3.32 \(\int \frac{\cos ^2(a+b x^2)}{\sqrt{x}} \, dx\)

Optimal. Leaf size=96 \[ -\frac{e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}+\sqrt{x} \]

[Out]

Sqrt[x] - (E^((2*I)*a)*Sqrt[x]*Gamma[1/4, (-2*I)*b*x^2])/(8*2^(1/4)*((-I)*b*x^2)^(1/4)) - (Sqrt[x]*Gamma[1/4,
(2*I)*b*x^2])/(8*2^(1/4)*E^((2*I)*a)*(I*b*x^2)^(1/4))

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Rubi [A]  time = 0.0681866, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3402, 3358, 3356, 2208} \[ -\frac{e^{2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}+\sqrt{x} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2/Sqrt[x],x]

[Out]

Sqrt[x] - (E^((2*I)*a)*Sqrt[x]*Gamma[1/4, (-2*I)*b*x^2])/(8*2^(1/4)*((-I)*b*x^2)^(1/4)) - (Sqrt[x]*Gamma[1/4,
(2*I)*b*x^2])/(8*2^(1/4)*E^((2*I)*a)*(I*b*x^2)^(1/4))

Rule 3402

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_.)*(x_))^(m_), x_Symbol] :> With[{k = Denominator[m
]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + b*Cos[c + (d*x^(k*n))/e^n])^p, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e}, x] && IntegerQ[p] && IGtQ[n, 0] && FractionQ[m]

Rule 3358

Int[((a_.) + Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)]*(b_.))^(p_), x_Symbol] :> Int[ExpandTrigReduce[(a +
b*Cos[c + d*(e + f*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 1] && IGtQ[n, 1]

Rule 3356

Int[Cos[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[1/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
 x] + Dist[1/2, Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]

Rule 2208

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> -Simp[(F^a*(c + d*x)*Gamma[1/n, -(b*(c + d*x)
^n*Log[F])])/(d*n*(-(b*(c + d*x)^n*Log[F]))^(1/n)), x] /; FreeQ[{F, a, b, c, d, n}, x] &&  !IntegerQ[2/n]

Rubi steps

\begin{align*} \int \frac{\cos ^2\left (a+b x^2\right )}{\sqrt{x}} \, dx &=2 \operatorname{Subst}\left (\int \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{1}{2} \cos \left (2 a+2 b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}+\operatorname{Subst}\left (\int \cos \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}+\frac{1}{2} \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} \, dx,x,\sqrt{x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} \, dx,x,\sqrt{x}\right )\\ &=\sqrt{x}-\frac{e^{2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac{e^{-2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.232448, size = 120, normalized size = 1.25 \[ -\frac{\sqrt{x} \left (2^{3/4} \sqrt [4]{-i b x^2} (\cos (2 a)-i \sin (2 a)) \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )+2^{3/4} \sqrt [4]{i b x^2} (\cos (2 a)+i \sin (2 a)) \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )-16 \sqrt [4]{b^2 x^4}\right )}{16 \sqrt [4]{b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2/Sqrt[x],x]

[Out]

-(Sqrt[x]*(-16*(b^2*x^4)^(1/4) + 2^(3/4)*((-I)*b*x^2)^(1/4)*Gamma[1/4, (2*I)*b*x^2]*(Cos[2*a] - I*Sin[2*a]) +
2^(3/4)*(I*b*x^2)^(1/4)*Gamma[1/4, (-2*I)*b*x^2]*(Cos[2*a] + I*Sin[2*a])))/(16*(b^2*x^4)^(1/4))

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Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}{\frac{1}{\sqrt{x}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.45108, size = 393, normalized size = 4.09 \begin{align*} -\frac{2^{\frac{3}{4}}{\left ({\left ({\left ({\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (-i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left ({\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )} \sqrt{x} - 16 \cdot 2^{\frac{1}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}} \sqrt{x}\right )}}{32 \, \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*2^(3/4)*((((gamma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*cos(1/8*pi + 1/4*arctan2(0, b)) + (gamma(1/4
, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) - (I*gamma(1/4, 2*I*b*x^2) - I*gamma(1
/4, -2*I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) - (-I*gamma(1/4, 2*I*b*x^2) + I*gamma(1/4, -2*I*b*x^2))*sin(-
1/8*pi + 1/4*arctan2(0, b)))*cos(2*a) - ((I*gamma(1/4, 2*I*b*x^2) - I*gamma(1/4, -2*I*b*x^2))*cos(1/8*pi + 1/4
*arctan2(0, b)) + (I*gamma(1/4, 2*I*b*x^2) - I*gamma(1/4, -2*I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + (gam
ma(1/4, 2*I*b*x^2) + gamma(1/4, -2*I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) - (gamma(1/4, 2*I*b*x^2) + gamma(
1/4, -2*I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(0, b)))*sin(2*a))*sqrt(x) - 16*2^(1/4)*(x^2*abs(b))^(1/4)*sqrt(x))
/(x^2*abs(b))^(1/4)

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Fricas [A]  time = 1.76987, size = 170, normalized size = 1.77 \begin{align*} \frac{i \, \left (2 i \, b\right )^{\frac{3}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) - i \, \left (-2 i \, b\right )^{\frac{3}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right ) + 16 \, b \sqrt{x}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(I*(2*I*b)^(3/4)*e^(-2*I*a)*gamma(1/4, 2*I*b*x^2) - I*(-2*I*b)^(3/4)*e^(2*I*a)*gamma(1/4, -2*I*b*x^2) + 1
6*b*sqrt(x))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x^2 + a)^2/sqrt(x), x)