3.283 \(\int \sqrt {x} (a+b x^2)^3 \, dx\)

Optimal. Leaf size=51 \[ \frac {2}{3} a^3 x^{3/2}+\frac {6}{7} a^2 b x^{7/2}+\frac {6}{11} a b^2 x^{11/2}+\frac {2}{15} b^3 x^{15/2} \]

[Out]

2/3*a^3*x^(3/2)+6/7*a^2*b*x^(7/2)+6/11*a*b^2*x^(11/2)+2/15*b^3*x^(15/2)

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Rubi [A]  time = 0.01, antiderivative size = 51, 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} \[ \frac {6}{7} a^2 b x^{7/2}+\frac {2}{3} a^3 x^{3/2}+\frac {6}{11} a b^2 x^{11/2}+\frac {2}{15} b^3 x^{15/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a + b*x^2)^3,x]

[Out]

(2*a^3*x^(3/2))/3 + (6*a^2*b*x^(7/2))/7 + (6*a*b^2*x^(11/2))/11 + (2*b^3*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 \sqrt {x} \left (a+b x^2\right )^3 \, dx &=\int \left (a^3 \sqrt {x}+3 a^2 b x^{5/2}+3 a b^2 x^{9/2}+b^3 x^{13/2}\right ) \, dx\\ &=\frac {2}{3} a^3 x^{3/2}+\frac {6}{7} a^2 b x^{7/2}+\frac {6}{11} a b^2 x^{11/2}+\frac {2}{15} b^3 x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.80 \[ \frac {2 x^{3/2} \left (385 a^3+495 a^2 b x^2+315 a b^2 x^4+77 b^3 x^6\right )}{1155} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a + b*x^2)^3,x]

[Out]

(2*x^(3/2)*(385*a^3 + 495*a^2*b*x^2 + 315*a*b^2*x^4 + 77*b^3*x^6))/1155

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fricas [A]  time = 0.92, size = 38, normalized size = 0.75 \[ \frac {2}{1155} \, {\left (77 \, b^{3} x^{7} + 315 \, a b^{2} x^{5} + 495 \, a^{2} b x^{3} + 385 \, a^{3} x\right )} \sqrt {x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*x^(1/2),x, algorithm="fricas")

[Out]

2/1155*(77*b^3*x^7 + 315*a*b^2*x^5 + 495*a^2*b*x^3 + 385*a^3*x)*sqrt(x)

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giac [A]  time = 0.58, size = 35, normalized size = 0.69 \[ \frac {2}{15} \, b^{3} x^{\frac {15}{2}} + \frac {6}{11} \, a b^{2} x^{\frac {11}{2}} + \frac {6}{7} \, a^{2} b x^{\frac {7}{2}} + \frac {2}{3} \, a^{3} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*x^(1/2),x, algorithm="giac")

[Out]

2/15*b^3*x^(15/2) + 6/11*a*b^2*x^(11/2) + 6/7*a^2*b*x^(7/2) + 2/3*a^3*x^(3/2)

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maple [A]  time = 0.01, size = 38, normalized size = 0.75 \[ \frac {2 \left (77 b^{3} x^{6}+315 a \,b^{2} x^{4}+495 a^{2} b \,x^{2}+385 a^{3}\right ) x^{\frac {3}{2}}}{1155} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3*x^(1/2),x)

[Out]

2/1155*x^(3/2)*(77*b^3*x^6+315*a*b^2*x^4+495*a^2*b*x^2+385*a^3)

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maxima [A]  time = 1.34, size = 35, normalized size = 0.69 \[ \frac {2}{15} \, b^{3} x^{\frac {15}{2}} + \frac {6}{11} \, a b^{2} x^{\frac {11}{2}} + \frac {6}{7} \, a^{2} b x^{\frac {7}{2}} + \frac {2}{3} \, a^{3} x^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3*x^(1/2),x, algorithm="maxima")

[Out]

2/15*b^3*x^(15/2) + 6/11*a*b^2*x^(11/2) + 6/7*a^2*b*x^(7/2) + 2/3*a^3*x^(3/2)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.69 \[ \frac {2\,a^3\,x^{3/2}}{3}+\frac {2\,b^3\,x^{15/2}}{15}+\frac {6\,a^2\,b\,x^{7/2}}{7}+\frac {6\,a\,b^2\,x^{11/2}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a + b*x^2)^3,x)

[Out]

(2*a^3*x^(3/2))/3 + (2*b^3*x^(15/2))/15 + (6*a^2*b*x^(7/2))/7 + (6*a*b^2*x^(11/2))/11

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sympy [A]  time = 2.17, size = 49, normalized size = 0.96 \[ \frac {2 a^{3} x^{\frac {3}{2}}}{3} + \frac {6 a^{2} b x^{\frac {7}{2}}}{7} + \frac {6 a b^{2} x^{\frac {11}{2}}}{11} + \frac {2 b^{3} x^{\frac {15}{2}}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3*x**(1/2),x)

[Out]

2*a**3*x**(3/2)/3 + 6*a**2*b*x**(7/2)/7 + 6*a*b**2*x**(11/2)/11 + 2*b**3*x**(15/2)/15

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