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

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

[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 = 0.25, 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 = {3435, 3417, 3415, 3409, 3385, 3386, 3353, 3352, 3351} \[ \frac {9 \sqrt {\frac {\pi }{2}} \sin (a) (c+d x)^{2/3} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}+\frac {9 \sqrt {\frac {\pi }{2}} \cos (a) (c+d x)^{2/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{4 b^{5/2} d e^2 (e (c+d x))^{2/3}}-\frac {9 \sqrt [3]{c+d x} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^2 (e (c+d x))^{2/3}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^2 \sqrt [3]{c+d x} (e (c+d x))^{2/3}} \]

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 3351

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

Rule 3352

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

Rule 3353

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 3385

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 3386

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 3409

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 3415

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 3417

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 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 \frac {\sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{(c e+d e x)^{8/3}} \, dx &=\frac {\operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 0.91, size = 165, normalized size = 0.70 \[ \frac {(c+d x)^{5/3} \left (9 \sqrt {2 \pi } \sin (a) (c+d x) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+9 \sqrt {2 \pi } \cos (a) (c+d x) S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )+6 \sqrt {b} \left (2 b \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )-3 (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )\right )}{8 b^{5/2} d (e (c+d x))^{8/3}} \]

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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fricas [F]  time = 1.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d e x + c e\right )}^{\frac {1}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

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

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

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*e*x + c*e)^(8/3), x)

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maple [F]  time = 0.08, 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]  time = 1.17, size = 414, normalized size = 1.75 \[ -\frac {{\left (d x + c\right )}^{\frac {1}{3}} {\left ({\left ({\left (-3 i \, \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + 3 i \, \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 (3 i \, \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \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 ) - 3 \, {\left (\Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \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 ) + 3 \, {\left (\Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \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 \relax (a) - {\left (3 \, {\left (\Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \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 ) + 3 \, {\left (\Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \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 (3 i \, \Gamma \left (\frac {5}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \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 (3 i \, \Gamma \left (\frac {5}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \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 \relax (a)\right )} e^{\frac {1}{3}}}{8 \, {\left (d^{3} e^{3} x^{2} + 2 \, c d^{2} e^{3} x + c^{2} d e^{3}\right )} \left (\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )^{\frac {5}{2}}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{8/3}} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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