+d\,e\,x\right )}^{1/3} \,d x \]

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

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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