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

Optimal. Leaf size=104 \[ -\frac{i e^{i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},-i b x^2\right )}{3 \sqrt [4]{-i b x^2}}+\frac{i e^{-i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},i b x^2\right )}{3 \sqrt [4]{i b x^2}}-\frac{2 \cos \left (a+b x^2\right )}{3 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0748721, antiderivative size = 104, 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} \[ -\frac{i e^{i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},-i b x^2\right )}{3 \sqrt [4]{-i b x^2}}+\frac{i e^{-i a} b \sqrt{x} \text{Gamma}\left (\frac{1}{4},i b x^2\right )}{3 \sqrt [4]{i b x^2}}-\frac{2 \cos \left (a+b x^2\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x^2]/x^(5/2),x]

[Out]

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

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^{5/2}} \, dx &=-\frac{2 \cos \left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (4 b) \int \frac{\sin \left (a+b x^2\right )}{\sqrt{x}} \, dx\\ &=-\frac{2 \cos \left (a+b x^2\right )}{3 x^{3/2}}-\frac{1}{3} (2 i b) \int \frac{e^{-i a-i b x^2}}{\sqrt{x}} \, dx+\frac{1}{3} (2 i b) \int \frac{e^{i a+i b x^2}}{\sqrt{x}} \, dx\\ &=-\frac{2 \cos \left (a+b x^2\right )}{3 x^{3/2}}-\frac{i b e^{i a} \sqrt{x} \Gamma \left (\frac{1}{4},-i b x^2\right )}{3 \sqrt [4]{-i b x^2}}+\frac{i b e^{-i a} \sqrt{x} \Gamma \left (\frac{1}{4},i b x^2\right )}{3 \sqrt [4]{i b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.156479, size = 117, normalized size = 1.12 \[ \frac{b x^2 \sqrt [4]{i b x^2} (\sin (a)-i \cos (a)) \text{Gamma}\left (\frac{1}{4},-i b x^2\right )+i \left (-i b x^2\right )^{5/4} (\sin (a)+i \cos (a)) \text{Gamma}\left (\frac{1}{4},i b x^2\right )-2 \sqrt [4]{b^2 x^4} \cos \left (a+b x^2\right )}{3 x^{3/2} \sqrt [4]{b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x^2]/x^(5/2),x]

[Out]

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

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Maple [C]  time = 0.069, size = 358, normalized size = 3.4 \begin{align*}{\frac{\cos \left ( a \right ) \sqrt{\pi }\sqrt [4]{2}}{8} \left ({b}^{2} \right ) ^{{\frac{3}{8}}} \left ( -4\,{\frac{{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{\sqrt{\pi }{x}^{7/2} \left ({b}^{2} \right ) ^{3/8}b} \left ({\frac{8\,{b}^{2}{x}^{4}}{15}}+2/3 \right ) }-{\frac{8\,{2}^{3/4} \left ( -16\,{b}^{2}{x}^{4}+5 \right ) \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{15\,\sqrt{\pi }b}{x}^{-{\frac{7}{2}}} \left ({b}^{2} \right ) ^{-{\frac{3}{8}}}}+{\frac{32\,{2}^{3/4}{b}^{3}\sin \left ( b{x}^{2} \right ) }{15\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{\it LommelS1} \left ({\frac{3}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ({b}^{2} \right ) ^{-{\frac{3}{8}}} \left ( b{x}^{2} \right ) ^{-{\frac{7}{4}}}}-{\frac{128\,{2}^{3/4}{b}^{3} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{15\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{\it LommelS1} \left ({\frac{7}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ({b}^{2} \right ) ^{-{\frac{3}{8}}} \left ( b{x}^{2} \right ) ^{-{\frac{11}{4}}}} \right ) }-{\frac{\sin \left ( a \right ) \sqrt{\pi }\sqrt [4]{2}}{8}{b}^{{\frac{3}{4}}} \left ( 12\,{\frac{{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{\sqrt{\pi }{x}^{3/2}{b}^{3/4}} \left ({\frac{32\,{b}^{2}{x}^{4}}{81}}+2/3 \right ) }+{\frac{32\,{2}^{3/4} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{3\,\sqrt{\pi }}{x}^{-{\frac{3}{2}}}{b}^{-{\frac{3}{4}}}}-{\frac{128\,{2}^{3/4}\sin \left ( b{x}^{2} \right ) }{27\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{b}^{{\frac{9}{4}}}{\it LommelS1} \left ({\frac{7}{4}},{\frac{3}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{7}{4}}}}-{\frac{32\,{2}^{3/4} \left ( \cos \left ( b{x}^{2} \right ) b{x}^{2}-\sin \left ( b{x}^{2} \right ) \right ) }{3\,\sqrt{\pi }}{x}^{{\frac{9}{2}}}{b}^{{\frac{9}{4}}}{\it LommelS1} \left ({\frac{3}{4}},{\frac{1}{2}},b{x}^{2} \right ) \left ( b{x}^{2} \right ) ^{-{\frac{11}{4}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*cos(a)*Pi^(1/2)*2^(1/4)*(b^2)^(3/8)*(-4/Pi^(1/2)/x^(7/2)*2^(3/4)/(b^2)^(3/8)*(8/15*b^2*x^4+2/3)*sin(b*x^2)
/b-8/15/Pi^(1/2)/x^(7/2)*2^(3/4)/(b^2)^(3/8)/b*(-16*b^2*x^4+5)*(cos(b*x^2)*b*x^2-sin(b*x^2))+32/15/Pi^(1/2)*x^
(9/2)/(b^2)^(3/8)*2^(3/4)*b^3/(b*x^2)^(7/4)*sin(b*x^2)*LommelS1(3/4,3/2,b*x^2)-128/15/Pi^(1/2)*x^(9/2)/(b^2)^(
3/8)*2^(3/4)*b^3/(b*x^2)^(11/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(7/4,1/2,b*x^2))-1/8*sin(a)*Pi^(1/2)*2^(
1/4)*b^(3/4)*(12/Pi^(1/2)/x^(3/2)*2^(3/4)/b^(3/4)*(32/81*b^2*x^4+2/3)*sin(b*x^2)+32/3/Pi^(1/2)/x^(3/2)*2^(3/4)
/b^(3/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))-128/27/Pi^(1/2)*x^(9/2)*b^(9/4)*2^(3/4)/(b*x^2)^(7/4)*sin(b*x^2)*Lommel
S1(7/4,3/2,b*x^2)-32/3/Pi^(1/2)*x^(9/2)*b^(9/4)*2^(3/4)/(b*x^2)^(11/4)*(cos(b*x^2)*b*x^2-sin(b*x^2))*LommelS1(
3/4,1/2,b*x^2))

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Maxima [B]  time = 1.44541, size = 355, normalized size = 3.41 \begin{align*} -\frac{\left (x^{2}{\left | b \right |}\right )^{\frac{3}{4}}{\left ({\left ({\left (\Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (i \, \Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) - i \, \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (a\right ) +{\left ({\left (-i \, \Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (-i \, \Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + i \, \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (\Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (\Gamma \left (-\frac{3}{4}, i \, b x^{2}\right ) + \Gamma \left (-\frac{3}{4}, -i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (a\right )\right )}}{8 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(x^2*abs(b))^(3/4)*(((gamma(-3/4, I*b*x^2) + gamma(-3/4, -I*b*x^2))*cos(3/8*pi + 3/4*arctan2(0, b)) + (ga
mma(-3/4, I*b*x^2) + gamma(-3/4, -I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b)) + (I*gamma(-3/4, I*b*x^2) - I*gam
ma(-3/4, -I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) + (-I*gamma(-3/4, I*b*x^2) + I*gamma(-3/4, -I*b*x^2))*sin(
-3/8*pi + 3/4*arctan2(0, b)))*cos(a) + ((-I*gamma(-3/4, I*b*x^2) + I*gamma(-3/4, -I*b*x^2))*cos(3/8*pi + 3/4*a
rctan2(0, b)) + (-I*gamma(-3/4, I*b*x^2) + I*gamma(-3/4, -I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b)) + (gamma(
-3/4, I*b*x^2) + gamma(-3/4, -I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) - (gamma(-3/4, I*b*x^2) + gamma(-3/4,
-I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*sin(a))/x^(3/2)

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Fricas [A]  time = 1.78771, size = 177, normalized size = 1.7 \begin{align*} \frac{\left (i \, b\right )^{\frac{3}{4}} x^{2} e^{\left (-i \, a\right )} \Gamma \left (\frac{1}{4}, i \, b x^{2}\right ) + \left (-i \, b\right )^{\frac{3}{4}} x^{2} e^{\left (i \, a\right )} \Gamma \left (\frac{1}{4}, -i \, b x^{2}\right ) - 2 \, \sqrt{x} \cos \left (b x^{2} + a\right )}{3 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x**2)/x**(5/2), 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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x^2 + a)/x^(5/2), x)