3.3.48 \(\int \frac {\sin (a+\frac {b}{\sqrt [3]{c+d x}})}{(c e+d e x)^{8/3}} \, dx\) [248]

Optimal. Leaf size=217 \[ -\frac {36 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}+\frac {72 (c+d x)^{2/3} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {72 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}} \]

[Out]

-36*cos(a+b/(d*x+c)^(1/3))/b^3/d/e^2/(e*(d*x+c))^(2/3)+3*cos(a+b/(d*x+c)^(1/3))/b/d/e^2/(d*x+c)^(2/3)/(e*(d*x+
c))^(2/3)+72*(d*x+c)^(2/3)*cos(a+b/(d*x+c)^(1/3))/b^5/d/e^2/(e*(d*x+c))^(2/3)-12*sin(a+b/(d*x+c)^(1/3))/b^2/d/
e^2/(d*x+c)^(1/3)/(e*(d*x+c))^(2/3)+72*(d*x+c)^(1/3)*sin(a+b/(d*x+c)^(1/3))/b^4/d/e^2/(e*(d*x+c))^(2/3)

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Rubi [A]
time = 0.13, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3512, 15, 3377, 2718} \begin {gather*} \frac {72 (c+d x)^{2/3} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}+\frac {72 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}-\frac {36 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

(-36*Cos[a + b/(c + d*x)^(1/3)])/(b^3*d*e^2*(e*(c + d*x))^(2/3)) + (3*Cos[a + b/(c + d*x)^(1/3)])/(b*d*e^2*(c
+ d*x)^(2/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(2/3)*Cos[a + b/(c + d*x)^(1/3)])/(b^5*d*e^2*(e*(c + d*x))^(
2/3)) - (12*Sin[a + b/(c + d*x)^(1/3)])/(b^2*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)) + (72*(c + d*x)^(1/3)*
Sin[a + b/(c + d*x)^(1/3)])/(b^4*d*e^2*(e*(c + d*x))^(2/3))

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 2718

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

Rule 3377

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 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+\frac {b}{\sqrt [3]{c+d x}}\right )}{(c e+d e x)^{8/3}} \, dx &=-\frac {3 \text {Subst}\left (\int \frac {\sin (a+b x)}{\left (\frac {e}{x^3}\right )^{8/3} x^4} \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d}\\ &=-\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x^4 \sin (a+b x) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac {\left (12 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x^3 \cos (a+b x) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{b d e^2 (e (c+d x))^{2/3}}\\ &=\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {\left (36 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x^2 \sin (a+b x) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac {36 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {\left (72 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x \cos (a+b x) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac {36 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {72 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}-\frac {\left (72 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \sin (a+b x) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}\\ &=-\frac {36 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^3 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b d e^2 (c+d x)^{2/3} (e (c+d x))^{2/3}}+\frac {72 (c+d x)^{2/3} \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^5 d e^2 (e (c+d x))^{2/3}}-\frac {12 \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^2 d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {72 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )}{b^4 d e^2 (e (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 112, normalized size = 0.52 \begin {gather*} \frac {(c+d x)^{4/3} \left (3 \left (b^4-12 b^2 (c+d x)^{2/3}+24 (c+d x)^{4/3}\right ) \cos \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )+12 b \left (6 c+6 d x-b^2 \sqrt [3]{c+d x}\right ) \sin \left (a+\frac {b}{\sqrt [3]{c+d x}}\right )\right )}{b^5 d (e (c+d x))^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(1/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

((c + d*x)^(4/3)*(3*(b^4 - 12*b^2*(c + d*x)^(2/3) + 24*(c + d*x)^(4/3))*Cos[a + b/(c + d*x)^(1/3)] + 12*b*(6*c
+ 6*d*x - b^2*(c + d*x)^(1/3))*Sin[a + b/(c + d*x)^(1/3)]))/(b^5*d*(e*(c + d*x))^(8/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {1}{3}}}\right )}{\left (d e x +c e \right )^{\frac {8}{3}}}\, dx\]

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)^(8/3),x)

[Out]

int(sin(a+b/(d*x+c)^(1/3))/(d*e*x+c*e)^(8/3),x)

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Maxima [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.08, size = 935, normalized size = 4.31 \begin {gather*} \text {Too large to display} \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)^(8/3),x, algorithm="maxima")

[Out]

3/20*(2*(b^5*cos((2*(d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3))*sin(a) - b^5*cos(b/(d*x + c)^(1/3))*sin(a) - b^5*c
os(a)*sin((2*(d*x + c)^(1/3)*a + b)/(d*x + c)^(1/3)) - b^5*cos(a)*sin(b/(d*x + c)^(1/3)))*(d*x + c)^(1/3)*e^(1
/3) + (((gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, I*b
/(d*x + c)^(1/3)) + gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)^3 + (-I*gamma(6, I*b*conjugate((d*x + c)^(-1/3))) +
I*gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(6, I*b/(d*x + c)^(1/3)) + I*gamma(6, -I*b/(d*x + c)^(1
/3)))*cos(a)^2*sin(a) + (gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, -I*b*conjugate((d*x + c)^(-1/3))
) + gamma(6, I*b/(d*x + c)^(1/3)) + gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)*sin(a)^2 + (-I*gamma(6, I*b*conjuga
te((d*x + c)^(-1/3))) + I*gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(6, I*b/(d*x + c)^(1/3)) + I*gam
ma(6, -I*b/(d*x + c)^(1/3)))*sin(a)^3)*d^2*x^2 + 2*((gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, -I*b
*conjugate((d*x + c)^(-1/3))) + gamma(6, I*b/(d*x + c)^(1/3)) + gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)^3 + (-I
*gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(6, I*b/(d*
x + c)^(1/3)) + I*gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)^2*sin(a) + (gamma(6, I*b*conjugate((d*x + c)^(-1/3)))
+ gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, I*b/(d*x + c)^(1/3)) + gamma(6, -I*b/(d*x + c)^(1/3))
)*cos(a)*sin(a)^2 + (-I*gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(6, -I*b*conjugate((d*x + c)^(-1/3)
)) - I*gamma(6, I*b/(d*x + c)^(1/3)) + I*gamma(6, -I*b/(d*x + c)^(1/3)))*sin(a)^3)*c*d*x + ((gamma(6, I*b*conj
ugate((d*x + c)^(-1/3))) + gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, I*b/(d*x + c)^(1/3)) + gamma(
6, -I*b/(d*x + c)^(1/3)))*cos(a)^3 + (-I*gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + I*gamma(6, -I*b*conjugate
((d*x + c)^(-1/3))) - I*gamma(6, I*b/(d*x + c)^(1/3)) + I*gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)^2*sin(a) + (g
amma(6, I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) + gamma(6, I*b/(d*x + c)
^(1/3)) + gamma(6, -I*b/(d*x + c)^(1/3)))*cos(a)*sin(a)^2 + (-I*gamma(6, I*b*conjugate((d*x + c)^(-1/3))) + I*
gamma(6, -I*b*conjugate((d*x + c)^(-1/3))) - I*gamma(6, I*b/(d*x + c)^(1/3)) + I*gamma(6, -I*b/(d*x + c)^(1/3)
))*sin(a)^3)*c^2)*e^(1/3))/((cos(a)^2*e^3 + e^3*sin(a)^2)*b^5*d^3*x^2 + 2*(cos(a)^2*e^3 + e^3*sin(a)^2)*b^5*c*
d^2*x + (cos(a)^2*e^3 + e^3*sin(a)^2)*b^5*c^2*d)

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Fricas [A]
time = 0.90, size = 172, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left ({\left ({\left (d x + c\right )}^{\frac {1}{3}} b^{4} - 12 \, b^{2} d x - 12 \, b^{2} c + 24 \, {\left (d x + c\right )}^{\frac {5}{3}}\right )} {\left (d x + c\right )}^{\frac {1}{3}} \cos \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right ) e^{\frac {1}{3}} - 4 \, {\left ({\left (d x + c\right )}^{\frac {2}{3}} b^{3} - 6 \, {\left (b d x + b c\right )} {\left (d x + c\right )}^{\frac {1}{3}}\right )} {\left (d x + c\right )}^{\frac {1}{3}} e^{\frac {1}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {2}{3}} b}{d x + c}\right )\right )} e^{\left (-3\right )}}{b^{5} d^{3} x^{2} + 2 \, b^{5} c d^{2} x + b^{5} c^{2} 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)^(8/3),x, algorithm="fricas")

[Out]

3*(((d*x + c)^(1/3)*b^4 - 12*b^2*d*x - 12*b^2*c + 24*(d*x + c)^(5/3))*(d*x + c)^(1/3)*cos((a*d*x + a*c + (d*x
+ c)^(2/3)*b)/(d*x + c))*e^(1/3) - 4*((d*x + c)^(2/3)*b^3 - 6*(b*d*x + b*c)*(d*x + c)^(1/3))*(d*x + c)^(1/3)*e
^(1/3)*sin((a*d*x + a*c + (d*x + c)^(2/3)*b)/(d*x + c)))*e^(-3)/(b^5*d^3*x^2 + 2*b^5*c*d^2*x + b^5*c^2*d)

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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)**(8/3),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8570 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)^(8/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(1/3))/(d*x*e + c*e)^(8/3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{1/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{8/3}} \,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)^(8/3),x)

[Out]

int(sin(a + b/(c + d*x)^(1/3))/(c*e + d*e*x)^(8/3), x)

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