3.1.29 \(\int x^{5/2} \cos ^2(a+b x^2) \, dx\) [29]

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

[Out]

1/7*x^(7/2)-3/128*I*exp(2*I*a)*x^(3/2)*GAMMA(3/4,-2*I*b*x^2)*2^(1/4)/b/(-I*b*x^2)^(3/4)+3/128*I*x^(3/2)*GAMMA(
3/4,2*I*b*x^2)*2^(1/4)/b/exp(2*I*a)/(I*b*x^2)^(3/4)+1/8*x^(3/2)*sin(2*b*x^2+2*a)/b

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Rubi [A]
time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 = {3483, 3485, 3467, 3470, 2250} \begin {gather*} -\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} \end {gather*}

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 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 3467

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 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]

Rule 3483

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 3485

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]

Rubi steps

\begin {align*} \int x^{5/2} \cos ^2\left (a+b x^2\right ) \, dx &=2 \text {Subst}\left (\int x^6 \cos ^2\left (a+b x^4\right ) \, dx,x,\sqrt {x}\right )\\ &=2 \text {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}+\text {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 \text {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) \text {Subst}\left (\int e^{-2 i a-2 i b x^4} x^2 \, dx,x,\sqrt {x}\right )}{16 b}+\frac {(3 i) \text {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*}

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Mathematica [A]
time = 0.27, size = 142, normalized size = 1.08 \begin {gather*} \frac {b x^{11/2} \left (21 \sqrt [4]{2} \left (i b x^2\right )^{3/4} \text {Gamma}\left (\frac {3}{4},-2 i b x^2\right ) (-i \cos (2 a)+\sin (2 a))+21 \sqrt [4]{2} \left (-i b x^2\right )^{3/4} \text {Gamma}\left (\frac {3}{4},2 i b x^2\right ) (i \cos (2 a)+\sin (2 a))+16 \left (b^2 x^4\right )^{3/4} \left (8 b x^2+7 \sin \left (2 \left (a+b x^2\right )\right )\right )\right )}{896 \left (b^2 x^4\right )^{7/4}} \end {gather*}

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))

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int x^{\frac {5}{2}} \left (\cos ^{2}\left (b \,x^{2}+a \right )\right )\, dx\]

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)

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Maxima [A]
time = 0.12, size = 171, normalized size = 1.30 \begin {gather*} \frac {256 \, b^{2} x^{4} + 224 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) + 21 \cdot 2^{\frac {1}{4}} \left (b x^{2}\right )^{\frac {1}{4}} {\left ({\left (\sqrt {\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} + \sqrt {-\sqrt {2} + 2} {\left (i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) - i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \cos \left (2 \, a\right ) + {\left (\sqrt {-\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} + \sqrt {\sqrt {2} + 2} {\left (-i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \sin \left (2 \, a\right )\right )}}{1792 \, b^{2} \sqrt {x}} \end {gather*}

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^2*x^4 + 224*b*x^2*sin(2*b*x^2 + 2*a) + 21*2^(1/4)*(b*x^2)^(1/4)*((sqrt(sqrt(2) + 2)*(gamma(3/4,
2*I*b*x^2) + gamma(3/4, -2*I*b*x^2)) + sqrt(-sqrt(2) + 2)*(I*gamma(3/4, 2*I*b*x^2) - I*gamma(3/4, -2*I*b*x^2))
)*cos(2*a) + (sqrt(-sqrt(2) + 2)*(gamma(3/4, 2*I*b*x^2) + gamma(3/4, -2*I*b*x^2)) + sqrt(sqrt(2) + 2)*(-I*gamm
a(3/4, 2*I*b*x^2) + I*gamma(3/4, -2*I*b*x^2)))*sin(2*a)))/(b^2*sqrt(x))

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Fricas [A]
time = 0.13, size = 78, normalized size = 0.59 \begin {gather*} \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 {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {5}{2}} \cos ^{2}{\left (a + b x^{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,{\cos \left (b\,x^2+a\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*cos(a + b*x^2)^2,x)

[Out]

int(x^(5/2)*cos(a + b*x^2)^2, x)

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