\(\int \frac {\cos (a+b x^2)}{x^{3/2}} \, dx\) [27]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F(-2)]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 98 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=-\frac {2 \cos \left (a+b x^2\right )}{\sqrt {x}}-\frac {i b e^{i a} x^{3/2} \Gamma \left (\frac {3}{4},-i b x^2\right )}{\left (-i b x^2\right )^{3/4}}+\frac {i b e^{-i a} x^{3/2} \Gamma \left (\frac {3}{4},i b x^2\right )}{\left (i b x^2\right )^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3469, 3470, 2250} \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=-\frac {2 \cos \left (a+b x^2\right )}{\sqrt {x}}-\frac {i e^{i a} b x^{3/2} \Gamma \left (\frac {3}{4},-i b x^2\right )}{\left (-i b x^2\right )^{3/4}}+\frac {i e^{-i a} b x^{3/2} \Gamma \left (\frac {3}{4},i b x^2\right )}{\left (i b x^2\right )^{3/4}} \]

[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 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 3469

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 3470

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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\frac {-2 \left (b^2 x^4\right )^{3/4} \cos \left (a+b x^2\right )+b x^2 \left (i b x^2\right )^{3/4} \Gamma \left (\frac {3}{4},-i b x^2\right ) (-i \cos (a)+\sin (a))+i \left (-i b x^2\right )^{7/4} \Gamma \left (\frac {3}{4},i b x^2\right ) (i \cos (a)+\sin (a))}{\sqrt {x} \left (b^2 x^4\right )^{3/4}} \]

[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))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.43 (sec) , antiderivative size = 338, normalized size of antiderivative = 3.45

method result size
meijerg \(\frac {\cos \left (a \right ) \sqrt {\pi }\, 2^{\frac {3}{4}} \left (b^{2}\right )^{\frac {1}{8}} \left (-\frac {12 \,2^{\frac {1}{4}} \left (\frac {8 x^{4} b^{2}}{21}+\frac {2}{3}\right ) \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{\frac {5}{2}} \left (b^{2}\right )^{\frac {1}{8}} b}-\frac {8 \,2^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{\sqrt {\pi }\, x^{\frac {5}{2}} \left (b^{2}\right )^{\frac {1}{8}} b}+\frac {32 x^{\frac {7}{2}} b^{2} 2^{\frac {1}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {5}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{7 \sqrt {\pi }\, \left (b^{2}\right )^{\frac {1}{8}} \left (b \,x^{2}\right )^{\frac {5}{4}}}+\frac {8 x^{\frac {7}{2}} b^{2} 2^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {1}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{\sqrt {\pi }\, \left (b^{2}\right )^{\frac {1}{8}} \left (b \,x^{2}\right )^{\frac {9}{4}}}\right )}{8}-\frac {\sin \left (a \right ) \sqrt {\pi }\, 2^{\frac {3}{4}} b^{\frac {1}{4}} \left (\frac {8 \,2^{\frac {1}{4}} \sin \left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \sqrt {x}\, b^{\frac {1}{4}}}+\frac {32 \,2^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{3 \sqrt {\pi }\, \sqrt {x}\, b^{\frac {1}{4}}}-\frac {8 x^{\frac {7}{2}} b^{\frac {7}{4}} 2^{\frac {1}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {1}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {5}{4}}}-\frac {32 x^{\frac {7}{2}} b^{\frac {7}{4}} 2^{\frac {1}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {5}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {9}{4}}}\right )}{8}\) \(338\)

[In]

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

[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*x^4*b^2+2/3)/b*sin(b*x
^2)-8/Pi^(1/2)/x^(5/2)*2^(1/4)/(b^2)^(1/8)/b*(cos(b*x^2)*x^2*b-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)*x^2*b-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)*x^2*b-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)*x^2*b-sin(b*x^2))*LommelS1(5/4,1/2,b*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\frac {{\left (x \cos \left (a\right ) - i \, x \sin \left (a\right )\right )} \left (i \, b\right )^{\frac {1}{4}} \Gamma \left (\frac {3}{4}, i \, b x^{2}\right ) + {\left (x \cos \left (a\right ) + i \, x \sin \left (a\right )\right )} \left (-i \, b\right )^{\frac {1}{4}} \Gamma \left (\frac {3}{4}, -i \, b x^{2}\right ) - 2 \, \sqrt {x} \cos \left (b x^{2} + a\right )}{x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\int \frac {\cos {\left (a + b x^{2} \right )}}{x^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral(cos(a + b*x**2)/x**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> Encountered operator mismatch in maxima-to-sr translation

Giac [F]

\[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\int { \frac {\cos \left (b x^{2} + a\right )}{x^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x^2 + a)/x^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos \left (a+b x^2\right )}{x^{3/2}} \, dx=\int \frac {\cos \left (b\,x^2+a\right )}{x^{3/2}} \,d x \]

[In]

int(cos(a + b*x^2)/x^(3/2),x)

[Out]

int(cos(a + b*x^2)/x^(3/2), x)