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

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

[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 = 0.17, 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 = {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 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]

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 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]

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*}

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Mathematica [A]  time = 0.40, size = 142, normalized size = 1.08 \[ \frac {b x^{11/2} \left (16 \left (b^2 x^4\right )^{3/4} \left (7 \sin \left (2 \left (a+b x^2\right )\right )+8 b x^2\right )+21 \sqrt [4]{2} \left (i b x^2\right )^{3/4} (\sin (2 a)-i \cos (2 a)) \Gamma \left (\frac {3}{4},-2 i b x^2\right )+21 \sqrt [4]{2} \left (-i b x^2\right )^{3/4} (\sin (2 a)+i \cos (2 a)) \Gamma \left (\frac {3}{4},2 i b x^2\right )\right )}{896 \left (b^2 x^4\right )^{7/4}} \]

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))

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fricas [A]  time = 1.05, size = 78, normalized size = 0.59 \[ \frac {21 \, \left (2 i \, b\right )^{\frac {1}{4}} e^{\left (-2 i \, a\right )} \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + 21 \, \left (-2 i \, b\right )^{\frac {1}{4}} e^{\left (2 i \, a\right )} \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right ) + 32 \, {\left (4 \, b^{2} x^{3} + 7 \, b x \cos \left (b x^{2} + a\right ) \sin \left (b x^{2} + a\right )\right )} \sqrt {x}}{896 \, b^{2}} \]

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

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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maple [F]  time = 0.20, 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 = 1.75, size = 174, normalized size = 1.32 \[ \frac {256 \, b^{2} x^{4} + 224 \, b x^{2} \sin \left (2 \, b x^{2} + 2 \, a\right ) + 2^{\frac {1}{4}} \left (b x^{2}\right )^{\frac {1}{4}} {\left ({\left (21 \, \sqrt {\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} - \sqrt {-\sqrt {2} + 2} {\left (-21 i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + 21 i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \cos \left (2 \, a\right ) + {\left (21 \, \sqrt {-\sqrt {2} + 2} {\left (\Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) + \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )} - \sqrt {\sqrt {2} + 2} {\left (21 i \, \Gamma \left (\frac {3}{4}, 2 i \, b x^{2}\right ) - 21 i \, \Gamma \left (\frac {3}{4}, -2 i \, b x^{2}\right )\right )}\right )} \sin \left (2 \, a\right )\right )}}{1792 \, b^{2} \sqrt {x}} \]

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

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

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

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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