3.4.80 \(\int x^3 (a+b x^2)^{5/2} \, dx\)

Optimal. Leaf size=38 \[ \frac {\left (a+b x^2\right )^{9/2}}{9 b^2}-\frac {a \left (a+b x^2\right )^{7/2}}{7 b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \begin {gather*} \frac {\left (a+b x^2\right )^{9/2}}{9 b^2}-\frac {a \left (a+b x^2\right )^{7/2}}{7 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(5/2),x]

[Out]

-(a*(a + b*x^2)^(7/2))/(7*b^2) + (a + b*x^2)^(9/2)/(9*b^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int x^3 \left (a+b x^2\right )^{5/2} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x (a+b x)^{5/2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {a (a+b x)^{5/2}}{b}+\frac {(a+b x)^{7/2}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {a \left (a+b x^2\right )^{7/2}}{7 b^2}+\frac {\left (a+b x^2\right )^{9/2}}{9 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 0.74 \begin {gather*} \frac {\left (a+b x^2\right )^{7/2} \left (7 b x^2-2 a\right )}{63 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(5/2),x]

[Out]

((a + b*x^2)^(7/2)*(-2*a + 7*b*x^2))/(63*b^2)

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IntegrateAlgebraic [A]  time = 0.03, size = 60, normalized size = 1.58 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2 a^4+a^3 b x^2+15 a^2 b^2 x^4+19 a b^3 x^6+7 b^4 x^8\right )}{63 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^3*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(-2*a^4 + a^3*b*x^2 + 15*a^2*b^2*x^4 + 19*a*b^3*x^6 + 7*b^4*x^8))/(63*b^2)

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fricas [A]  time = 0.91, size = 56, normalized size = 1.47 \begin {gather*} \frac {{\left (7 \, b^{4} x^{8} + 19 \, a b^{3} x^{6} + 15 \, a^{2} b^{2} x^{4} + a^{3} b x^{2} - 2 \, a^{4}\right )} \sqrt {b x^{2} + a}}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/63*(7*b^4*x^8 + 19*a*b^3*x^6 + 15*a^2*b^2*x^4 + a^3*b*x^2 - 2*a^4)*sqrt(b*x^2 + a)/b^2

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giac [A]  time = 0.99, size = 29, normalized size = 0.76 \begin {gather*} \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} - 9 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/63*(7*(b*x^2 + a)^(9/2) - 9*(b*x^2 + a)^(7/2)*a)/b^2

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maple [A]  time = 0.00, size = 25, normalized size = 0.66 \begin {gather*} -\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} \left (-7 b \,x^{2}+2 a \right )}{63 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(5/2),x)

[Out]

-1/63*(b*x^2+a)^(7/2)*(-7*b*x^2+2*a)/b^2

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maxima [A]  time = 1.37, size = 33, normalized size = 0.87 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} x^{2}}{9 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a}{63 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/9*(b*x^2 + a)^(7/2)*x^2/b - 2/63*(b*x^2 + a)^(7/2)*a/b^2

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mupad [B]  time = 4.74, size = 53, normalized size = 1.39 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {5\,a^2\,x^4}{21}-\frac {2\,a^4}{63\,b^2}+\frac {b^2\,x^8}{9}+\frac {a^3\,x^2}{63\,b}+\frac {19\,a\,b\,x^6}{63}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(5/2),x)

[Out]

(a + b*x^2)^(1/2)*((5*a^2*x^4)/21 - (2*a^4)/(63*b^2) + (b^2*x^8)/9 + (a^3*x^2)/(63*b) + (19*a*b*x^6)/63)

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sympy [A]  time = 3.83, size = 109, normalized size = 2.87 \begin {gather*} \begin {cases} - \frac {2 a^{4} \sqrt {a + b x^{2}}}{63 b^{2}} + \frac {a^{3} x^{2} \sqrt {a + b x^{2}}}{63 b} + \frac {5 a^{2} x^{4} \sqrt {a + b x^{2}}}{21} + \frac {19 a b x^{6} \sqrt {a + b x^{2}}}{63} + \frac {b^{2} x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\frac {a^{\frac {5}{2}} x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(b*x**2+a)**(5/2),x)

[Out]

Piecewise((-2*a**4*sqrt(a + b*x**2)/(63*b**2) + a**3*x**2*sqrt(a + b*x**2)/(63*b) + 5*a**2*x**4*sqrt(a + b*x**
2)/21 + 19*a*b*x**6*sqrt(a + b*x**2)/63 + b**2*x**8*sqrt(a + b*x**2)/9, Ne(b, 0)), (a**(5/2)*x**4/4, True))

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