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

Optimal. Leaf size=89 \[ \frac {3 \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{2 b^2 d \sqrt [3]{c+d x}}-\frac {3 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3435, 3381, 3379, 3296, 2637} \[ \frac {3 \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{2 b^2 d \sqrt [3]{c+d x}}-\frac {3 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 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 \sqrt [3]{c e+d e x} \sin \left (a+b (c+d x)^{2/3}\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \sqrt [3]{e x} \sin \left (a+b x^{2/3}\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac {\sqrt [3]{e (c+d x)} \operatorname {Subst}\left (\int \sqrt [3]{x} \sin \left (a+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 ) \operatorname {Subst}\left (\int x \sin (a+b x) \, dx,x,(c+d x)^{2/3}\right )}{2 d \sqrt [3]{c+d x}}\\ &=-\frac {3 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac {\left (3 \sqrt [3]{e (c+d x)}\right ) \operatorname {Subst}\left (\int \cos (a+b x) \, dx,x,(c+d x)^{2/3}\right )}{2 b d \sqrt [3]{c+d x}}\\ &=-\frac {3 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)} \cos \left (a+b (c+d x)^{2/3}\right )}{2 b d}+\frac {3 \sqrt [3]{e (c+d x)} \sin \left (a+b (c+d x)^{2/3}\right )}{2 b^2 d \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 72, normalized size = 0.81 \[ -\frac {3 \sqrt [3]{e (c+d x)} \left (b (c+d x)^{2/3} \cos \left (a+b (c+d x)^{2/3}\right )-\sin \left (a+b (c+d x)^{2/3}\right )\right )}{2 b^2 d \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 1.56, size = 89, normalized size = 1.00 \[ -\frac {3 \, {\left ({\left (b d x + b c\right )} {\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {1}{3}} \cos \left ({\left (d x + c\right )}^{\frac {2}{3}} b + a\right ) - {\left (d e x + c e\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \sin \left ({\left (d x + c\right )}^{\frac {2}{3}} b + a\right )\right )}}{2 \, {\left (b^{2} d^{2} x + b^{2} c d\right )}} \]

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)^(2/3)),x, algorithm="fricas")

[Out]

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

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giac [C]  time = 1.85, size = 265, normalized size = 2.98 \[ -\frac {3 \, {\left ({\left (-\frac {i \, \sqrt {\pi } \operatorname {erf}\left (-{\left (d x e + c e\right )}^{\frac {1}{3}} \sqrt {-i \, b e^{\left (-\frac {2}{3}\right )}}\right ) e^{\left (i \, a\right )}}{\sqrt {-i \, b e^{\left (-\frac {2}{3}\right )}}} + \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-{\left (d x e + c e\right )}^{\frac {1}{3}} \sqrt {i \, b e^{\left (-\frac {2}{3}\right )}}\right ) e^{\left (-i \, a\right )}}{\sqrt {i \, b e^{\left (-\frac {2}{3}\right )}}}\right )} c + {\left (\frac {i \, \sqrt {\pi } c \operatorname {erf}\left (-{\left (d x e + c e\right )}^{\frac {1}{3}} \sqrt {-i \, b e^{\left (-\frac {2}{3}\right )}}\right ) e^{\left (i \, a + 1\right )}}{\sqrt {-i \, b e^{\left (-\frac {2}{3}\right )}}} - \frac {i \, \sqrt {\pi } c \operatorname {erf}\left (-{\left (d x e + c e\right )}^{\frac {1}{3}} \sqrt {i \, b e^{\left (-\frac {2}{3}\right )}}\right ) e^{\left (-i \, a + 1\right )}}{\sqrt {i \, b e^{\left (-\frac {2}{3}\right )}}} - \frac {i \, {\left (i \, {\left (d x e + c e\right )}^{\frac {2}{3}} b e^{\left (-\frac {2}{3}\right )} - 1\right )} e^{\left (i \, {\left (d x e + c e\right )}^{\frac {2}{3}} b e^{\left (-\frac {2}{3}\right )} + i \, a + \frac {4}{3}\right )}}{b^{2}} - \frac {i \, {\left (i \, {\left (d x e + c e\right )}^{\frac {2}{3}} b e^{\left (-\frac {2}{3}\right )} + 1\right )} e^{\left (-i \, {\left (d x e + c e\right )}^{\frac {2}{3}} b e^{\left (-\frac {2}{3}\right )} - i \, a + \frac {4}{3}\right )}}{b^{2}}\right )} e^{\left (-1\right )}\right )}}{4 \, d} \]

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)^(2/3)),x, algorithm="giac")

[Out]

-3/4*((-I*sqrt(pi)*erf(-(d*x*e + c*e)^(1/3)*sqrt(-I*b*e^(-2/3)))*e^(I*a)/sqrt(-I*b*e^(-2/3)) + I*sqrt(pi)*erf(
-(d*x*e + c*e)^(1/3)*sqrt(I*b*e^(-2/3)))*e^(-I*a)/sqrt(I*b*e^(-2/3)))*c + (I*sqrt(pi)*c*erf(-(d*x*e + c*e)^(1/
3)*sqrt(-I*b*e^(-2/3)))*e^(I*a + 1)/sqrt(-I*b*e^(-2/3)) - I*sqrt(pi)*c*erf(-(d*x*e + c*e)^(1/3)*sqrt(I*b*e^(-2
/3)))*e^(-I*a + 1)/sqrt(I*b*e^(-2/3)) - I*(I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) - 1)*e^(I*(d*x*e + c*e)^(2/3)*b*e^
(-2/3) + I*a + 4/3)/b^2 - I*(I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) + 1)*e^(-I*(d*x*e + c*e)^(2/3)*b*e^(-2/3) - I*a
+ 4/3)/b^2)*e^(-1))/d

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{\frac {1}{3}} \sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )\, dx \]

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]  time = 1.34, size = 129, normalized size = 1.45 \[ \frac {{\left ({\left (3 i \, \Gamma \left (2, i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) - 3 i \, \Gamma \left (2, -i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + 3 i \, \Gamma \left (2, i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right ) - 3 i \, \Gamma \left (2, -i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right )\right )} \cos \relax (a) + 3 \, {\left (\Gamma \left (2, i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (2, -i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (2, i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right ) + \Gamma \left (2, -i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right )\right )} \sin \relax (a)\right )} e^{\frac {1}{3}}}{8 \, b^{2} d} \]

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)^(2/3)),x, algorithm="maxima")

[Out]

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

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

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

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)**(2/3)),x)

[Out]

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

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