Integrand size = 16, antiderivative size = 116 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=-\frac {2 \cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac {i b e^{2 i a} \sqrt {x} \Gamma \left (\frac {1}{4},-2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{-i b x^2}}+\frac {i b e^{-2 i a} \sqrt {x} \Gamma \left (\frac {1}{4},2 i b x^2\right )}{3 \sqrt [4]{2} \sqrt [4]{i b x^2}} \]
-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)/e xp(2*I*a)/(I*b*x^2)^(1/4)
Time = 0.57 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\frac {-4 \sqrt [4]{b^2 x^4} \cos ^2\left (a+b x^2\right )+2^{3/4} b x^2 \sqrt [4]{i b x^2} \Gamma \left (\frac {1}{4},-2 i b x^2\right ) (-i \cos (2 a)+\sin (2 a))+i 2^{3/4} \left (-i b x^2\right )^{5/4} \Gamma \left (\frac {1}{4},2 i b x^2\right ) (i \cos (2 a)+\sin (2 a))}{6 x^{3/2} \sqrt [4]{b^2 x^4}} \]
(-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))
Time = 0.39 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3883, 3875, 5084, 3854, 3836, 2637}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx\) |
\(\Big \downarrow \) 3883 |
\(\displaystyle 2 \int \frac {\cos ^2\left (b x^2+a\right )}{x^2}d\sqrt {x}\) |
\(\Big \downarrow \) 3875 |
\(\displaystyle 2 \left (-\frac {8}{3} b \int \cos \left (b x^2+a\right ) \sin \left (b x^2+a\right )d\sqrt {x}-\frac {\cos ^2\left (a+b x^2\right )}{3 x^{3/2}}\right )\) |
\(\Big \downarrow \) 5084 |
\(\displaystyle 2 \left (-\frac {4}{3} b \int \sin \left (2 \left (b x^2+a\right )\right )d\sqrt {x}-\frac {\cos ^2\left (a+b x^2\right )}{3 x^{3/2}}\right )\) |
\(\Big \downarrow \) 3854 |
\(\displaystyle 2 \left (-\frac {4}{3} b \int \sin \left (2 b x^2+2 a\right )d\sqrt {x}-\frac {\cos ^2\left (a+b x^2\right )}{3 x^{3/2}}\right )\) |
\(\Big \downarrow \) 3836 |
\(\displaystyle 2 \left (-\frac {\cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac {4}{3} b \left (\frac {1}{2} i \int e^{-2 i b x^2-2 i a}d\sqrt {x}-\frac {1}{2} i \int e^{2 i b x^2+2 i a}d\sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 2637 |
\(\displaystyle 2 \left (-\frac {\cos ^2\left (a+b x^2\right )}{3 x^{3/2}}-\frac {4}{3} b \left (\frac {i e^{2 i a} \sqrt {x} \Gamma \left (\frac {1}{4},-2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{-i b x^2}}-\frac {i e^{-2 i a} \sqrt {x} \Gamma \left (\frac {1}{4},2 i b x^2\right )}{8 \sqrt [4]{2} \sqrt [4]{i b x^2}}\right )\right )\) |
2*(-1/3*Cos[a + b*x^2]^2/x^(3/2) - (4*b*(((I/8)*E^((2*I)*a)*Sqrt[x]*Gamma[ 1/4, (-2*I)*b*x^2])/(2^(1/4)*((-I)*b*x^2)^(1/4)) - ((I/8)*Sqrt[x]*Gamma[1/ 4, (2*I)*b*x^2])/(2^(1/4)*E^((2*I)*a)*(I*b*x^2)^(1/4))))/3)
3.1.34.3.1 Defintions of rubi rules used
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]
Int[Sin[(c_.) + (d_.)*((e_.) + (f_.)*(x_))^(n_)], x_Symbol] :> Simp[I/2 I nt[E^((-c)*I - d*I*(e + f*x)^n), x], x] - Simp[I/2 Int[E^(c*I + d*I*(e + f*x)^n), x], x] /; FreeQ[{c, d, e, f}, x] && IGtQ[n, 2]
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]
Int[Cos[(a_.) + (b_.)*(x_)^(n_)]^(p_)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(Cos[a + b*x^n]^p/(m + 1)), x] + Simp[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]
Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_.)*(x_))^(m_), x _Symbol] :> With[{k = Denominator[m]}, Simp[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]] /; FreeQ[{a , b, c, d, e}, x] && IntegerQ[p] && IGtQ[n, 0] && FractionQ[m]
Int[Cos[w_]^(p_.)*(u_.)*Sin[v_]^(p_.), x_Symbol] :> Simp[1/2^p Int[u*Sin[ 2*v]^p, x], x] /; EqQ[w, v] && IntegerQ[p]
\[\int \frac {\cos ^{2}\left (b \,x^{2}+a \right )}{x^{\frac {5}{2}}}d x\]
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\frac {{\left (x^{2} \cos \left (2 \, a\right ) - i \, x^{2} \sin \left (2 \, a\right )\right )} \left (2 i \, b\right )^{\frac {3}{4}} \Gamma \left (\frac {1}{4}, 2 i \, b x^{2}\right ) + {\left (x^{2} \cos \left (2 \, a\right ) + i \, x^{2} \sin \left (2 \, a\right )\right )} \left (-2 i \, b\right )^{\frac {3}{4}} \Gamma \left (\frac {1}{4}, -2 i \, b x^{2}\right ) - 4 \, \sqrt {x} \cos \left (b x^{2} + a\right )^{2}}{6 \, x^{2}} \]
1/6*((x^2*cos(2*a) - I*x^2*sin(2*a))*(2*I*b)^(3/4)*gamma(1/4, 2*I*b*x^2) + (x^2*cos(2*a) + I*x^2*sin(2*a))*(-2*I*b)^(3/4)*gamma(1/4, -2*I*b*x^2) - 4 *sqrt(x)*cos(b*x^2 + a)^2)/x^2
\[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\int \frac {\cos ^{2}{\left (a + b x^{2} \right )}}{x^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\int { \frac {\cos \left (b x^{2} + a\right )^{2}}{x^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2\left (a+b x^2\right )}{x^{5/2}} \, dx=\int \frac {{\cos \left (b\,x^2+a\right )}^2}{x^{5/2}} \,d x \]