3.354 \(\int x^5 \sqrt {a+b x^2} \, dx\)

Optimal. Leaf size=59 \[ \frac {a^2 \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {\left (a+b x^2\right )^{7/2}}{7 b^3}-\frac {2 a \left (a+b x^2\right )^{5/2}}{5 b^3} \]

[Out]

1/3*a^2*(b*x^2+a)^(3/2)/b^3-2/5*a*(b*x^2+a)^(5/2)/b^3+1/7*(b*x^2+a)^(7/2)/b^3

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Rubi [A]  time = 0.03, antiderivative size = 59, 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} \[ \frac {a^2 \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {\left (a+b x^2\right )^{7/2}}{7 b^3}-\frac {2 a \left (a+b x^2\right )^{5/2}}{5 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*Sqrt[a + b*x^2],x]

[Out]

(a^2*(a + b*x^2)^(3/2))/(3*b^3) - (2*a*(a + b*x^2)^(5/2))/(5*b^3) + (a + b*x^2)^(7/2)/(7*b^3)

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

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.66 \[ \frac {\left (a+b x^2\right )^{3/2} \left (8 a^2-12 a b x^2+15 b^2 x^4\right )}{105 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*Sqrt[a + b*x^2],x]

[Out]

((a + b*x^2)^(3/2)*(8*a^2 - 12*a*b*x^2 + 15*b^2*x^4))/(105*b^3)

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fricas [A]  time = 0.55, size = 46, normalized size = 0.78 \[ \frac {{\left (15 \, b^{3} x^{6} + 3 \, a b^{2} x^{4} - 4 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*b^3*x^6 + 3*a*b^2*x^4 - 4*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a)/b^3

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giac [A]  time = 0.59, size = 43, normalized size = 0.73 \[ \frac {15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} - 42 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/105*(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^3

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maple [A]  time = 0.01, size = 36, normalized size = 0.61 \[ \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (15 b^{2} x^{4}-12 a b \,x^{2}+8 a^{2}\right )}{105 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^(1/2),x)

[Out]

1/105*(b*x^2+a)^(3/2)*(15*b^2*x^4-12*a*b*x^2+8*a^2)/b^3

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maxima [A]  time = 1.32, size = 53, normalized size = 0.90 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{4}}{7 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}}{35 \, b^{2}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}}{105 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(b*x^2 + a)^(3/2)*x^4/b - 4/35*(b*x^2 + a)^(3/2)*a*x^2/b^2 + 8/105*(b*x^2 + a)^(3/2)*a^2/b^3

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mupad [B]  time = 4.66, size = 44, normalized size = 0.75 \[ \sqrt {b\,x^2+a}\,\left (\frac {x^6}{7}+\frac {8\,a^3}{105\,b^3}+\frac {a\,x^4}{35\,b}-\frac {4\,a^2\,x^2}{105\,b^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2)^(1/2),x)

[Out]

(a + b*x^2)^(1/2)*(x^6/7 + (8*a^3)/(105*b^3) + (a*x^4)/(35*b) - (4*a^2*x^2)/(105*b^2))

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sympy [A]  time = 0.70, size = 87, normalized size = 1.47 \[ \begin {cases} \frac {8 a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((8*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*a**2*x**2*sqrt(a + b*x**2)/(105*b**2) + a*x**4*sqrt(a + b*x*
*2)/(35*b) + x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (sqrt(a)*x**6/6, True))

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