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

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

[Out]

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

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Rubi [A]  time = 0.0236686, antiderivative size = 38, normalized size of antiderivative = 1., 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} \[ \frac{\left (a+b x^2\right )^{9/2}}{9 b^2}-\frac{a \left (a+b x^2\right )^{7/2}}{7 b^2} \]

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 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]]

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])

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*}

Mathematica [A]  time = 0.0161245, size = 28, normalized size = 0.74 \[ \frac{\left (a+b x^2\right )^{7/2} \left (7 b x^2-2 a\right )}{63 b^2} \]

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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Maple [A]  time = 0.004, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-7\,b{x}^{2}+2\,a}{63\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56116, size = 122, normalized size = 3.21 \begin{align*} \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{align*}

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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Sympy [A]  time = 3.53274, size = 109, normalized size = 2.87 \begin{align*} \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{align*}

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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Giac [B]  time = 2.6514, size = 186, normalized size = 4.89 \begin{align*} \frac{\frac{21 \,{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} a^{2}}{b} + \frac{6 \,{\left (15 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} - 42 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a + 35 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2}\right )} a}{b} + \frac{35 \,{\left (b x^{2} + a\right )}^{\frac{9}{2}} - 135 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} a + 189 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} a^{2} - 105 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{3}}{b}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(21*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)*a^2/b + 6*(15*(b*x^2 + a)^(7/2) - 42*(b*x^2 + a)^(5/2)
*a + 35*(b*x^2 + a)^(3/2)*a^2)*a/b + (35*(b*x^2 + a)^(9/2) - 135*(b*x^2 + a)^(7/2)*a + 189*(b*x^2 + a)^(5/2)*a
^2 - 105*(b*x^2 + a)^(3/2)*a^3)/b)/b