3.252 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{\sqrt [3]{c e+d e x}} \, dx\)

Optimal. Leaf size=122 \[ -\frac {3 b \cos (a) \sqrt [3]{c+d x} \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 b \sin (a) \sqrt [3]{c+d x} \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3435, 3381, 3379, 3297, 3303, 3299, 3302} \[ -\frac {3 b \cos (a) \sqrt [3]{c+d x} \text {CosIntegral}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 b \sin (a) \sqrt [3]{c+d x} \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}}+\frac {3 (c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(1/3),x]

[Out]

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

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 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3381

Int[((e_)*(x_))^(m_)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[(e^IntPart[m]*(e*x)
^FracPart[m])/x^FracPart[m], Int[x^m*(a + b*Sin[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] &&
IntegerQ[Simplify[(m + 1)/n]]

Rule 3435

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 90, normalized size = 0.74 \[ \frac {3 \left (-b \cos (a) \sqrt [3]{c+d x} \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right )+b \sin (a) \sqrt [3]{c+d x} \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )+(c+d x) \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )}{2 d \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(1/3),x]

[Out]

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

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fricas [F]  time = 1.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{{\left (d e x + c e\right )}^{\frac {1}{3}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.66, size = 128, normalized size = 1.05 \[ -\frac {{\left (3 \, {\left (\Gamma \left (-1, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-1, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-1, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (-1, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \relax (a) - {\left (3 i \, \Gamma \left (-1, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \Gamma \left (-1, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + 3 i \, \Gamma \left (-1, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) - 3 i \, \Gamma \left (-1, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \relax (a)\right )} b}{8 \, d e^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(1/3),x, algorithm="maxima")

[Out]

-1/8*(3*(gamma(-1, I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(-1,
I*b/(d*x + c)^(2/3)) + gamma(-1, -I*b/(d*x + c)^(2/3)))*cos(a) - (3*I*gamma(-1, I*b*conjugate((d*x + c)^(-2/3)
)) - 3*I*gamma(-1, -I*b*conjugate((d*x + c)^(-2/3))) + 3*I*gamma(-1, I*b/(d*x + c)^(2/3)) - 3*I*gamma(-1, -I*b
/(d*x + c)^(2/3)))*sin(a))*b/(d*e^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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