3.258 \(\int \frac {\sin (a+\frac {b}{(c+d x)^{2/3}})}{(c e+d e x)^{10/3}} \, dx\)

Optimal. Leaf size=277 \[ \frac {45 \sqrt {\pi } \cos (a) \sqrt [3]{c+d x} C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {45 \sqrt {\pi } \sin (a) \sqrt [3]{c+d x} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}} \]

[Out]

-45/8*cos(a+b/(d*x+c)^(2/3))/b^3/d/e^3/(e*(d*x+c))^(1/3)+3/2*cos(a+b/(d*x+c)^(2/3))/b/d/e^3/(d*x+c)^(4/3)/(e*(
d*x+c))^(1/3)-15/4*sin(a+b/(d*x+c)^(2/3))/b^2/d/e^3/(d*x+c)^(2/3)/(e*(d*x+c))^(1/3)+45/16*(d*x+c)^(1/3)*cos(a)
*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*Pi^(1/2)/b^(7/2)/d/e^3/(e*(d*x+c))^(1/3)*2^(1/2)-45/16*(d*x+
c)^(1/3)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)/(d*x+c)^(1/3))*sin(a)*Pi^(1/2)/b^(7/2)/d/e^3/(e*(d*x+c))^(1/3)*2^(1
/2)

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, 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, 3354, 3352, 3351} \[ \frac {45 \sqrt {\pi } \cos (a) \sqrt [3]{c+d x} \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {b}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {45 \sqrt {\pi } \sin (a) \sqrt [3]{c+d x} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(10/3),x]

[Out]

(-45*Cos[a + b/(c + d*x)^(2/3)])/(8*b^3*d*e^3*(e*(c + d*x))^(1/3)) + (3*Cos[a + b/(c + d*x)^(2/3)])/(2*b*d*e^3
*(c + d*x)^(4/3)*(e*(c + d*x))^(1/3)) + (45*Sqrt[Pi]*(c + d*x)^(1/3)*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c +
d*x)^(1/3)])/(8*Sqrt[2]*b^(7/2)*d*e^3*(e*(c + d*x))^(1/3)) - (45*Sqrt[Pi]*(c + d*x)^(1/3)*FresnelS[(Sqrt[b]*S
qrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a])/(8*Sqrt[2]*b^(7/2)*d*e^3*(e*(c + d*x))^(1/3)) - (15*Sin[a + b/(c + d*x)^(2
/3)])/(4*b^2*d*e^3*(c + d*x)^(2/3)*(e*(c + d*x))^(1/3))

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 3354

Int[Cos[(c_) + (d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Dist[Cos[c], Int[Cos[d*(e + f*x)^2], x], x] - Dist[
Sin[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)^{10/3}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{(e x)^{10/3}} \, dx,x,c+d x\right )}{d}\\ &=\frac {\sqrt [3]{c+d x} \operatorname {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^{2/3}}\right )}{x^{10/3}} \, dx,x,c+d x\right )}{d e^3 \sqrt [3]{e (c+d x)}}\\ &=\frac {\left (3 \sqrt [3]{c+d x}\right ) \operatorname {Subst}\left (\int \frac {\sin \left (a+\frac {b}{x^2}\right )}{x^8} \, dx,x,\sqrt [3]{c+d x}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\\ &=-\frac {\left (3 \sqrt [3]{c+d x}\right ) \operatorname {Subst}\left (\int x^6 \sin \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{d e^3 \sqrt [3]{e (c+d x)}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}-\frac {\left (15 \sqrt [3]{c+d x}\right ) \operatorname {Subst}\left (\int x^4 \cos \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{2 b d e^3 \sqrt [3]{e (c+d x)}}\\ &=\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {\left (45 \sqrt [3]{c+d x}\right ) \operatorname {Subst}\left (\int x^2 \sin \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{4 b^2 d e^3 \sqrt [3]{e (c+d x)}}\\ &=-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {\left (45 \sqrt [3]{c+d x}\right ) \operatorname {Subst}\left (\int \cos \left (a+b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}\\ &=-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}+\frac {\left (45 \sqrt [3]{c+d x} \cos (a)\right ) \operatorname {Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}-\frac {\left (45 \sqrt [3]{c+d x} \sin (a)\right ) \operatorname {Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\frac {1}{\sqrt [3]{c+d x}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}\\ &=-\frac {45 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{8 b^3 d e^3 \sqrt [3]{e (c+d x)}}+\frac {3 \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{2 b d e^3 (c+d x)^{4/3} \sqrt [3]{e (c+d x)}}+\frac {45 \sqrt {\pi } \sqrt [3]{c+d x} \cos (a) C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {45 \sqrt {\pi } \sqrt [3]{c+d x} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right ) \sin (a)}{8 \sqrt {2} b^{7/2} d e^3 \sqrt [3]{e (c+d x)}}-\frac {15 \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )}{4 b^2 d e^3 (c+d x)^{2/3} \sqrt [3]{e (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.27, size = 192, normalized size = 0.69 \[ \frac {(e (c+d x))^{2/3} \left (-6 \sqrt {b} \left (\left (15 (c+d x)^{4/3}-4 b^2\right ) \cos \left (a+\frac {b}{(c+d x)^{2/3}}\right )+10 b (c+d x)^{2/3} \sin \left (a+\frac {b}{(c+d x)^{2/3}}\right )\right )+45 \sqrt {2 \pi } \cos (a) (c+d x)^{5/3} C\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )-45 \sqrt {2 \pi } \sin (a) (c+d x)^{5/3} S\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }}}{\sqrt [3]{c+d x}}\right )\right )}{16 b^{7/2} d e^4 (c+d x)^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b/(c + d*x)^(2/3)]/(c*e + d*e*x)^(10/3),x]

[Out]

((e*(c + d*x))^(2/3)*(45*Sqrt[2*Pi]*(c + d*x)^(5/3)*Cos[a]*FresnelC[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)] - 45
*Sqrt[2*Pi]*(c + d*x)^(5/3)*FresnelS[(Sqrt[b]*Sqrt[2/Pi])/(c + d*x)^(1/3)]*Sin[a] - 6*Sqrt[b]*((-4*b^2 + 15*(c
+ d*x)^(4/3))*Cos[a + b/(c + d*x)^(2/3)] + 10*b*(c + d*x)^(2/3)*Sin[a + b/(c + d*x)^(2/3)])))/(16*b^(7/2)*d*e
^4*(c + d*x)^(7/3))

________________________________________________________________________________________

fricas [F]  time = 1.36, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (d e x + c e\right )}^{\frac {2}{3}} \sin \left (\frac {a d x + a c + {\left (d x + c\right )}^{\frac {1}{3}} b}{d x + c}\right )}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

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)^(10/3),x, algorithm="fricas")

[Out]

integral((d*e*x + c*e)^(2/3)*sin((a*d*x + a*c + (d*x + c)^(1/3)*b)/(d*x + c))/(d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 +
6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a + \frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )}{{\left (d e x + c e\right )}^{\frac {10}{3}}}\,{d x} \]

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)^(10/3),x, algorithm="giac")

[Out]

integrate(sin(a + b/(d*x + c)^(2/3))/(d*e*x + c*e)^(10/3), x)

________________________________________________________________________________________

maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (a +\frac {b}{\left (d x +c \right )^{\frac {2}{3}}}\right )}{\left (d e x +c e \right )^{\frac {10}{3}}}\, dx \]

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)^(10/3),x)

[Out]

int(sin(a+b/(d*x+c)^(2/3))/(d*e*x+c*e)^(10/3),x)

________________________________________________________________________________________

maxima [C]  time = 0.83, size = 411, normalized size = 1.48 \[ -\frac {{\left ({\left (-3 i \, \Gamma \left (\frac {7}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + 3 i \, \Gamma \left (\frac {7}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) + {\left (3 i \, \Gamma \left (\frac {7}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \Gamma \left (\frac {7}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) - 3 \, {\left (\Gamma \left (\frac {7}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (\frac {7}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) + 3 \, {\left (\Gamma \left (\frac {7}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (\frac {7}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \cos \relax (a) - {\left (3 \, {\left (\Gamma \left (\frac {7}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (\frac {7}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) + 3 \, {\left (\Gamma \left (\frac {7}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) + \Gamma \left (\frac {7}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (3 i \, \Gamma \left (\frac {7}{2}, i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \Gamma \left (\frac {7}{2}, -\frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right ) - {\left (3 i \, \Gamma \left (\frac {7}{2}, -i \, b \overline {\frac {1}{{\left (d x + c\right )}^{\frac {2}{3}}}}\right ) - 3 i \, \Gamma \left (\frac {7}{2}, \frac {i \, b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )\right )} \sin \left (-\frac {7}{4} \, \pi + \frac {7}{3} \, \arctan \left (0, d x + c\right )\right )\right )} \sin \relax (a)}{8 \, {\left (d^{3} e^{\frac {10}{3}} x^{2} + 2 \, c d^{2} e^{\frac {10}{3}} x + c^{2} d e^{\frac {10}{3}}\right )} {\left (d x + c\right )}^{\frac {1}{3}} \left (\frac {b}{{\left (d x + c\right )}^{\frac {2}{3}}}\right )^{\frac {7}{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)^(10/3),x, algorithm="maxima")

[Out]

-1/8*(((-3*I*gamma(7/2, I*b*conjugate((d*x + c)^(-2/3))) + 3*I*gamma(7/2, -I*b/(d*x + c)^(2/3)))*cos(7/4*pi +
7/3*arctan2(0, d*x + c)) + (3*I*gamma(7/2, -I*b*conjugate((d*x + c)^(-2/3))) - 3*I*gamma(7/2, I*b/(d*x + c)^(2
/3)))*cos(-7/4*pi + 7/3*arctan2(0, d*x + c)) - 3*(gamma(7/2, I*b*conjugate((d*x + c)^(-2/3))) + gamma(7/2, -I*
b/(d*x + c)^(2/3)))*sin(7/4*pi + 7/3*arctan2(0, d*x + c)) + 3*(gamma(7/2, -I*b*conjugate((d*x + c)^(-2/3))) +
gamma(7/2, I*b/(d*x + c)^(2/3)))*sin(-7/4*pi + 7/3*arctan2(0, d*x + c)))*cos(a) - (3*(gamma(7/2, I*b*conjugate
((d*x + c)^(-2/3))) + gamma(7/2, -I*b/(d*x + c)^(2/3)))*cos(7/4*pi + 7/3*arctan2(0, d*x + c)) + 3*(gamma(7/2,
-I*b*conjugate((d*x + c)^(-2/3))) + gamma(7/2, I*b/(d*x + c)^(2/3)))*cos(-7/4*pi + 7/3*arctan2(0, d*x + c)) -
(3*I*gamma(7/2, I*b*conjugate((d*x + c)^(-2/3))) - 3*I*gamma(7/2, -I*b/(d*x + c)^(2/3)))*sin(7/4*pi + 7/3*arct
an2(0, d*x + c)) - (3*I*gamma(7/2, -I*b*conjugate((d*x + c)^(-2/3))) - 3*I*gamma(7/2, I*b/(d*x + c)^(2/3)))*si
n(-7/4*pi + 7/3*arctan2(0, d*x + c)))*sin(a))/((d^3*e^(10/3)*x^2 + 2*c*d^2*e^(10/3)*x + c^2*d*e^(10/3))*(d*x +
c)^(1/3)*(b/(d*x + c)^(2/3))^(7/2))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (a+\frac {b}{{\left (c+d\,x\right )}^{2/3}}\right )}{{\left (c\,e+d\,e\,x\right )}^{10/3}} \,d x \]

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)^(10/3),x)

[Out]

int(sin(a + b/(c + d*x)^(2/3))/(c*e + d*e*x)^(10/3), x)

________________________________________________________________________________________

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)**(10/3),x)

[Out]

Timed out

________________________________________________________________________________________