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

Optimal. Leaf size=132 \[ -\frac{3 i e^{2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{64\ 2^{3/4} b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{64\ 2^{3/4} b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}+\frac{x^{7/2}}{7} \]

[Out]

x^(7/2)/7 - (((3*I)/64)*E^((2*I)*a)*x^(3/2)*Gamma[3/4, (-2*I)*b*x^2])/(2^(3/4)*b*((-I)*b*x^2)^(3/4)) + (((3*I)
/64)*x^(3/2)*Gamma[3/4, (2*I)*b*x^2])/(2^(3/4)*b*E^((2*I)*a)*(I*b*x^2)^(3/4)) + (x^(3/2)*Sin[2*(a + b*x^2)])/(
8*b)

________________________________________________________________________________________

Rubi [A]  time = 0.174658, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3402, 3404, 3386, 3389, 2218} \[ -\frac{3 i e^{2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )}{64\ 2^{3/4} b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-2 i a} x^{3/2} \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )}{64\ 2^{3/4} b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}+\frac{x^{7/2}}{7} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Cos[a + b*x^2]^2,x]

[Out]

x^(7/2)/7 - (((3*I)/64)*E^((2*I)*a)*x^(3/2)*Gamma[3/4, (-2*I)*b*x^2])/(2^(3/4)*b*((-I)*b*x^2)^(3/4)) + (((3*I)
/64)*x^(3/2)*Gamma[3/4, (2*I)*b*x^2])/(2^(3/4)*b*E^((2*I)*a)*(I*b*x^2)^(3/4)) + (x^(3/2)*Sin[2*(a + b*x^2)])/(
8*b)

Rule 3402

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_.)*(x_))^(m_), x_Symbol] :> With[{k = Denominator[m
]}, Dist[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]] /; Free
Q[{a, b, c, d, e}, x] && IntegerQ[p] && IGtQ[n, 0] && FractionQ[m]

Rule 3404

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(e
*x)^m, (a + b*Cos[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rule 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, 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 x^{5/2} \cos ^2\left (a+b x^2\right ) \, dx &=2 \operatorname{Subst}\left (\int x^6 \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{x^6}{2}+\frac{1}{2} x^6 \cos \left (2 a+2 b x^4\right )\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{7/2}}{7}+\operatorname{Subst}\left (\int x^6 \cos \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{7/2}}{7}+\frac{x^{3/2} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}-\frac{3 \operatorname{Subst}\left (\int x^2 \sin \left (2 a+2 b x^4\right ) \, dx,x,\sqrt{x}\right )}{8 b}\\ &=\frac{x^{7/2}}{7}+\frac{x^{3/2} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}-\frac{(3 i) \operatorname{Subst}\left (\int e^{-2 i a-2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )}{16 b}+\frac{(3 i) \operatorname{Subst}\left (\int e^{2 i a+2 i b x^4} x^2 \, dx,x,\sqrt{x}\right )}{16 b}\\ &=\frac{x^{7/2}}{7}-\frac{3 i e^{2 i a} x^{3/2} \Gamma \left (\frac{3}{4},-2 i b x^2\right )}{64\ 2^{3/4} b \left (-i b x^2\right )^{3/4}}+\frac{3 i e^{-2 i a} x^{3/2} \Gamma \left (\frac{3}{4},2 i b x^2\right )}{64\ 2^{3/4} b \left (i b x^2\right )^{3/4}}+\frac{x^{3/2} \sin \left (2 \left (a+b x^2\right )\right )}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.458192, size = 142, normalized size = 1.08 \[ \frac{b x^{11/2} \left (21 \sqrt [4]{2} \left (i b x^2\right )^{3/4} (\sin (2 a)-i \cos (2 a)) \text{Gamma}\left (\frac{3}{4},-2 i b x^2\right )+21 \sqrt [4]{2} \left (-i b x^2\right )^{3/4} (\sin (2 a)+i \cos (2 a)) \text{Gamma}\left (\frac{3}{4},2 i b x^2\right )+16 \left (b^2 x^4\right )^{3/4} \left (7 \sin \left (2 \left (a+b x^2\right )\right )+8 b x^2\right )\right )}{896 \left (b^2 x^4\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*Cos[a + b*x^2]^2,x]

[Out]

(b*x^(11/2)*(21*2^(1/4)*(I*b*x^2)^(3/4)*Gamma[3/4, (-2*I)*b*x^2]*((-I)*Cos[2*a] + Sin[2*a]) + 21*2^(1/4)*((-I)
*b*x^2)^(3/4)*Gamma[3/4, (2*I)*b*x^2]*(I*Cos[2*a] + Sin[2*a]) + 16*(b^2*x^4)^(3/4)*(8*b*x^2 + 7*Sin[2*(a + b*x
^2)])))/(896*(b^2*x^4)^(7/4))

________________________________________________________________________________________

Maple [F]  time = 0.094, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{5}{2}}} \left ( \cos \left ( b{x}^{2}+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.49486, size = 421, normalized size = 3.19 \begin{align*} \frac{256 \, b x^{4}{\left | b \right |} + 224 \, x^{2}{\left | b \right |} \sin \left (2 \, b x^{2} + 2 \, a\right ) - 2^{\frac{1}{4}} \left (x^{2}{\left | b \right |}\right )^{\frac{1}{4}}{\left ({\left ({\left (-21 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + 21 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) +{\left (-21 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + 21 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) - 21 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 21 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \cos \left (2 \, a\right ) -{\left (21 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) + 21 \,{\left (\Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \cos \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (21 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) - 21 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right ) -{\left (-21 i \, \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + 21 i \, \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right )\right )} \sin \left (-\frac{3}{8} \, \pi + \frac{3}{4} \, \arctan \left (0, b\right )\right )\right )} \sin \left (2 \, a\right )\right )}}{1792 \, b \sqrt{x}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1792*(256*b*x^4*abs(b) + 224*x^2*abs(b)*sin(2*b*x^2 + 2*a) - 2^(1/4)*(x^2*abs(b))^(1/4)*(((-21*I*gamma(3/4,
2*I*b*x^2) + 21*I*gamma(3/4, -2*I*b*x^2))*cos(3/8*pi + 3/4*arctan2(0, b)) + (-21*I*gamma(3/4, 2*I*b*x^2) + 21*
I*gamma(3/4, -2*I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b)) - 21*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2
))*sin(3/8*pi + 3/4*arctan2(0, b)) + 21*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*sin(-3/8*pi + 3/4*arc
tan2(0, b)))*cos(2*a) - (21*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*cos(3/8*pi + 3/4*arctan2(0, b)) +
21*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2))*cos(-3/8*pi + 3/4*arctan2(0, b)) - (21*I*gamma(3/4, 2*I*b
*x^2) - 21*I*gamma(3/4, -2*I*b*x^2))*sin(3/8*pi + 3/4*arctan2(0, b)) - (-21*I*gamma(3/4, 2*I*b*x^2) + 21*I*gam
ma(3/4, -2*I*b*x^2))*sin(-3/8*pi + 3/4*arctan2(0, b)))*sin(2*a)))/(b*sqrt(x)*abs(b))

________________________________________________________________________________________

Fricas [A]  time = 1.81022, size = 242, normalized size = 1.83 \begin{align*} \frac{21 \, \left (2 i \, b\right )^{\frac{1}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac{3}{4}, 2 i \, b x^{2}\right ) + 21 \, \left (-2 i \, b\right )^{\frac{1}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac{3}{4}, -2 i \, b x^{2}\right ) + 32 \,{\left (4 \, b^{2} x^{3} + 7 \, b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \sqrt{x}}{896 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/896*(21*(2*I*b)^(1/4)*e^(-2*I*a)*gamma(3/4, 2*I*b*x^2) + 21*(-2*I*b)^(1/4)*e^(2*I*a)*gamma(3/4, -2*I*b*x^2)
+ 32*(4*b^2*x^3 + 7*b*x*cos(b*x^2 + a)*sin(b*x^2 + a))*sqrt(x))/b^2

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{5}{2}} \cos \left (b x^{2} + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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