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

Optimal. Leaf size=100 \[ -\frac{e^{2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{8\ 2^{3/4} \left (-i b x^2\right )^{3/4}}-\frac{e^{-2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{8\ 2^{3/4} \left (i b x^2\right )^{3/4}}+\frac{x^{3/2}}{3} \]

[Out]

x^(3/2)/3 - (E^((2*I)*a)*x^(3/2)*Gamma[3/4, (-2*I)*b*x^2])/(8*2^(3/4)*((-I)*b*x^2)^(3/4)) - (x^(3/2)*Gamma[3/4
, (2*I)*b*x^2])/(8*2^(3/4)*E^((2*I)*a)*(I*b*x^2)^(3/4))

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Rubi [A]  time = 0.130385, antiderivative size = 100, 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, 3404, 3390, 2218} \[ -\frac{e^{2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{8\ 2^{3/4} \left (-i b x^2\right )^{3/4}}-\frac{e^{-2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{8\ 2^{3/4} \left (i b x^2\right )^{3/4}}+\frac{x^{3/2}}{3} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Cos[a + b*x^2]^2,x]

[Out]

x^(3/2)/3 - (E^((2*I)*a)*x^(3/2)*Gamma[3/4, (-2*I)*b*x^2])/(8*2^(3/4)*((-I)*b*x^2)^(3/4)) - (x^(3/2)*Gamma[3/4
, (2*I)*b*x^2])/(8*2^(3/4)*E^((2*I)*a)*(I*b*x^2)^(3/4))

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 3404

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 3390

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Dist[1/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] + Dist[1/2, Int[(e*x)^m*E^(c*I + d*I*x^n), x], x] /; FreeQ[{c, d, e, m}, x] && IGtQ[n, 0]

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \sqrt{x} \cos ^2\left (a+b x^2\right ) \, dx &=2 \operatorname{Subst}\left (\int x^2 \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{x^2}{2}+\frac{1}{2} x^2 \cos \left (2 a+2 b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2}}{3}+\operatorname{Subst}\left (\int x^2 \cos \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2}}{3}+\frac{1}{2} \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2}}{3}-\frac{e^{2 i a} x^{3/2} \Gamma \left (\frac{3}{4},-2 i b x^2\right )}{8\ 2^{3/4} \left (-i b x^2\right )^{3/4}}-\frac{e^{-2 i a} x^{3/2} \Gamma \left (\frac{3}{4},2 i b x^2\right )}{8\ 2^{3/4} \left (i b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.226213, size = 122, normalized size = 1.22 \[ \frac{x^{3/2} \left (-3 \sqrt [4]{2} \left (-i b x^2\right )^{3/4} (\cos (2 a)-i \sin (2 a)) \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )-3 \sqrt [4]{2} \left (i b x^2\right )^{3/4} (\cos (2 a)+i \sin (2 a)) \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )+16 \left (b^2 x^4\right )^{3/4}\right )}{48 \left (b^2 x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Cos[a + b*x^2]^2,x]

[Out]

(x^(3/2)*(16*(b^2*x^4)^(3/4) - 3*2^(1/4)*((-I)*b*x^2)^(3/4)*Gamma[3/4, (2*I)*b*x^2]*(Cos[2*a] - I*Sin[2*a]) -
3*2^(1/4)*(I*b*x^2)^(3/4)*Gamma[3/4, (-2*I)*b*x^2]*(Cos[2*a] + I*Sin[2*a])))/(48*(b^2*x^4)^(3/4))

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Maple [F]  time = 0.085, size = 0, normalized size = 0. \begin{align*} \int \sqrt{x} \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.4219, size = 392, normalized size = 3.92 \begin{align*} \frac{32 \, x^{2}{\left | b \right |} - 2^{\frac{1}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}{\left ({\left (3 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 3 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (3 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (-3 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + 3 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left ({\left (3 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (3 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 3 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) - 3 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )}}{96 \, \sqrt{x}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*x^2*abs(b) - 2^(1/4)*(x^2*abs(b))^(1/4)*((3*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*cos(3/8*
pi + 3/4*arctan2(0, b)) + 3*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b))
- (3*I*gamma(3/4, 2*I*b*x^2) - 3*I*gamma(3/4, -2*I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) - (-3*I*gamma(3/4,
2*I*b*x^2) + 3*I*gamma(3/4, -2*I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*cos(2*a) - ((3*I*gamma(3/4, 2*I*b*x
^2) - 3*I*gamma(3/4, -2*I*b*x^2))*cos(3/8*pi + 3/4*arctan2(0, b)) + (3*I*gamma(3/4, 2*I*b*x^2) - 3*I*gamma(3/4
, -2*I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b)) + 3*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*sin(3/8*p
i + 3/4*arctan2(0, b)) - 3*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*
sin(2*a)))/(sqrt(x)*abs(b))

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Fricas [A]  time = 1.7438, size = 176, normalized size = 1.76 \begin{align*} \frac{16 \, b x^{\frac{3}{2}} + 3 i \, \left (2 i \, b\right )^{\frac{1}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \left (-2 i \, b\right )^{\frac{1}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*b*x^(3/2) + 3*I*(2*I*b)^(1/4)*e^(-2*I*a)*gamma(3/4, 2*I*b*x^2) - 3*I*(-2*I*b)^(1/4)*e^(2*I*a)*gamma(3
/4, -2*I*b*x^2))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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