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

Optimal. Leaf size=36 \[ \frac {2}{7} a^2 x^{7/2}+\frac {4}{11} a b x^{11/2}+\frac {2}{15} b^2 x^{15/2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {270} \[ \frac {2}{7} a^2 x^{7/2}+\frac {4}{11} a b x^{11/2}+\frac {2}{15} b^2 x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(a + b*x^2)^2,x]

[Out]

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

Rule 270

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.83 \[ \frac {2 x^{7/2} \left (165 a^2+210 a b x^2+77 b^2 x^4\right )}{1155} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(a + b*x^2)^2,x]

[Out]

(2*x^(7/2)*(165*a^2 + 210*a*b*x^2 + 77*b^2*x^4))/1155

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fricas [A]  time = 1.02, size = 29, normalized size = 0.81 \[ \frac {2}{1155} \, {\left (77 \, b^{2} x^{7} + 210 \, a b x^{5} + 165 \, a^{2} x^{3}\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(77*b^2*x^7 + 210*a*b*x^5 + 165*a^2*x^3)*sqrt(x)

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giac [A]  time = 0.63, size = 24, normalized size = 0.67 \[ \frac {2}{15} \, b^{2} x^{\frac {15}{2}} + \frac {4}{11} \, a b x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 27, normalized size = 0.75 \[ \frac {2 \left (77 b^{2} x^{4}+210 a b \,x^{2}+165 a^{2}\right ) x^{\frac {7}{2}}}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*x^(7/2)*(77*b^2*x^4+210*a*b*x^2+165*a^2)

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maxima [A]  time = 1.26, size = 24, normalized size = 0.67 \[ \frac {2}{15} \, b^{2} x^{\frac {15}{2}} + \frac {4}{11} \, a b x^{\frac {11}{2}} + \frac {2}{7} \, a^{2} x^{\frac {7}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 26, normalized size = 0.72 \[ \frac {2\,x^{7/2}\,\left (165\,a^2+210\,a\,b\,x^2+77\,b^2\,x^4\right )}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(a + b*x^2)^2,x)

[Out]

(2*x^(7/2)*(165*a^2 + 77*b^2*x^4 + 210*a*b*x^2))/1155

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sympy [A]  time = 5.51, size = 34, normalized size = 0.94 \[ \frac {2 a^{2} x^{\frac {7}{2}}}{7} + \frac {4 a b x^{\frac {11}{2}}}{11} + \frac {2 b^{2} x^{\frac {15}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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