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

Optimal. Leaf size=102 \[ -\frac{3 \sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{3 \sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )}{8 b^{3/2}}+\frac{3 \sqrt [6]{x} \sin \left (2 \left (a+b \sqrt [3]{x}\right )\right )}{4 b}+\sqrt{x} \]

[Out]

Sqrt[x] - (3*Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x^(1/6))/Sqrt[Pi]])/(8*b^(3/2)) - (3*Sqrt[Pi]*FresnelC[(2*S
qrt[b]*x^(1/6))/Sqrt[Pi]]*Sin[2*a])/(8*b^(3/2)) + (3*x^(1/6)*Sin[2*(a + b*x^(1/3))])/(4*b)

________________________________________________________________________________________

Rubi [A]  time = 0.169637, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {3416, 3312, 3296, 3306, 3305, 3351, 3304, 3352} \[ -\frac{3 \sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )}{8 b^{3/2}}-\frac{3 \sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )}{8 b^{3/2}}+\frac{3 \sqrt [6]{x} \sin \left (2 \left (a+b \sqrt [3]{x}\right )\right )}{4 b}+\sqrt{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^(1/3)]^2/Sqrt[x],x]

[Out]

Sqrt[x] - (3*Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x^(1/6))/Sqrt[Pi]])/(8*b^(3/2)) - (3*Sqrt[Pi]*FresnelC[(2*S
qrt[b]*x^(1/6))/Sqrt[Pi]]*Sin[2*a])/(8*b^(3/2)) + (3*x^(1/6)*Sin[2*(a + b*x^(1/3))])/(4*b)

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 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 3296

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

Mathematica [A]  time = 0.189478, size = 103, normalized size = 1.01 \[ \frac{-3 \sqrt{\pi } \sin (2 a) \text{FresnelC}\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )-3 \sqrt{\pi } \cos (2 a) S\left (\frac{2 \sqrt{b} \sqrt [6]{x}}{\sqrt{\pi }}\right )+2 \sqrt{b} \sqrt [6]{x} \left (3 \sin \left (2 \left (a+b \sqrt [3]{x}\right )\right )+4 b \sqrt [3]{x}\right )}{8 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[Pi]*Cos[2*a]*FresnelS[(2*Sqrt[b]*x^(1/6))/Sqrt[Pi]] - 3*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*x^(1/6))/Sqrt[Pi
]]*Sin[2*a] + 2*Sqrt[b]*x^(1/6)*(4*b*x^(1/3) + 3*Sin[2*(a + b*x^(1/3))]))/(8*b^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.039, size = 67, normalized size = 0.7 \begin{align*} \sqrt{x}+{\frac{3}{4\,b}\sqrt [6]{x}\sin \left ( 2\,a+2\,b\sqrt [3]{x} \right ) }-{\frac{3\,\sqrt{\pi }}{8} \left ( \cos \left ( 2\,a \right ){\it FresnelS} \left ( 2\,{\frac{\sqrt [6]{x}\sqrt{b}}{\sqrt{\pi }}} \right ) +\sin \left ( 2\,a \right ){\it FresnelC} \left ( 2\,{\frac{\sqrt [6]{x}\sqrt{b}}{\sqrt{\pi }}} \right ) \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/2)+3/4*x^(1/6)*sin(2*a+2*b*x^(1/3))/b-3/8/b^(3/2)*Pi^(1/2)*(cos(2*a)*FresnelS(2*x^(1/6)*b^(1/2)/Pi^(1/2))
+sin(2*a)*FresnelC(2*x^(1/6)*b^(1/2)/Pi^(1/2)))

________________________________________________________________________________________

Maxima [C]  time = 2.96517, size = 392, normalized size = 3.84 \begin{align*} \frac{\sqrt{2} \sqrt{\pi }{\left ({\left ({\left (-3 i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (3 \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{2 i \, b} x^{\frac{1}{6}}\right ) +{\left ({\left (3 i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (3 \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) + 3 i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right ) - 3 i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \operatorname{erf}\left (\sqrt{-2 i \, b} x^{\frac{1}{6}}\right )\right )} \sqrt{{\left | b \right |}} + 64 \, b \sqrt{x}{\left | b \right |} + 48 \, x^{\frac{1}{6}}{\left | b \right |} \sin \left (2 \, b x^{\frac{1}{3}} + 2 \, a\right )}{64 \, b{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(2)*sqrt(pi)*(((-3*I*cos(1/4*pi + 1/2*arctan2(0, b)) - 3*I*cos(-1/4*pi + 1/2*arctan2(0, b)) - 3*sin(
1/4*pi + 1/2*arctan2(0, b)) + 3*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) - (3*cos(1/4*pi + 1/2*arctan2(0, b)
) + 3*cos(-1/4*pi + 1/2*arctan2(0, b)) - 3*I*sin(1/4*pi + 1/2*arctan2(0, b)) + 3*I*sin(-1/4*pi + 1/2*arctan2(0
, b)))*sin(2*a))*erf(sqrt(2*I*b)*x^(1/6)) + ((3*I*cos(1/4*pi + 1/2*arctan2(0, b)) + 3*I*cos(-1/4*pi + 1/2*arct
an2(0, b)) - 3*sin(1/4*pi + 1/2*arctan2(0, b)) + 3*sin(-1/4*pi + 1/2*arctan2(0, b)))*cos(2*a) - (3*cos(1/4*pi
+ 1/2*arctan2(0, b)) + 3*cos(-1/4*pi + 1/2*arctan2(0, b)) + 3*I*sin(1/4*pi + 1/2*arctan2(0, b)) - 3*I*sin(-1/4
*pi + 1/2*arctan2(0, b)))*sin(2*a))*erf(sqrt(-2*I*b)*x^(1/6)))*sqrt(abs(b)) + 64*b*sqrt(x)*abs(b) + 48*x^(1/6)
*abs(b)*sin(2*b*x^(1/3) + 2*a))/(b*abs(b))

________________________________________________________________________________________

Fricas [A]  time = 2.08258, size = 271, normalized size = 2.66 \begin{align*} -\frac{3 \, \pi \sqrt{\frac{b}{\pi }} \cos \left (2 \, a\right ) \operatorname{S}\left (2 \, x^{\frac{1}{6}} \sqrt{\frac{b}{\pi }}\right ) + 3 \, \pi \sqrt{\frac{b}{\pi }} \operatorname{C}\left (2 \, x^{\frac{1}{6}} \sqrt{\frac{b}{\pi }}\right ) \sin \left (2 \, a\right ) - 12 \, b x^{\frac{1}{6}} \cos \left (b x^{\frac{1}{3}} + a\right ) \sin \left (b x^{\frac{1}{3}} + a\right ) - 8 \, b^{2} \sqrt{x}}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(3*pi*sqrt(b/pi)*cos(2*a)*fresnel_sin(2*x^(1/6)*sqrt(b/pi)) + 3*pi*sqrt(b/pi)*fresnel_cos(2*x^(1/6)*sqrt(
b/pi))*sin(2*a) - 12*b*x^(1/6)*cos(b*x^(1/3) + a)*sin(b*x^(1/3) + a) - 8*b^2*sqrt(x))/b^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [C]  time = 1.17301, size = 167, normalized size = 1.64 \begin{align*} \sqrt{x} - \frac{3 i \, x^{\frac{1}{6}} e^{\left (2 i \, b x^{\frac{1}{3}} + 2 i \, a\right )}}{8 \, b} + \frac{3 i \, x^{\frac{1}{6}} e^{\left (-2 i \, b x^{\frac{1}{3}} - 2 i \, a\right )}}{8 \, b} - \frac{3 i \, \sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x^{\frac{1}{6}}{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (2 i \, a\right )}}{16 \, b^{\frac{3}{2}}{\left (-\frac{i \, b}{{\left | b \right |}} + 1\right )}} + \frac{3 i \, \sqrt{\pi } \operatorname{erf}\left (-\sqrt{b} x^{\frac{1}{6}}{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}\right ) e^{\left (-2 i \, a\right )}}{16 \, b^{\frac{3}{2}}{\left (\frac{i \, b}{{\left | b \right |}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x) - 3/8*I*x^(1/6)*e^(2*I*b*x^(1/3) + 2*I*a)/b + 3/8*I*x^(1/6)*e^(-2*I*b*x^(1/3) - 2*I*a)/b - 3/16*I*sqrt
(pi)*erf(-sqrt(b)*x^(1/6)*(-I*b/abs(b) + 1))*e^(2*I*a)/(b^(3/2)*(-I*b/abs(b) + 1)) + 3/16*I*sqrt(pi)*erf(-sqrt
(b)*x^(1/6)*(I*b/abs(b) + 1))*e^(-2*I*a)/(b^(3/2)*(I*b/abs(b) + 1))