3.1.34 \(\int \frac {\cos ^2(a+b x^2)}{x^{5/2}} \, dx\) [34]

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

[Out]

-2/3*cos(b*x^2+a)^2/x^(3/2)-1/6*I*b*exp(2*I*a)*GAMMA(1/4,-2*I*b*x^2)*x^(1/2)*2^(3/4)/(-I*b*x^2)^(1/4)+1/6*I*b*
GAMMA(1/4,2*I*b*x^2)*x^(1/2)*2^(3/4)/exp(2*I*a)/(I*b*x^2)^(1/4)

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Rubi [A]
time = 0.06, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 = {3483, 3475, 4669, 3454, 3436, 2239} \begin {gather*} -\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}} \end {gather*}

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 2239

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]

Rule 3436

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 3454

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 3475

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 3483

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 4669

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]

Rubi steps

\begin {align*} \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx &=2 \text {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) \text {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) \text {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) \text {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) \text {Subst}\left (\int e^{-2 i a-2 i b x^4} \, dx,x,\sqrt {x}\right )+\frac {1}{3} (4 i b) \text {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*}

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Mathematica [A]
time = 0.24, size = 137, normalized size = 1.18 \begin {gather*} \frac {-4 \sqrt [4]{b^2 x^4} \cos ^2\left (a+b x^2\right )+2^{3/4} b x^2 \sqrt [4]{i b x^2} \text {Gamma}\left (\frac {1}{4},-2 i b x^2\right ) (-i \cos (2 a)+\sin (2 a))+i 2^{3/4} \left (-i b x^2\right )^{5/4} \text {Gamma}\left (\frac {1}{4},2 i b x^2\right ) (i \cos (2 a)+\sin (2 a))}{6 x^{3/2} \sqrt [4]{b^2 x^4}} \end {gather*}

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 = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{2}\left (b \,x^{2}+a \right )}{x^{\frac {5}{2}}}\, dx\]

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 [A]
time = 0.13, size = 146, normalized size = 1.26 \begin {gather*} -\frac {3 \cdot 2^{\frac {3}{4}} \left (b x^{2}\right )^{\frac {3}{4}} {\left ({\left (\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 (-i \, \Gamma \left (-\frac {3}{4}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (-\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \cos \left (2 \, a\right ) + {\left (\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 (i \, \Gamma \left (-\frac {3}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (-\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \sin \left (2 \, a\right )\right )} + 16}{48 \, x^{\frac {3}{2}}} \end {gather*}

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*(3*2^(3/4)*(b*x^2)^(3/4)*((sqrt(-sqrt(2) + 2)*(gamma(-3/4, 2*I*b*x^2) + gamma(-3/4, -2*I*b*x^2)) - sqrt(
sqrt(2) + 2)*(-I*gamma(-3/4, 2*I*b*x^2) + I*gamma(-3/4, -2*I*b*x^2)))*cos(2*a) + (sqrt(sqrt(2) + 2)*(gamma(-3/
4, 2*I*b*x^2) + gamma(-3/4, -2*I*b*x^2)) - sqrt(-sqrt(2) + 2)*(I*gamma(-3/4, 2*I*b*x^2) - I*gamma(-3/4, -2*I*b
*x^2)))*sin(2*a)) + 16)/x^(3/2)

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Fricas [A]
time = 0.11, size = 63, normalized size = 0.54 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos ^{2}{\left (a + b x^{2} \right )}}{x^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\cos \left (b\,x^2+a\right )}^2}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x^2)^2/x^(5/2),x)

[Out]

int(cos(a + b*x^2)^2/x^(5/2), x)

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