3.3.51 \(\int \sqrt [3]{c e+d e x} \sin (a+\frac {b}{(c+d x)^{2/3}}) \, dx\) [251]

Optimal. Leaf size=168 \[ \frac {3 b \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d}+\frac {3 b^2 \sqrt [3]{e (c+d x)} \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right ) \sin (a)}{4 d \sqrt [3]{c+d x}}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d}+\frac {3 b^2 \sqrt [3]{e (c+d x)} \cos (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d \sqrt [3]{c+d x}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3516, 3462, 3460, 3378, 3384, 3380, 3383} \begin {gather*} \frac {3 b^2 \sin (a) \sqrt [3]{e (c+d x)} \text {CosIntegral}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d \sqrt [3]{c+d x}}+\frac {3 b^2 \cos (a) \sqrt [3]{e (c+d x)} \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )}{4 d \sqrt [3]{c+d x}}+\frac {3 (c+d x) \sqrt [3]{e (c+d x)} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d}+\frac {3 b \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3460

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 3462

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 3516

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

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Mathematica [A]
time = 0.20, size = 113, normalized size = 0.67 \begin {gather*} \frac {3 \sqrt [3]{e (c+d x)} \left (b (c+d x)^{2/3} \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+b^2 \text {Ci}\left (\frac {b}{(c+d x)^{2/3}}\right ) \sin (a)+(c+d x)^{4/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )+b^2 \cos (a) \text {Si}\left (\frac {b}{(c+d x)^{2/3}}\right )\right )}{4 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)^(2/3)],x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (d e x +c e \right )^{\frac {1}{3}} \sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )\, dx\]

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)^(2/3)),x)

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.51, size = 128, normalized size = 0.76 \begin {gather*} \frac {3 \, {\left ({\left (-i \, \Gamma \left (-2, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, \Gamma \left (-2, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - i \, \Gamma \left (-2, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + i \, \Gamma \left (-2, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (-2, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-2, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (-2, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (-2, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (a\right )\right )} b^{2} e^{\frac {1}{3}}}{8 \, 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)^(2/3)),x, algorithm="maxima")

[Out]

3/8*((-I*gamma(-2, I*b*conjugate((d*x + c)^(-2/3))) + I*gamma(-2, -I*b*conjugate((d*x + c)^(-2/3))) - I*gamma(
-2, I*b/(d*x + c)^(2/3)) + I*gamma(-2, -I*b/(d*x + c)^(2/3)))*cos(a) - (gamma(-2, I*b*conjugate((d*x + c)^(-2/
3))) + gamma(-2, -I*b*conjugate((d*x + c)^(-2/3))) + gamma(-2, I*b/(d*x + c)^(2/3)) + gamma(-2, -I*b/(d*x + c)
^(2/3)))*sin(a))*b^2*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)^(2/3)),x, algorithm="fricas")

[Out]

integral((d*x + c)^(1/3)*e^(1/3)*sin((a*d*x + a*c + (d*x + c)^(1/3)*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}{\left (c + d x\right )^{\frac {2}{3}}} \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)**(2/3)),x)

[Out]

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

[Out]

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

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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 )}^{2/3}}\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)^(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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