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

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

[Out]

-2/3*cos(a+b*x^(1/3))/x^(3/2)+8/105*b^2*cos(a+b*x^(1/3))/x^(5/6)-32/315*b^4*cos(a+b*x^(1/3))/x^(1/6)+4/21*b*si
n(a+b*x^(1/3))/x^(7/6)-32/315*b^(9/2)*cos(a)*FresnelS(x^(1/6)*b^(1/2)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)-32/31
5*b^(9/2)*FresnelC(x^(1/6)*b^(1/2)*2^(1/2)/Pi^(1/2))*sin(a)*2^(1/2)*Pi^(1/2)-16/315*b^3*sin(a+b*x^(1/3))/x^(1/
2)

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Rubi [A]  time = 0.24, antiderivative size = 184, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.25, size = 180, normalized size = 0.98 \[ -\frac {2 \left (16 \sqrt {2 \pi } b^{9/2} x^{3/2} \sin (a) C\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [6]{x}\right )+16 \sqrt {2 \pi } b^{9/2} x^{3/2} \cos (a) S\left (\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt [6]{x}\right )+16 b^4 x^{4/3} \cos \left (a+b \sqrt [3]{x}\right )+8 b^3 x \sin \left (a+b \sqrt [3]{x}\right )-12 b^2 x^{2/3} \cos \left (a+b \sqrt [3]{x}\right )-30 b \sqrt [3]{x} \sin \left (a+b \sqrt [3]{x}\right )+105 \cos \left (a+b \sqrt [3]{x}\right )\right )}{315 x^{3/2}} \]

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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fricas [A]  time = 0.81, size = 134, normalized size = 0.73 \[ -\frac {2 \, {\left (16 \, \sqrt {2} \pi b^{4} x^{2} \sqrt {\frac {b}{\pi }} \cos \relax (a) \operatorname {S}\left (\sqrt {2} x^{\frac {1}{6}} \sqrt {\frac {b}{\pi }}\right ) + 16 \, \sqrt {2} \pi b^{4} x^{2} \sqrt {\frac {b}{\pi }} \operatorname {C}\left (\sqrt {2} x^{\frac {1}{6}} \sqrt {\frac {b}{\pi }}\right ) \sin \relax (a) + {\left (16 \, b^{4} x^{\frac {11}{6}} - 12 \, b^{2} x^{\frac {7}{6}} + 105 \, \sqrt {x}\right )} \cos \left (b x^{\frac {1}{3}} + a\right ) + 2 \, {\left (4 \, b^{3} x^{\frac {3}{2}} - 15 \, b x^{\frac {5}{6}}\right )} \sin \left (b x^{\frac {1}{3}} + a\right )\right )}}{315 \, x^{2}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x^{\frac {1}{3}} + a\right )}{x^{\frac {5}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 129, normalized size = 0.70 \[ -\frac {2 \cos \left (a +b \,x^{\frac {1}{3}}\right )}{3 x^{\frac {3}{2}}}-\frac {4 b \left (-\frac {\sin \left (a +b \,x^{\frac {1}{3}}\right )}{7 x^{\frac {7}{6}}}+\frac {2 b \left (-\frac {\cos \left (a +b \,x^{\frac {1}{3}}\right )}{5 x^{\frac {5}{6}}}-\frac {2 b \left (-\frac {\sin \left (a +b \,x^{\frac {1}{3}}\right )}{3 \sqrt {x}}+\frac {2 b \left (-\frac {\cos \left (a +b \,x^{\frac {1}{3}}\right )}{x^{\frac {1}{6}}}-\sqrt {b}\, \sqrt {2}\, \sqrt {\pi }\, \left (\cos \relax (a ) \mathrm {S}\left (\frac {x^{\frac {1}{6}} \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )+\sin \relax (a ) \FresnelC \left (\frac {x^{\frac {1}{6}} \sqrt {b}\, \sqrt {2}}{\sqrt {\pi }}\right )\right )\right )}{3}\right )}{5}\right )}{7}\right )}{3} \]

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*(-cos(a+b*x^(1/3))/x^(1/6)-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 = 1.62, size = 76, normalized size = 0.41 \[ \frac {3 \, {\left ({\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {9}{2}, i \, b x^{\frac {1}{3}}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {9}{2}, -i \, b x^{\frac {1}{3}}\right )\right )} \cos \relax (a) + {\left (\left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {9}{2}, i \, b x^{\frac {1}{3}}\right ) - \left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {9}{2}, -i \, b x^{\frac {1}{3}}\right )\right )} \sin \relax (a)\right )} \sqrt {b x^{\frac {1}{3}}} b^{4}}{4 \, x^{\frac {1}{6}}} \]

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*((-(I + 1)*sqrt(2)*gamma(-9/2, I*b*x^(1/3)) + (I - 1)*sqrt(2)*gamma(-9/2, -I*b*x^(1/3)))*cos(a) + ((I - 1)
*sqrt(2)*gamma(-9/2, I*b*x^(1/3)) - (I + 1)*sqrt(2)*gamma(-9/2, -I*b*x^(1/3)))*sin(a))*sqrt(b*x^(1/3))*b^4/x^(
1/6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x^{1/3}\right )}{x^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^(1/3))/x^(5/2),x)

[Out]

int(cos(a + b*x^(1/3))/x^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + b \sqrt [3]{x} \right )}}{x^{\frac {5}{2}}}\, dx \]

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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