3.34 \(\int \frac{\cos ^2(a+b x^2)}{x^{5/2}} \, dx\)

Optimal. Leaf size=116 \[ -\frac{i e^{2 i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{-i b x^2}}+\frac{i e^{-2 i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{i b x^2}}-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0997398, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3402, 3394, 4573, 3373, 3355, 2208} \[ -\frac{i e^{2 i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{-i b x^2}}+\frac{i e^{-2 i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{i b x^2}}-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]^2/x^(5/2),x]

[Out]

(-2*Cos[a + b*x^2]^2)/(3*x^(3/2)) - ((I/3)*b*E^((2*I)*a)*Sqrt[x]*Gamma[1/4, (-2*I)*b*x^2])/(2^(1/4)*((-I)*b*x^
2)^(1/4)) + ((I/3)*b*Sqrt[x]*Gamma[1/4, (2*I)*b*x^2])/(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 3394

Int[Cos[(a_.) + (b_.)*(x_)^(n_)]^(p_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*Cos[a + b*x^n]^p)/(m + 1), x] +
Dist[(b*n*p)/(m + 1), Int[Cos[a + b*x^n]^(p - 1)*Sin[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 1] &&
EqQ[m + n, 0] && NeQ[n, 1] && IntegerQ[n]

Rule 4573

Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sin[2*v]^p, x], x] /; EqQ[w, v] && Integ
erQ[p]

Rule 3373

Int[((a_.) + (b_.)*Sin[u_])^(p_.), x_Symbol] :> Int[(a + b*Sin[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p}, x
] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rule 3355

Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Dist[I/2, Int[E^(-(c*I) - d*I*(e + f*x)^n), x],
x] - Dist[I/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 )}{x^{5/2}} \, dx &=2 \operatorname{Subst}\left (\int \frac{\cos ^2\left (a+b x^4\right )}{x^4} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (16 b) \operatorname{Subst}\left (\int \cos \left (a+b x^4\right ) \sin \left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (8 b) \operatorname{Subst}\left (\int \sin \left (2 \left (a+b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (8 b) \operatorname{Subst}\left (\int \sin \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (4 i b) \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} \, dx,x,\sqrt{x}\right )+\frac{1}{3} (4 i b) \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac{i b e^{2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},-2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{-i b x^2}}+\frac{i b e^{-2 i a} \sqrt{x} \Gamma \left (\frac{1}{4},2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{i b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.368087, size = 137, normalized size = 1.18 \[ \frac{2^{3/4} b x^2 \sqrt [4]{i b x^2} (\sin (2 a)-i \cos (2 a)) \text{Gamma}\left (\frac{1}{4},-2 i b x^2\right )+i 2^{3/4} \left (-i b x^2\right )^{5/4} (\sin (2 a)+i \cos (2 a)) \text{Gamma}\left (\frac{1}{4},2 i b x^2\right )-4 \sqrt [4]{b^2 x^4} \cos ^2\left (a+b x^2\right )}{6 x^{3/2} \sqrt [4]{b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]^2/x^(5/2),x]

[Out]

(-4*(b^2*x^4)^(1/4)*Cos[a + b*x^2]^2 + 2^(3/4)*b*x^2*(I*b*x^2)^(1/4)*Gamma[1/4, (-2*I)*b*x^2]*((-I)*Cos[2*a] +
Sin[2*a]) + I*2^(3/4)*((-I)*b*x^2)^(5/4)*Gamma[1/4, (2*I)*b*x^2]*(I*Cos[2*a] + Sin[2*a]))/(6*x^(3/2)*(b^2*x^4
)^(1/4))

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}{x}^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x^2+a)^2/x^(5/2),x)

[Out]

int(cos(b*x^2+a)^2/x^(5/2),x)

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Maxima [B]  time = 1.44658, size = 373, normalized size = 3.22 \begin{align*} -\frac{2^{\frac{3}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{3}{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 )} + 16}{48 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x^(5/2),x, algorithm="maxima")

[Out]

-1/48*(2^(3/4)*(x^2*abs(b))^(3/4)*((3*(gamma(-3/4, 2*I*b*x^2) + gamma(-3/4, -2*I*b*x^2))*cos(3/8*pi + 3/4*arct
an2(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*gamm
a(-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*
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*pi + 3/4*arctan2(0, b)
))*sin(2*a)) + 16)/x^(3/2)

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Fricas [A]  time = 1.74185, size = 196, normalized size = 1.69 \begin{align*} \frac{\left (2 i \, b\right )^{\frac{3}{4}} x^{2} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{1}{4}, 2 i \, b x^{2}\right ) + \left (-2 i \, b\right )^{\frac{3}{4}} x^{2} e^{\left (2 i \, a\right )} \Gamma \left (\frac{1}{4}, -2 i \, b x^{2}\right ) - 4 \, \sqrt{x} \cos \left (b x^{2} + a\right )^{2}}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x**2+a)**2/x**(5/2),x)

[Out]

Timed out

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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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x^2+a)^2/x^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x^2 + a)^2/x^(5/2), x)