3.33 \(\int \frac{\cos ^2(a+b x^2)}{x^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{i e^{2 i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{2^{3/4} \left (-i b x^2\right )^{3/4}}+\frac{i e^{-2 i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{2^{3/4} \left (i b x^2\right )^{3/4}}-\frac{\cos \left (2 \left (a+b x^2\right )\right )}{\sqrt{x}}-\frac{1}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.163031, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3402, 3404, 3388, 3389, 2218} \[ -\frac{i e^{2 i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{2^{3/4} \left (-i b x^2\right )^{3/4}}+\frac{i e^{-2 i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{2^{3/4} \left (i b x^2\right )^{3/4}}-\frac{\cos \left (2 \left (a+b x^2\right )\right )}{\sqrt{x}}-\frac{1}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2/x^(3/2),x]

[Out]

-(1/Sqrt[x]) - Cos[2*(a + b*x^2)]/Sqrt[x] - (I*b*E^((2*I)*a)*x^(3/2)*Gamma[3/4, (-2*I)*b*x^2])/(2^(3/4)*((-I)*
b*x^2)^(3/4)) + (I*b*x^(3/2)*Gamma[3/4, (2*I)*b*x^2])/(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 3388

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

Rule 3389

Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[I/2, Int[(e*x)^m*E^(-(c*I) - d*I*x^n),
x], x] - Dist[I/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 \frac{\cos ^2\left (a+b x^2\right )}{x^{3/2}} \, dx &=2 \operatorname{Subst}\left (\int \frac{\cos ^2\left (a+b x^4\right )}{x^2} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{1}{2 x^2}+\frac{\cos \left (2 a+2 b x^4\right )}{2 x^2}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{\sqrt{x}}+\operatorname{Subst}\left (\int \frac{\cos \left (2 a+2 b x^4\right )}{x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{\sqrt{x}}-\frac{\cos \left (2 \left (a+b x^2\right )\right )}{\sqrt{x}}-(8 b) \operatorname{Subst}\left (\int x^2 \sin \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{\sqrt{x}}-\frac{\cos \left (2 \left (a+b x^2\right )\right )}{\sqrt{x}}-(4 i b) \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )+(4 i b) \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{\sqrt{x}}-\frac{\cos \left (2 \left (a+b x^2\right )\right )}{\sqrt{x}}-\frac{i b e^{2 i a} x^{3/2} \Gamma \left (\frac{3}{4},-2 i b x^2\right )}{2^{3/4} \left (-i b x^2\right )^{3/4}}+\frac{i b e^{-2 i a} x^{3/2} \Gamma \left (\frac{3}{4},2 i b x^2\right )}{2^{3/4} \left (i b x^2\right )^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.339617, size = 137, normalized size = 1.17 \[ \frac{\sqrt [4]{2} b x^2 \left (i b x^2\right )^{3/4} (\sin (2 a)-i \cos (2 a)) \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )+i \sqrt [4]{2} \left (-i b x^2\right )^{7/4} (\sin (2 a)+i \cos (2 a)) \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )-4 \left (b^2 x^4\right )^{3/4} \cos ^2\left (a+b x^2\right )}{2 \sqrt{x} \left (b^2 x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2/x^(3/2),x]

[Out]

(-4*(b^2*x^4)^(3/4)*Cos[a + b*x^2]^2 + 2^(1/4)*b*x^2*(I*b*x^2)^(3/4)*Gamma[3/4, (-2*I)*b*x^2]*((-I)*Cos[2*a] +
Sin[2*a]) + I*2^(1/4)*((-I)*b*x^2)^(7/4)*Gamma[3/4, (2*I)*b*x^2]*(I*Cos[2*a] + Sin[2*a]))/(2*Sqrt[x]*(b^2*x^4
)^(3/4))

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}{x}^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.34728, size = 369, normalized size = 3.15 \begin{align*} -\frac{2^{\frac{1}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}{\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 )} + 16}{16 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2^(1/4)*(x^2*abs(b))^(1/4)*(((gamma(-1/4, 2*I*b*x^2) + gamma(-1/4, -2*I*b*x^2))*cos(1/8*pi + 1/4*arctan
2(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*gam
ma(-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)) + (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)
) - (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)) + 16)/sqrt(
x)

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Fricas [A]  time = 1.76632, size = 188, normalized size = 1.61 \begin{align*} \frac{\left (2 i \, b\right )^{\frac{1}{4}} x e^{\left (-2 i \, a\right )} \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \left (-2 i \, b\right )^{\frac{1}{4}} x e^{\left (2 i \, a\right )} \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right ) - 4 \, \sqrt{x} \cos \left (b x^{2} + a\right )^{2}}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}}{x^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x**2)**2/x**(3/2), 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}}{x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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