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

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

[Out]

1/3*x^(3/2)-1/16*exp(2*I*a)*x^(3/2)*GAMMA(3/4,-2*I*b*x^2)*2^(1/4)/(-I*b*x^2)^(3/4)-1/16*x^(3/2)*GAMMA(3/4,2*I*
b*x^2)*2^(1/4)/exp(2*I*a)/(I*b*x^2)^(3/4)

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Rubi [A]  time = 0.13, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.22, size = 122, normalized size = 1.22 \[ \frac {x^{3/2} \left (-3 \sqrt [4]{2} \left (-i b x^2\right )^{3/4} (\cos (2 a)-i \sin (2 a)) \Gamma \left (\frac {3}{4},2 i b x^2\right )-3 \sqrt [4]{2} \left (i b x^2\right )^{3/4} (\cos (2 a)+i \sin (2 a)) \Gamma \left (\frac {3}{4},-2 i b x^2\right )+16 \left (b^2 x^4\right )^{3/4}\right )}{48 \left (b^2 x^4\right )^{3/4}} \]

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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fricas [A]  time = 0.73, size = 50, normalized size = 0.50 \[ \frac {16 \, b x^{\frac {3}{2}} + 3 i \, \left (2 i \, b\right )^{\frac {1}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \left (-2 i \, b\right )^{\frac {1}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )}{48 \, b} \]

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

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)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \sqrt {x}\, \left (\cos ^{2}\left (b \,x^{2}+a \right )\right )\, dx \]

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 = 2.60, size = 156, normalized size = 1.56 \[ \frac {32 \, b x^{2} - 2^{\frac {1}{4}} \left (b x^{2}\right )^{\frac {1}{4}} {\left ({\left (3 \, \sqrt {-\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} - \sqrt {\sqrt {2} + 2} {\left (3 i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \cos \left (2 \, a\right ) - {\left (3 \, \sqrt {\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} + \sqrt {-\sqrt {2} + 2} {\left (3 i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) - 3 i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \sin \left (2 \, a\right )\right )}}{96 \, b \sqrt {x}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*cos(a + b*x^2)^2,x)

[Out]

int(x^(1/2)*cos(a + b*x^2)^2, x)

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

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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