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

Optimal. Leaf size=38 \[ \frac{\left (a+b x^2\right )^{7/2}}{7 b^2}-\frac{a \left (a+b x^2\right )^{5/2}}{5 b^2} \]

[Out]

-(a*(a + b*x^2)^(5/2))/(5*b^2) + (a + b*x^2)^(7/2)/(7*b^2)

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Rubi [A]  time = 0.0243898, 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} \[ \frac{\left (a+b x^2\right )^{7/2}}{7 b^2}-\frac{a \left (a+b x^2\right )^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0137865, size = 28, normalized size = 0.74 \[ \frac{\left (a+b x^2\right )^{5/2} \left (5 b x^2-2 a\right )}{35 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(3/2),x]

[Out]

((a + b*x^2)^(5/2)*(-2*a + 5*b*x^2))/(35*b^2)

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*} -{\frac{-5\,b{x}^{2}+2\,a}{35\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(b*x^2+a)^(5/2)*(-5*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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55671, size = 97, normalized size = 2.55 \begin{align*} \frac{{\left (5 \, b^{3} x^{6} + 8 \, a b^{2} x^{4} + a^{2} b x^{2} - 2 \, a^{3}\right )} \sqrt{b x^{2} + a}}{35 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*b^3*x^6 + 8*a*b^2*x^4 + a^2*b*x^2 - 2*a^3)*sqrt(b*x^2 + a)/b^2

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Sympy [A]  time = 1.16344, size = 85, normalized size = 2.24 \begin{align*} \begin{cases} - \frac{2 a^{3} \sqrt{a + b x^{2}}}{35 b^{2}} + \frac{a^{2} x^{2} \sqrt{a + b x^{2}}}{35 b} + \frac{8 a x^{4} \sqrt{a + b x^{2}}}{35} + \frac{b x^{6} \sqrt{a + b x^{2}}}{7} & \text{for}\: b \neq 0 \\\frac{a^{\frac{3}{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)**(3/2),x)

[Out]

Piecewise((-2*a**3*sqrt(a + b*x**2)/(35*b**2) + a**2*x**2*sqrt(a + b*x**2)/(35*b) + 8*a*x**4*sqrt(a + b*x**2)/
35 + b*x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (a**(3/2)*x**4/4, True))

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Giac [B]  time = 1.44151, size = 105, normalized size = 2.76 \begin{align*} \frac{\frac{7 \,{\left (3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a\right )} a}{b} + \frac{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}}{b}}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(7*(3*(b*x^2 + a)^(5/2) - 5*(b*x^2 + a)^(3/2)*a)*a/b + (15*(b*x^2 + a)^(7/2) - 42*(b*x^2 + a)^(5/2)*a +
35*(b*x^2 + a)^(3/2)*a^2)/b)/b