3.53 \(\int \frac{\cos (a+b \sqrt [3]{x})}{x^{5/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac{32}{315} \sqrt{2 \pi } b^{9/2} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [6]{x}\right )-\frac{32}{315} \sqrt{2 \pi } b^{9/2} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [6]{x}\right )+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}} \]

[Out]

(-2*Cos[a + b*x^(1/3)])/(3*x^(3/2)) + (8*b^2*Cos[a + b*x^(1/3)])/(105*x^(5/6)) - (32*b^4*Cos[a + b*x^(1/3)])/(
315*x^(1/6)) - (32*b^(9/2)*Sqrt[2*Pi]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x^(1/6)])/315 - (32*b^(9/2)*Sqrt[2*Pi
]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x^(1/6)]*Sin[a])/315 + (4*b*Sin[a + b*x^(1/3)])/(21*x^(7/6)) - (16*b^3*Sin[a + b
*x^(1/3)])/(315*Sqrt[x])

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Rubi [A]  time = 0.237778, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3416, 3297, 3306, 3305, 3351, 3304, 3352} \[ -\frac{32}{315} \sqrt{2 \pi } b^{9/2} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [6]{x}\right )-\frac{32}{315} \sqrt{2 \pi } b^{9/2} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [6]{x}\right )+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^(1/3)]/x^(5/2),x]

[Out]

(-2*Cos[a + b*x^(1/3)])/(3*x^(3/2)) + (8*b^2*Cos[a + b*x^(1/3)])/(105*x^(5/6)) - (32*b^4*Cos[a + b*x^(1/3)])/(
315*x^(1/6)) - (32*b^(9/2)*Sqrt[2*Pi]*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x^(1/6)])/315 - (32*b^(9/2)*Sqrt[2*Pi
]*FresnelC[Sqrt[b]*Sqrt[2/Pi]*x^(1/6)]*Sin[a])/315 + (4*b*Sin[a + b*x^(1/3)])/(21*x^(7/6)) - (16*b^3*Sin[a + b
*x^(1/3)])/(315*Sqrt[x])

Rule 3416

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

Rule 3297

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 3306

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

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 3304

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[(f*x^2)/d],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

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]

Rubi steps

\begin{align*} \int \frac{\cos \left (a+b \sqrt [3]{x}\right )}{x^{5/2}} \, dx &=3 \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^{11/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}-\frac{1}{3} (2 b) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^{9/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{1}{21} \left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^{7/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}+\frac{1}{105} \left (8 b^3\right ) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{x^{5/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}+\frac{1}{315} \left (16 b^4\right ) \operatorname{Subst}\left (\int \frac{\cos (a+b x)}{x^{3/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}-\frac{1}{315} \left (32 b^5\right ) \operatorname{Subst}\left (\int \frac{\sin (a+b x)}{\sqrt{x}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}-\frac{1}{315} \left (32 b^5 \cos (a)\right ) \operatorname{Subst}\left (\int \frac{\sin (b x)}{\sqrt{x}} \, dx,x,\sqrt [3]{x}\right )-\frac{1}{315} \left (32 b^5 \sin (a)\right ) \operatorname{Subst}\left (\int \frac{\cos (b x)}{\sqrt{x}} \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}-\frac{1}{315} \left (64 b^5 \cos (a)\right ) \operatorname{Subst}\left (\int \sin \left (b x^2\right ) \, dx,x,\sqrt [6]{x}\right )-\frac{1}{315} \left (64 b^5 \sin (a)\right ) \operatorname{Subst}\left (\int \cos \left (b x^2\right ) \, dx,x,\sqrt [6]{x}\right )\\ &=-\frac{2 \cos \left (a+b \sqrt [3]{x}\right )}{3 x^{3/2}}+\frac{8 b^2 \cos \left (a+b \sqrt [3]{x}\right )}{105 x^{5/6}}-\frac{32 b^4 \cos \left (a+b \sqrt [3]{x}\right )}{315 \sqrt [6]{x}}-\frac{32}{315} b^{9/2} \sqrt{2 \pi } \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [6]{x}\right )-\frac{32}{315} b^{9/2} \sqrt{2 \pi } C\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [6]{x}\right ) \sin (a)+\frac{4 b \sin \left (a+b \sqrt [3]{x}\right )}{21 x^{7/6}}-\frac{16 b^3 \sin \left (a+b \sqrt [3]{x}\right )}{315 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.238466, size = 180, normalized size = 0.98 \[ -\frac{2 \left (16 \sqrt{2 \pi } b^{9/2} x^{3/2} \sin (a) \text{FresnelC}\left (\sqrt{\frac{2}{\pi }} \sqrt{b} \sqrt [6]{x}\right )+16 \sqrt{2 \pi } b^{9/2} x^{3/2} \cos (a) S\left (\sqrt{b} \sqrt{\frac{2}{\pi }} \sqrt [6]{x}\right )+16 b^4 x^{4/3} \cos \left (a+b \sqrt [3]{x}\right )-12 b^2 x^{2/3} \cos \left (a+b \sqrt [3]{x}\right )+8 b^3 x \sin \left (a+b \sqrt [3]{x}\right )-30 b \sqrt [3]{x} \sin \left (a+b \sqrt [3]{x}\right )+105 \cos \left (a+b \sqrt [3]{x}\right )\right )}{315 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^(1/3)]/x^(5/2),x]

[Out]

(-2*(105*Cos[a + b*x^(1/3)] - 12*b^2*x^(2/3)*Cos[a + b*x^(1/3)] + 16*b^4*x^(4/3)*Cos[a + b*x^(1/3)] + 16*b^(9/
2)*Sqrt[2*Pi]*x^(3/2)*Cos[a]*FresnelS[Sqrt[b]*Sqrt[2/Pi]*x^(1/6)] + 16*b^(9/2)*Sqrt[2*Pi]*x^(3/2)*FresnelC[Sqr
t[b]*Sqrt[2/Pi]*x^(1/6)]*Sin[a] - 30*b*x^(1/3)*Sin[a + b*x^(1/3)] + 8*b^3*x*Sin[a + b*x^(1/3)]))/(315*x^(3/2))

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Maple [A]  time = 0.03, size = 129, normalized size = 0.7 \begin{align*} -{\frac{2}{3}\cos \left ( a+b\sqrt [3]{x} \right ){x}^{-{\frac{3}{2}}}}-{\frac{4\,b}{3} \left ( -{\frac{1}{7}\sin \left ( a+b\sqrt [3]{x} \right ){x}^{-{\frac{7}{6}}}}+{\frac{2\,b}{7} \left ( -{\frac{1}{5}\cos \left ( a+b\sqrt [3]{x} \right ){x}^{-{\frac{5}{6}}}}-{\frac{2\,b}{5} \left ( -{\frac{1}{3}\sin \left ( a+b\sqrt [3]{x} \right ){\frac{1}{\sqrt{x}}}}+{\frac{2\,b}{3} \left ( -{\cos \left ( a+b\sqrt [3]{x} \right ){\frac{1}{\sqrt [6]{x}}}}-\sqrt{b}\sqrt{2}\sqrt{\pi } \left ( \cos \left ( a \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }}\sqrt [6]{x}\sqrt{b}} \right ) +\sin \left ( a \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }}\sqrt [6]{x}\sqrt{b}} \right ) \right ) \right ) } \right ) } \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a+b*x^(1/3))/x^(5/2),x)

[Out]

-2/3*cos(a+b*x^(1/3))/x^(3/2)-4/3*b*(-1/7/x^(7/6)*sin(a+b*x^(1/3))+2/7*b*(-1/5/x^(5/6)*cos(a+b*x^(1/3))-2/5*b*
(-1/3/x^(1/2)*sin(a+b*x^(1/3))+2/3*b*(-1/x^(1/6)*cos(a+b*x^(1/3))-b^(1/2)*2^(1/2)*Pi^(1/2)*(cos(a)*FresnelS(x^
(1/6)*b^(1/2)*2^(1/2)/Pi^(1/2))+sin(a)*FresnelC(x^(1/6)*b^(1/2)*2^(1/2)/Pi^(1/2)))))))

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Maxima [C]  time = 1.50286, size = 359, normalized size = 1.95 \begin{align*} -\frac{3 \,{\left ({\left ({\left (\Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \cos \left (\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \cos \left (-\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) - i \, \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \sin \left (\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + i \, \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \sin \left (-\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) +{\left ({\left (-i \, \Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + i \, \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \cos \left (\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + i \, \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \cos \left (-\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \sin \left (\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (-\frac{9}{2}, i \, b x^{\frac{1}{3}}\right ) + \Gamma \left (-\frac{9}{2}, -i \, b x^{\frac{1}{3}}\right )\right )} \sin \left (-\frac{9}{4} \, \pi + \frac{9}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )} \sqrt{x^{\frac{1}{3}}{\left | b \right |}} b^{4}}{4 \, x^{\frac{1}{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*x^(1/3))/x^(5/2),x, algorithm="maxima")

[Out]

-3/4*(((gamma(-9/2, I*b*x^(1/3)) + gamma(-9/2, -I*b*x^(1/3)))*cos(9/4*pi + 9/2*arctan2(0, b)) + (gamma(-9/2, I
*b*x^(1/3)) + gamma(-9/2, -I*b*x^(1/3)))*cos(-9/4*pi + 9/2*arctan2(0, b)) + (I*gamma(-9/2, I*b*x^(1/3)) - I*ga
mma(-9/2, -I*b*x^(1/3)))*sin(9/4*pi + 9/2*arctan2(0, b)) + (-I*gamma(-9/2, I*b*x^(1/3)) + I*gamma(-9/2, -I*b*x
^(1/3)))*sin(-9/4*pi + 9/2*arctan2(0, b)))*cos(a) + ((-I*gamma(-9/2, I*b*x^(1/3)) + I*gamma(-9/2, -I*b*x^(1/3)
))*cos(9/4*pi + 9/2*arctan2(0, b)) + (-I*gamma(-9/2, I*b*x^(1/3)) + I*gamma(-9/2, -I*b*x^(1/3)))*cos(-9/4*pi +
9/2*arctan2(0, b)) + (gamma(-9/2, I*b*x^(1/3)) + gamma(-9/2, -I*b*x^(1/3)))*sin(9/4*pi + 9/2*arctan2(0, b)) -
(gamma(-9/2, I*b*x^(1/3)) + gamma(-9/2, -I*b*x^(1/3)))*sin(-9/4*pi + 9/2*arctan2(0, b)))*sin(a))*sqrt(x^(1/3)
*abs(b))*b^4/x^(1/6)

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Fricas [A]  time = 2.07561, size = 405, normalized size = 2.2 \begin{align*} -\frac{2 \,{\left (16 \, \sqrt{2} \pi b^{4} x^{2} \sqrt{\frac{b}{\pi }} \cos \left (a\right ) \operatorname{S}\left (\sqrt{2} x^{\frac{1}{6}} \sqrt{\frac{b}{\pi }}\right ) + 16 \, \sqrt{2} \pi b^{4} x^{2} \sqrt{\frac{b}{\pi }} \operatorname{C}\left (\sqrt{2} x^{\frac{1}{6}} \sqrt{\frac{b}{\pi }}\right ) \sin \left (a\right ) +{\left (16 \, b^{4} x^{\frac{11}{6}} - 12 \, b^{2} x^{\frac{7}{6}} + 105 \, \sqrt{x}\right )} \cos \left (b x^{\frac{1}{3}} + a\right ) + 2 \,{\left (4 \, b^{3} x^{\frac{3}{2}} - 15 \, b x^{\frac{5}{6}}\right )} \sin \left (b x^{\frac{1}{3}} + a\right )\right )}}{315 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*x^(1/3))/x^(5/2),x, algorithm="fricas")

[Out]

-2/315*(16*sqrt(2)*pi*b^4*x^2*sqrt(b/pi)*cos(a)*fresnel_sin(sqrt(2)*x^(1/6)*sqrt(b/pi)) + 16*sqrt(2)*pi*b^4*x^
2*sqrt(b/pi)*fresnel_cos(sqrt(2)*x^(1/6)*sqrt(b/pi))*sin(a) + (16*b^4*x^(11/6) - 12*b^2*x^(7/6) + 105*sqrt(x))
*cos(b*x^(1/3) + a) + 2*(4*b^3*x^(3/2) - 15*b*x^(5/6))*sin(b*x^(1/3) + a))/x^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (a + b \sqrt [3]{x} \right )}}{x^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*x**(1/3))/x**(5/2),x)

[Out]

Integral(cos(a + b*x**(1/3))/x**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x^{\frac{1}{3}} + a\right )}{x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(a+b*x^(1/3))/x^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x^(1/3) + a)/x^(5/2), x)