3.3.34 \(\int \frac {\sin (a+b \sqrt [3]{c+d x})}{(c e+d e x)^{7/3}} \, dx\) [234]

Optimal. Leaf size=267 \[ \frac {b^3 \cos \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {b^4 \sqrt [3]{c+d x} \text {Ci}\left (b \sqrt [3]{c+d x}\right ) \sin (a)}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}+\frac {b^4 \sqrt [3]{c+d x} \cos (a) \text {Si}\left (b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 267, 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]{c+d x} \text {CosIntegral}\left (b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}+\frac {b^4 \cos (a) \sqrt [3]{c+d x} \text {Si}\left (b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}+\frac {b^3 \cos \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*(c + d*x)^(1/3)]/(c*e + d*e*x)^(7/3),x]

[Out]

(b^3*Cos[a + b*(c + d*x)^(1/3)])/(8*d*e^2*(e*(c + d*x))^(1/3)) - (b*Cos[a + b*(c + d*x)^(1/3)])/(4*d*e^2*(c +
d*x)^(2/3)*(e*(c + d*x))^(1/3)) + (b^4*(c + d*x)^(1/3)*CosIntegral[b*(c + d*x)^(1/3)]*Sin[a])/(8*d*e^2*(e*(c +
d*x))^(1/3)) - (3*Sin[a + b*(c + d*x)^(1/3)])/(4*d*e^2*(c + d*x)*(e*(c + d*x))^(1/3)) + (b^2*Sin[a + b*(c + d
*x)^(1/3)])/(8*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(1/3)) + (b^4*(c + d*x)^(1/3)*Cos[a]*SinIntegral[b*(c + d*x
)^(1/3)])/(8*d*e^2*(e*(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 \frac {\sin \left (a+b \sqrt [3]{c+d x}\right )}{(c e+d e x)^{7/3}} \, dx &=\frac {3 \text {Subst}\left (\int \frac {x^2 \sin (a+b x)}{\left (e x^3\right )^{7/3}} \, dx,x,\sqrt [3]{c+d x}\right )}{d}\\ &=\frac {\left (3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x^5} \, dx,x,\sqrt [3]{c+d x}\right )}{d e^2 \sqrt [3]{e (c+d x)}}\\ &=-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {\left (3 b \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \frac {\cos (a+b x)}{x^4} \, dx,x,\sqrt [3]{c+d x}\right )}{4 d e^2 \sqrt [3]{e (c+d x)}}\\ &=-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}-\frac {\left (b^2 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x^3} \, dx,x,\sqrt [3]{c+d x}\right )}{4 d e^2 \sqrt [3]{e (c+d x)}}\\ &=-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}-\frac {\left (b^3 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \frac {\cos (a+b x)}{x^2} \, dx,x,\sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}\\ &=\frac {b^3 \cos \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}+\frac {\left (b^4 \sqrt [3]{c+d x}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x} \, dx,x,\sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}\\ &=\frac {b^3 \cos \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}+\frac {\left (b^4 \sqrt [3]{c+d x} \cos (a)\right ) \text {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,\sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}+\frac {\left (b^4 \sqrt [3]{c+d x} \sin (a)\right ) \text {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,\sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}\\ &=\frac {b^3 \cos \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {b \cos \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {b^4 \sqrt [3]{c+d x} \text {Ci}\left (b \sqrt [3]{c+d x}\right ) \sin (a)}{8 d e^2 \sqrt [3]{e (c+d x)}}-\frac {3 \sin \left (a+b \sqrt [3]{c+d x}\right )}{4 d e^2 (c+d x) \sqrt [3]{e (c+d x)}}+\frac {b^2 \sin \left (a+b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{c+d x} \sqrt [3]{e (c+d x)}}+\frac {b^4 \sqrt [3]{c+d x} \cos (a) \text {Si}\left (b \sqrt [3]{c+d x}\right )}{8 d e^2 \sqrt [3]{e (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 184, normalized size = 0.69 \begin {gather*} \frac {b^3 c \cos \left (a+b \sqrt [3]{c+d x}\right )+b^3 d x \cos \left (a+b \sqrt [3]{c+d x}\right )-2 b \sqrt [3]{c+d x} \cos \left (a+b \sqrt [3]{c+d x}\right )+b^4 (c+d x)^{4/3} \text {Ci}\left (b \sqrt [3]{c+d x}\right ) \sin (a)-6 \sin \left (a+b \sqrt [3]{c+d x}\right )+b^2 (c+d x)^{2/3} \sin \left (a+b \sqrt [3]{c+d x}\right )+b^4 (c+d x)^{4/3} \cos (a) \text {Si}\left (b \sqrt [3]{c+d x}\right )}{8 d e (e (c+d x))^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*(c + d*x)^(1/3)]/(c*e + d*e*x)^(7/3),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +b \left (d x +c \right )^{\frac {1}{3}}\right )}{\left (d e x +c e \right )^{\frac {7}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b*(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x)

[Out]

int(sin(a+b*(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x)

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.48, size = 128, normalized size = 0.48 \begin {gather*} \frac {3 \, {\left ({\left (-i \, \Gamma \left (-4, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + i \, \Gamma \left (-4, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) - i \, \Gamma \left (-4, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + i \, \Gamma \left (-4, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \cos \left (a\right ) - {\left (\Gamma \left (-4, i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-4, -i \, b \overline {{\left (d x + c\right )}^{\frac {1}{3}}}\right ) + \Gamma \left (-4, i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right ) + \Gamma \left (-4, -i \, {\left (d x + c\right )}^{\frac {1}{3}} b\right )\right )} \sin \left (a\right )\right )} b^{4} e^{\left (-\frac {7}{3}\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b*(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),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*(d*x + c)^(1/3)*b) + I*gamma(-4, -I*(d*x + c)^(1/3)*b))*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*(d*x + c)^(1/3)*b) + gamma(-4, -I*(d*x + c)^(1/3)
*b))*sin(a))*b^4*e^(-7/3)/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(sin(a+b*(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x, algorithm="fricas")

[Out]

integral((d*x + c)^(2/3)*e^(-7/3)*sin((d*x + c)^(1/3)*b + a)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(a+b*(d*x+c)**(1/3))/(d*e*x+c*e)**(7/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

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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(sin(a+b*(d*x+c)^(1/3))/(d*e*x+c*e)^(7/3),x, algorithm="giac")

[Out]

integrate(sin((d*x + c)^(1/3)*b + a)/(d*x*e + c*e)^(7/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sin \left (a+b\,{\left (c+d\,x\right )}^{1/3}\right )}{{\left (c\,e+d\,e\,x\right )}^{7/3}} \,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)^(7/3),x)

[Out]

int(sin(a + b*(c + d*x)^(1/3))/(c*e + d*e*x)^(7/3), x)

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