3.3.57 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{8/3}} \, dx\) [257]

Optimal. Leaf size=237 \[ \frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}} \]

[Out]

3/2*cos(a+b/(d*x+c)^(2/3))/b/d/e^2/(d*x+c)^(1/3)/(e*(d*x+c))^(2/3)-9/4*(d*x+c)^(1/3)*sin(a+b/(d*x+c)^(2/3))/b^
2/d/e^2/(e*(d*x+c))^(2/3)+9/8*(d*x+c)^(2/3)*cos(a)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*2^(1/2)*Pi
^(1/2)/b^(5/2)/d/e^2/(e*(d*x+c))^(2/3)+9/8*(d*x+c)^(2/3)*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(
a)*2^(1/2)*Pi^(1/2)/b^(5/2)/d/e^2/(e*(d*x+c))^(2/3)

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Rubi [A]
time = 0.17, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3516, 3498, 3496, 3490, 3466, 3467, 3434, 3433, 3432} \begin {gather*} \frac {9 \sqrt {\frac {\pi }{2}} \sin (a) (c+d x)^{2/3} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} \cos (a) (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (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)^(8/3),x]

[Out]

(3*Cos[a + b/(c + d*x)^(2/3)])/(2*b*d*e^2*(c + d*x)^(1/3)*(e*(c + d*x))^(2/3)) + (9*Sqrt[Pi/2]*(c + d*x)^(2/3)
*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) + (9*Sqrt[Pi/2]*
(c + d*x)^(2/3)*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(4*b^(5/2)*d*e^2*(e*(c + d*x))^(2/3)) -
(9*(c + d*x)^(1/3)*Sin[a + b/(c + d*x)^(2/3)])/(4*b^2*d*e^2*(e*(c + d*x))^(2/3))

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3434

Int[Sin[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Sin[c], Int[Cos[d*(e + f*x)^2], x], x] + Dist[
Cos[c], Int[Sin[d*(e + f*x)^2], x], x] /; FreeQ[{c, d, e, f}, x]

Rule 3466

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^(n - 1))*(e*x)^(m - n + 1)*(Cos[c +
d*x^n]/(d*n)), x] + Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3467

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*
x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 3490

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sin[c + d/x^
n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m] && EqQ[n, -2
]

Rule 3496

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Module[{k = Denominator[n]}, D
ist[k, Subst[Int[x^(k*(m + 1) - 1)*(a + b*Sin[c + d*x^(k*n)])^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, m}
, x] && IntegerQ[p] && FractionQ[n]

Rule 3498

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}, x] && Integ
erQ[p] && FractionQ[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+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx &=\frac {\text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{(e x)^{8/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x)^{2/3} \text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{x^{8/3}} \, dx,x,c+d x\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^6} \, dx,x,\sqrt [3]{c+d x}\right )}{d e^2 (e (c+d x))^{2/3}}\\ &=-\frac {\left (3 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x^4 \sin \left (a+b x^2\right ) \, 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}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac {\left (9 (c+d x)^{2/3}\right ) \text {Subst}\left (\int x^2 \cos \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{2 b d e^2 (e (c+d x))^{2/3}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac {\left (9 (c+d x)^{2/3}\right ) \text {Subst}\left (\int \sin \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac {\left (9 (c+d x)^{2/3} \cos (a)\right ) \text {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac {\left (9 (c+d x)^{2/3} \sin (a)\right ) \text {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} (c+d x)^{2/3} C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 165, normalized size = 0.70 \begin {gather*} \frac {(c+d x)^{5/3} \left (9 \sqrt {2 \pi } (c+d x) \cos (a) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+9 \sqrt {2 \pi } (c+d x) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)+6 \sqrt {b} \left (2 b \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )-3 (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )\right )}{8 b^{5/2} d (e (c+d x))^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(8/3),x]

[Out]

((c + d*x)^(5/3)*(9*Sqrt[2*Pi]*(c + d*x)*Cos[a]*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] + 9*Sqrt[2*Pi]*
(c + d*x)*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] + 6*Sqrt[b]*(2*b*Cos[a + b/(c + d*x)^(2/3)] -
3*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)])))/(8*b^(5/2)*d*(e*(c + d*x))^(8/3))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {8}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(8/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(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 4.
time = 0.61, size = 442, normalized size = 1.86 \begin {gather*} \frac {3 \, {\left (d x + c\right )}^{\frac {1}{3}} {\left ({\left ({\left (i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (-i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \cos \left (a\right ) + {\left ({\left (e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (-i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (-i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + i \, e^{\frac {1}{3}} \Gamma \left (\frac {5}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \sin \left (a\right )\right )}}{8 \, {\left (d^{3} x^{2} e^{3} + 2 \, c d^{2} x e^{3} + c^{2} d e^{3}\right )} \left (\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )^{\frac {5}{2}}} \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)^(8/3),x, algorithm="maxima")

[Out]

3/8*(d*x + c)^(1/3)*(((I*e^(1/3)*gamma(5/2, I*b*conjugate((d*x + c)^(-2/3))) - I*e^(1/3)*gamma(5/2, -I*b/(d*x
+ c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (-I*e^(1/3)*gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3)))
+ I*e^(1/3)*gamma(5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (e^(1/3)*gamma(5/2, I*b*
conjugate((d*x + c)^(-2/3))) + e^(1/3)*gamma(5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan2(0, d*x + c))
- (e^(1/3)*gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3))) + e^(1/3)*gamma(5/2, I*b/(d*x + c)^(2/3)))*sin(-5/4*p
i + 5/3*arctan2(0, d*x + c)))*cos(a) + ((e^(1/3)*gamma(5/2, I*b*conjugate((d*x + c)^(-2/3))) + e^(1/3)*gamma(5
/2, -I*b/(d*x + c)^(2/3)))*cos(5/4*pi + 5/3*arctan2(0, d*x + c)) + (e^(1/3)*gamma(5/2, -I*b*conjugate((d*x + c
)^(-2/3))) + e^(1/3)*gamma(5/2, I*b/(d*x + c)^(2/3)))*cos(-5/4*pi + 5/3*arctan2(0, d*x + c)) + (-I*e^(1/3)*gam
ma(5/2, I*b*conjugate((d*x + c)^(-2/3))) + I*e^(1/3)*gamma(5/2, -I*b/(d*x + c)^(2/3)))*sin(5/4*pi + 5/3*arctan
2(0, d*x + c)) + (-I*e^(1/3)*gamma(5/2, -I*b*conjugate((d*x + c)^(-2/3))) + I*e^(1/3)*gamma(5/2, I*b/(d*x + c)
^(2/3)))*sin(-5/4*pi + 5/3*arctan2(0, d*x + c)))*sin(a))/((d^3*x^2*e^3 + 2*c*d^2*x*e^3 + c^2*d*e^3)*(b/(d*x +
c)^(2/3))^(5/2))

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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)^(8/3),x, algorithm="fricas")

[Out]

integral((d*x + c)^(1/3)*e^(-8/3)*sin((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))/(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)**(2/3))/(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)^(2/3))/(d*e*x+c*e)^(8/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3))/(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 )}^{2/3}}\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)^(2/3))/(c*e + d*e*x)^(8/3),x)

[Out]

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(8/3), x)

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