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

Optimal. Leaf size=98 \[ -\frac{i e^{i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},-i b x^2\right )}{\left (-i b x^2\right )^{3/4}}+\frac{i e^{-i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},i b x^2\right )}{\left (i b x^2\right )^{3/4}}-\frac{2 \cos \left (a+b x^2\right )}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0758502, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3388, 3389, 2218} \[ -\frac{i e^{i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},-i b x^2\right )}{\left (-i b x^2\right )^{3/4}}+\frac{i e^{-i a} b x^{3/2} \text{Gamma}\left (\frac{3}{4},i b x^2\right )}{\left (i b x^2\right )^{3/4}}-\frac{2 \cos \left (a+b x^2\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/x^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 0.182243, size = 114, normalized size = 1.16 \[ \frac{b x^2 \left (i b x^2\right )^{3/4} (\sin (a)-i \cos (a)) \text{Gamma}\left (\frac{3}{4},-i b x^2\right )+i \left (-i b x^2\right )^{7/4} (\sin (a)+i \cos (a)) \text{Gamma}\left (\frac{3}{4},i b x^2\right )-2 \left (b^2 x^4\right )^{3/4} \cos \left (a+b x^2\right )}{\sqrt{x} \left (b^2 x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x^(3/2),x]

[Out]

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

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Maple [C]  time = 0.067, size = 338, normalized size = 3.5 \begin{align*}{\frac{\cos \left ( a \right ) \sqrt{\pi }{2}^{{\frac{3}{4}}}}{8}\sqrt [8]{{b}^{2}} \left ( -12\,{\frac{\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{\sqrt{\pi }{x}^{5/2}\sqrt [8]{{b}^{2}}b} \left ({\frac{8\,{b}^{2}{x}^{4}}{21}}+2/3 \right ) }-8\,{\frac{\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{\sqrt{\pi }{x}^{5/2}\sqrt [8]{{b}^{2}}b}}+{\frac{32\,{b}^{2}\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{7\,\sqrt{\pi }}{x}^{{\frac{7}{2}}}{\it LommelS1} \left ({\frac{5}{4}},{\frac{3}{2}},b{x}^{2} \right ){\frac{1}{\sqrt [8]{{b}^{2}}}} \left ( b{x}^{2} \right ) ^{-{\frac{5}{4}}}}+8\,{\frac{{x}^{7/2}{b}^{2}\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ){\it LommelS1} \left ( 1/4,1/2,b{x}^{2} \right ) }{\sqrt{\pi }\sqrt [8]{{b}^{2}} \left ( b{x}^{2} \right ) ^{9/4}}} \right ) }-{\frac{\sin \left ( a \right ) \sqrt{\pi }{2}^{{\frac{3}{4}}}}{8}\sqrt [4]{b} \left ({\frac{8\,\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{3\,\sqrt{\pi }}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt [4]{b}}}}+{\frac{32\,\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{3\,\sqrt{\pi }}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt [4]{b}}}}-{\frac{8\,\sqrt [4]{2}\sin \left ( b{x}^{2} \right ) }{3\,\sqrt{\pi }}{x}^{{\frac{7}{2}}}{b}^{{\frac{7}{4}}}{\it LommelS1} \left ({\frac{1}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{5}{4}}}}-{\frac{32\,\sqrt [4]{2} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{3\,\sqrt{\pi }}{x}^{{\frac{7}{2}}}{b}^{{\frac{7}{4}}}{\it LommelS1} \left ({\frac{5}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{9}{4}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*cos(a)*Pi^(1/2)*2^(3/4)*(b^2)^(1/8)*(-12/Pi^(1/2)/x^(5/2)*2^(1/4)/(b^2)^(1/8)*(8/21*b^2*x^4+2/3)*sin(b*x^2
)/b-8/Pi^(1/2)/x^(5/2)*2^(1/4)/(b^2)^(1/8)/b*(cos(b*x^2)*b*x^2-sin(b*x^2))+32/7/Pi^(1/2)*x^(7/2)/(b^2)^(1/8)*b
^2*2^(1/4)/(b*x^2)^(5/4)*sin(b*x^2)*LommelS1(5/4,3/2,b*x^2)+8/Pi^(1/2)*x^(7/2)/(b^2)^(1/8)*b^2*2^(1/4)/(b*x^2)
^(9/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(1/4,1/2,b*x^2))-1/8*sin(a)*Pi^(1/2)*2^(3/4)*b^(1/4)*(8/3/Pi^(1/2
)/x^(1/2)*2^(1/4)/b^(1/4)*sin(b*x^2)+32/3/Pi^(1/2)/x^(1/2)*2^(1/4)/b^(1/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))-8/3/P
i^(1/2)*x^(7/2)*b^(7/4)*2^(1/4)/(b*x^2)^(5/4)*sin(b*x^2)*LommelS1(1/4,3/2,b*x^2)-32/3/Pi^(1/2)*x^(7/2)*b^(7/4)
*2^(1/4)/(b*x^2)^(9/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(5/4,1/2,b*x^2))

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Maxima [B]  time = 1.41277, size = 355, normalized size = 3.62 \begin{align*} -\frac{\left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}{\left ({\left ({\left (\Gamma \left (-\frac{1}{4}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) - i \, \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) +{\left ({\left (-i \, \Gamma \left (-\frac{1}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{4}, -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}, i \, b x^{2}\right ) + \Gamma \left (-\frac{1}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{1}{8} \, \pi + \frac{1}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )}}{8 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(x^2*abs(b))^(1/4)*(((gamma(-1/4, I*b*x^2) + gamma(-1/4, -I*b*x^2))*cos(1/8*pi + 1/4*arctan2(0, b)) + (ga
mma(-1/4, I*b*x^2) + gamma(-1/4, -I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + (I*gamma(-1/4, I*b*x^2) - I*gam
ma(-1/4, -I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) + (-I*gamma(-1/4, I*b*x^2) + I*gamma(-1/4, -I*b*x^2))*sin(
-1/8*pi + 1/4*arctan2(0, b)))*cos(a) + ((-I*gamma(-1/4, I*b*x^2) + I*gamma(-1/4, -I*b*x^2))*cos(1/8*pi + 1/4*a
rctan2(0, b)) + (-I*gamma(-1/4, I*b*x^2) + I*gamma(-1/4, -I*b*x^2))*cos(-1/8*pi + 1/4*arctan2(0, b)) + (gamma(
-1/4, I*b*x^2) + gamma(-1/4, -I*b*x^2))*sin(1/8*pi + 1/4*arctan2(0, b)) - (gamma(-1/4, I*b*x^2) + gamma(-1/4,
-I*b*x^2))*sin(-1/8*pi + 1/4*arctan2(0, b)))*sin(a))/sqrt(x)

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Fricas [A]  time = 1.69499, size = 163, normalized size = 1.66 \begin{align*} \frac{\left (i \, b\right )^{\frac{1}{4}} x e^{\left (-i \, a\right )} \Gamma \left (\frac{3}{4}, i \, b x^{2}\right ) + \left (-i \, b\right )^{\frac{1}{4}} x e^{\left (i \, a\right )} \Gamma \left (\frac{3}{4}, -i \, b x^{2}\right ) - 2 \, \sqrt{x} \cos \left (b x^{2} + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\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)/x**(3/2),x)

[Out]

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

[Out]

integrate(cos(b*x^2 + a)/x^(3/2), x)