\(\int \frac {\cos (a+b x^2)}{x^{5/2}} \, dx\) [28]

Optimal result

Integrand size = 14, antiderivative size = 104 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=-\frac {2 \cos \left (a+b x^2\right )}{3 x^{3/2}}-\frac {i b e^{i a} \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{3 \sqrt [4]{-i b x^2}}+\frac {i b e^{-i a} \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{3 \sqrt [4]{i b x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 104, 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} \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=-\frac {i e^{i a} b \sqrt {x} \Gamma \left (\frac {1}{4},-i b x^2\right )}{3 \sqrt [4]{-i b x^2}}+\frac {i e^{-i a} b \sqrt {x} \Gamma \left (\frac {1}{4},i b x^2\right )}{3 \sqrt [4]{i b x^2}}-\frac {2 \cos \left (a+b x^2\right )}{3 x^{3/2}} \]

[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 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 )}{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] (verified)

Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=\frac {-2 \sqrt [4]{b^2 x^4} \cos \left (a+b x^2\right )+b x^2 \sqrt [4]{i b x^2} \Gamma \left (\frac {1}{4},-i b x^2\right ) (-i \cos (a)+\sin (a))+i \left (-i b x^2\right )^{5/4} \Gamma \left (\frac {1}{4},i b x^2\right ) (i \cos (a)+\sin (a))}{3 x^{3/2} \sqrt [4]{b^2 x^4}} \]

[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))

Maple [C] (verified)

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

Time = 0.46 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.44

method	result	size
meijerg	\(\frac {\cos \left (a \right ) \sqrt {\pi }\, 2^{\frac {1}{4}} \left (b^{2}\right )^{\frac {3}{8}} \left (-\frac {4 \,2^{\frac {3}{4}} \left (\frac {8 x^{4} b^{2}}{15}+\frac {2}{3}\right ) \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{\frac {7}{2}} \left (b^{2}\right )^{\frac {3}{8}} b}-\frac {8 \,2^{\frac {3}{4}} \left (-16 x^{4} b^{2}+5\right ) \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{15 \sqrt {\pi }\, x^{\frac {7}{2}} \left (b^{2}\right )^{\frac {3}{8}} b}+\frac {32 x^{\frac {9}{2}} 2^{\frac {3}{4}} b^{3} \sin \left (b \,x^{2}\right ) s_{\frac {3}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{15 \sqrt {\pi }\, \left (b^{2}\right )^{\frac {3}{8}} \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {128 x^{\frac {9}{2}} 2^{\frac {3}{4}} b^{3} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {7}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{15 \sqrt {\pi }\, \left (b^{2}\right )^{\frac {3}{8}} \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{8}-\frac {\sin \left (a \right ) \sqrt {\pi }\, 2^{\frac {1}{4}} b^{\frac {3}{4}} \left (\frac {12 \,2^{\frac {3}{4}} \left (\frac {32 x^{4} b^{2}}{81}+\frac {2}{3}\right ) \sin \left (b \,x^{2}\right )}{\sqrt {\pi }\, x^{\frac {3}{2}} b^{\frac {3}{4}}}+\frac {32 \,2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right )}{3 \sqrt {\pi }\, x^{\frac {3}{2}} b^{\frac {3}{4}}}-\frac {128 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \sin \left (b \,x^{2}\right ) s_{\frac {7}{4},\frac {3}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{27 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {7}{4}}}-\frac {32 x^{\frac {9}{2}} b^{\frac {9}{4}} 2^{\frac {3}{4}} \left (\cos \left (b \,x^{2}\right ) x^{2} b -\sin \left (b \,x^{2}\right )\right ) s_{\frac {3}{4},\frac {1}{2}}^{\left (+\right )}\left (b \,x^{2}\right )}{3 \sqrt {\pi }\, \left (b \,x^{2}\right )^{\frac {11}{4}}}\right )}{8}\)	\(358\)

[In]

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

[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*x^4*b^2+2/3)/b*sin(b*x^
2)-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)*x^2*b-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)*x^2*b-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*x^4*b^2+2/3)*sin(b*x^2)+32/3/Pi^(1/2)/x^(3/2)*2^(3/4)
/b^(3/4)*(cos(b*x^2)*x^2*b-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)*x^2*b-sin(b*x^2))*LommelS1(
3/4,1/2,b*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.72 \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=\frac {{\left (x^{2} \cos \left (a\right ) - i \, x^{2} \sin \left (a\right )\right )} \left (i \, b\right )^{\frac {3}{4}} \Gamma \left (\frac {1}{4}, i \, b x^{2}\right ) + {\left (x^{2} \cos \left (a\right ) + i \, x^{2} \sin \left (a\right )\right )} \left (-i \, b\right )^{\frac {3}{4}} \Gamma \left (\frac {1}{4}, -i \, b x^{2}\right ) - 2 \, \sqrt {x} \cos \left (b x^{2} + a\right )}{3 \, x^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(cos(a + b*x**2)/x**(5/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=\int { \frac {\cos \left (b x^{2} + a\right )}{x^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x^2 + a)/x^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos \left (a+b x^2\right )}{x^{5/2}} \, dx=\int \frac {\cos \left (b\,x^2+a\right )}{x^{5/2}} \,d x \]

[In]

int(cos(a + b*x^2)/x^(5/2),x)

[Out]

int(cos(a + b*x^2)/x^(5/2), x)