3.3.41 \(\int \frac {\sin (a+b (c+d x)^{2/3})}{(c e+d e x)^{5/3}} \, dx\) [241]

Optimal. Leaf size=126 \[ \frac {3 b (c+d x)^{2/3} \cos (a) \text {Ci}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 \sin \left (a+b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 b (c+d x)^{2/3} \sin (a) \text {Si}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}} \]

[Out]

3/2*b*(d*x+c)^(2/3)*Ci(b*(d*x+c)^(2/3))*cos(a)/d/e/(e*(d*x+c))^(2/3)-3/2*b*(d*x+c)^(2/3)*Si(b*(d*x+c)^(2/3))*s
in(a)/d/e/(e*(d*x+c))^(2/3)-3/2*sin(a+b*(d*x+c)^(2/3))/d/e/(e*(d*x+c))^(2/3)

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Rubi [A]
time = 0.10, antiderivative size = 126, 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 = {3516, 3462, 3460, 3378, 3384, 3380, 3383} \begin {gather*} \frac {3 b \cos (a) (c+d x)^{2/3} \text {CosIntegral}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 b \sin (a) (c+d x)^{2/3} \text {Si}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 \sin \left (a+b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*(c + d*x)^(2/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

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

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 \frac {\sin \left (a+b (c+d x)^{2/3}\right )}{(c e+d e x)^{5/3}} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin \left (a+b x^{2/3}\right )}{(e x)^{5/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x)^{2/3} \text {Subst}\left (\int \frac {\sin \left (a+b x^{2/3}\right )}{x^{5/3}} \, dx,x,c+d x\right )}{d e (e (c+d x))^{2/3}}\\ &=\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \frac {\sin (a+b x)}{x^2} \, dx,x,(c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}\\ &=-\frac {3 \sin \left (a+b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}+\frac {\left (3 b (c+d x)^{2/3}\right ) \text {Subst}\left (\int \frac {\cos (a+b x)}{x} \, dx,x,(c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}\\ &=-\frac {3 \sin \left (a+b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}+\frac {\left (3 b (c+d x)^{2/3} \cos (a)\right ) \text {Subst}\left (\int \frac {\cos (b x)}{x} \, dx,x,(c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {\left (3 b (c+d x)^{2/3} \sin (a)\right ) \text {Subst}\left (\int \frac {\sin (b x)}{x} \, dx,x,(c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}\\ &=\frac {3 b (c+d x)^{2/3} \cos (a) \text {Ci}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 \sin \left (a+b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}-\frac {3 b (c+d x)^{2/3} \sin (a) \text {Si}\left (b (c+d x)^{2/3}\right )}{2 d e (e (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 87, normalized size = 0.69 \begin {gather*} -\frac {3 \left (-b (c+d x)^{2/3} \cos (a) \text {Ci}\left (b (c+d x)^{2/3}\right )+\sin \left (a+b (c+d x)^{2/3}\right )+b (c+d x)^{2/3} \sin (a) \text {Si}\left (b (c+d x)^{2/3}\right )\right )}{2 d e (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*(c + d*x)^(2/3)]/(c*e + d*e*x)^(5/3),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +b \left (d x +c \right )^{\frac {2}{3}}\right )}{\left (d e x +c e \right )^{\frac {5}{3}}}\, dx\]

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)^(5/3),x)

[Out]

int(sin(a+b*(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x)

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.50, size = 125, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left ({\left (\Gamma \left (-1, i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (-1, -i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + \Gamma \left (-1, i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right ) + \Gamma \left (-1, -i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right )\right )} \cos \left (a\right ) + {\left (-i \, \Gamma \left (-1, i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) + i \, \Gamma \left (-1, -i \, b \overline {{\left (d x + c\right )}^{\frac {2}{3}}}\right ) - i \, \Gamma \left (-1, i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right ) + i \, \Gamma \left (-1, -i \, {\left (d x + c\right )}^{\frac {2}{3}} b\right )\right )} \sin \left (a\right )\right )} b e^{\left (-\frac {5}{3}\right )}}{8 \, d} \end {gather*}

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)^(5/3),x, algorithm="maxima")

[Out]

3/8*((gamma(-1, I*b*conjugate((d*x + c)^(2/3))) + gamma(-1, -I*b*conjugate((d*x + c)^(2/3))) + gamma(-1, I*(d*
x + c)^(2/3)*b) + gamma(-1, -I*(d*x + c)^(2/3)*b))*cos(a) + (-I*gamma(-1, I*b*conjugate((d*x + c)^(2/3))) + I*
gamma(-1, -I*b*conjugate((d*x + c)^(2/3))) - I*gamma(-1, I*(d*x + c)^(2/3)*b) + I*gamma(-1, -I*(d*x + c)^(2/3)
*b))*sin(a))*b*e^(-5/3)/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)^(2/3))/(d*e*x+c*e)^(5/3),x, algorithm="fricas")

[Out]

integral((d*x + c)^(1/3)*e^(-5/3)*sin((d*x + c)^(2/3)*b + a)/(d^2*x^2 + 2*c*d*x + c^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (a + b \left (c + d x\right )^{\frac {2}{3}} \right )}}{\left (e \left (c + d x\right )\right )^{\frac {5}{3}}}\, dx \end {gather*}

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)**(5/3),x)

[Out]

Integral(sin(a + b*(c + d*x)**(2/3))/(e*(c + d*x))**(5/3), 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(sin(a+b*(d*x+c)^(2/3))/(d*e*x+c*e)^(5/3),x, algorithm="giac")

[Out]

integrate(sin((d*x + c)^(2/3)*b + a)/(d*x*e + c*e)^(5/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sin \left (a+b\,{\left (c+d\,x\right )}^{2/3}\right )}{{\left (c\,e+d\,e\,x\right )}^{5/3}} \,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)^(5/3),x)

[Out]

int(sin(a + b*(c + d*x)^(2/3))/(c*e + d*e*x)^(5/3), x)

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