3.9.15 \(\int x^{5/2} (a+b x^2+c x^4)^2 \, dx\)

Optimal. Leaf size=64 \[ \frac {2}{7} a^2 x^{7/2}+\frac {2}{15} x^{15/2} \left (2 a c+b^2\right )+\frac {4}{11} a b x^{11/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {1108} \begin {gather*} \frac {2}{7} a^2 x^{7/2}+\frac {2}{15} x^{15/2} \left (2 a c+b^2\right )+\frac {4}{11} a b x^{11/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x^2 + c*x^4)^2,x]

[Out]

(2*a^2*x^(7/2))/7 + (4*a*b*x^(11/2))/11 + (2*(b^2 + 2*a*c)*x^(15/2))/15 + (4*b*c*x^(19/2))/19 + (2*c^2*x^(23/2
))/23

Rule 1108

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^m*(a
+ b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] &&  !IntegerQ[(m + 1)/2]

Rubi steps

\begin {align*} \int x^{5/2} \left (a+b x^2+c x^4\right )^2 \, dx &=\int \left (a^2 x^{5/2}+2 a b x^{9/2}+\left (b^2+2 a c\right ) x^{13/2}+2 b c x^{17/2}+c^2 x^{21/2}\right ) \, dx\\ &=\frac {2}{7} a^2 x^{7/2}+\frac {4}{11} a b x^{11/2}+\frac {2}{15} \left (b^2+2 a c\right ) x^{15/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2}\\ \end {align*}

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Mathematica [A]  time = 3.70, size = 64, normalized size = 1.00 \begin {gather*} \frac {2}{7} a^2 x^{7/2}+\frac {2}{15} x^{15/2} \left (2 a c+b^2\right )+\frac {4}{11} a b x^{11/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x^2 + c*x^4)^2,x]

[Out]

(2*a^2*x^(7/2))/7 + (4*a*b*x^(11/2))/11 + (2*(b^2 + 2*a*c)*x^(15/2))/15 + (4*b*c*x^(19/2))/19 + (2*c^2*x^(23/2
))/23

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IntegrateAlgebraic [A]  time = 0.03, size = 62, normalized size = 0.97 \begin {gather*} \frac {2 \left (72105 a^2 x^{7/2}+91770 a b x^{11/2}+67298 a c x^{15/2}+33649 b^2 x^{15/2}+53130 b c x^{19/2}+21945 c^2 x^{23/2}\right )}{504735} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(a + b*x^2 + c*x^4)^2,x]

[Out]

(2*(72105*a^2*x^(7/2) + 91770*a*b*x^(11/2) + 33649*b^2*x^(15/2) + 67298*a*c*x^(15/2) + 53130*b*c*x^(19/2) + 21
945*c^2*x^(23/2)))/504735

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fricas [A]  time = 0.93, size = 49, normalized size = 0.77 \begin {gather*} \frac {2}{504735} \, {\left (21945 \, c^{2} x^{11} + 53130 \, b c x^{9} + 33649 \, {\left (b^{2} + 2 \, a c\right )} x^{7} + 91770 \, a b x^{5} + 72105 \, a^{2} x^{3}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

2/504735*(21945*c^2*x^11 + 53130*b*c*x^9 + 33649*(b^2 + 2*a*c)*x^7 + 91770*a*b*x^5 + 72105*a^2*x^3)*sqrt(x)

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giac [A]  time = 0.21, size = 46, normalized size = 0.72 \begin {gather*} \frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{19} \, b c x^{\frac {19}{2}} + \frac {2}{15} \, b^{2} x^{\frac {15}{2}} + \frac {4}{15} \, a c x^{\frac {15}{2}} + \frac {4}{11} \, a b x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*b^2*x^(15/2) + 4/15*a*c*x^(15/2) + 4/11*a*b*x^(11/2) + 2/7*a^2*x^
(7/2)

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maple [A]  time = 0.01, size = 49, normalized size = 0.77 \begin {gather*} \frac {2 \left (21945 c^{2} x^{8}+53130 b c \,x^{6}+67298 a c \,x^{4}+33649 b^{2} x^{4}+91770 a b \,x^{2}+72105 a^{2}\right ) x^{\frac {7}{2}}}{504735} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(c*x^4+b*x^2+a)^2,x)

[Out]

2/504735*x^(7/2)*(21945*c^2*x^8+53130*b*c*x^6+67298*a*c*x^4+33649*b^2*x^4+91770*a*b*x^2+72105*a^2)

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maxima [A]  time = 1.03, size = 44, normalized size = 0.69 \begin {gather*} \frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{19} \, b c x^{\frac {19}{2}} + \frac {2}{15} \, {\left (b^{2} + 2 \, a c\right )} x^{\frac {15}{2}} + \frac {4}{11} \, a b x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*(b^2 + 2*a*c)*x^(15/2) + 4/11*a*b*x^(11/2) + 2/7*a^2*x^(7/2)

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mupad [B]  time = 4.41, size = 45, normalized size = 0.70 \begin {gather*} x^{15/2}\,\left (\frac {2\,b^2}{15}+\frac {4\,a\,c}{15}\right )+\frac {2\,a^2\,x^{7/2}}{7}+\frac {2\,c^2\,x^{23/2}}{23}+\frac {4\,a\,b\,x^{11/2}}{11}+\frac {4\,b\,c\,x^{19/2}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x^2 + c*x^4)^2,x)

[Out]

x^(15/2)*((4*a*c)/15 + (2*b^2)/15) + (2*a^2*x^(7/2))/7 + (2*c^2*x^(23/2))/23 + (4*a*b*x^(11/2))/11 + (4*b*c*x^
(19/2))/19

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sympy [A]  time = 22.35, size = 70, normalized size = 1.09 \begin {gather*} \frac {2 a^{2} x^{\frac {7}{2}}}{7} + \frac {4 a b x^{\frac {11}{2}}}{11} + \frac {4 a c x^{\frac {15}{2}}}{15} + \frac {2 b^{2} x^{\frac {15}{2}}}{15} + \frac {4 b c x^{\frac {19}{2}}}{19} + \frac {2 c^{2} x^{\frac {23}{2}}}{23} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(c*x**4+b*x**2+a)**2,x)

[Out]

2*a**2*x**(7/2)/7 + 4*a*b*x**(11/2)/11 + 4*a*c*x**(15/2)/15 + 2*b**2*x**(15/2)/15 + 4*b*c*x**(19/2)/19 + 2*c**
2*x**(23/2)/23

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