3.4.70 \(\int x^5 (a+b x^2)^{3/2} \, dx\) [370]

Optimal. Leaf size=59 \[ \frac {a^2 \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {2 a \left (a+b x^2\right )^{7/2}}{7 b^3}+\frac {\left (a+b x^2\right )^{9/2}}{9 b^3} \]

[Out]

1/5*a^2*(b*x^2+a)^(5/2)/b^3-2/7*a*(b*x^2+a)^(7/2)/b^3+1/9*(b*x^2+a)^(9/2)/b^3

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Rubi [A]
time = 0.02, 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 = {272, 45} \begin {gather*} \frac {a^2 \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {\left (a+b x^2\right )^{9/2}}{9 b^3}-\frac {2 a \left (a+b x^2\right )^{7/2}}{7 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^(3/2),x]

[Out]

(a^2*(a + b*x^2)^(5/2))/(5*b^3) - (2*a*(a + b*x^2)^(7/2))/(7*b^3) + (a + b*x^2)^(9/2)/(9*b^3)

Rule 45

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 272

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

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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.66 \begin {gather*} \frac {\left (a+b x^2\right )^{5/2} \left (8 a^2-20 a b x^2+35 b^2 x^4\right )}{315 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^(3/2),x]

[Out]

((a + b*x^2)^(5/2)*(8*a^2 - 20*a*b*x^2 + 35*b^2*x^4))/(315*b^3)

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Maple [A]
time = 0.04, size = 58, normalized size = 0.98

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (35 b^{2} x^{4}-20 a b \,x^{2}+8 a^{2}\right )}{315 b^{3}}\)	\(36\)
default	\(\frac {x^{4} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{9 b}-\frac {4 a \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )}{9 b}\)	\(58\)
trager	\(\frac {\left (35 b^{4} x^{8}+50 a \,b^{3} x^{6}+3 a^{2} b^{2} x^{4}-4 a^{3} b \,x^{2}+8 a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 b^{3}}\)	\(58\)
risch	\(\frac {\left (35 b^{4} x^{8}+50 a \,b^{3} x^{6}+3 a^{2} b^{2} x^{4}-4 a^{3} b \,x^{2}+8 a^{4}\right ) \sqrt {b \,x^{2}+a}}{315 b^{3}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^4*(b*x^2+a)^(5/2)/b-4/9*a/b*(1/7*x^2*(b*x^2+a)^(5/2)/b-2/35*a*(b*x^2+a)^(5/2)/b^2)

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Maxima [A]
time = 0.27, size = 53, normalized size = 0.90 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} x^{4}}{9 \, b} - \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a x^{2}}{63 \, b^{2}} + \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}}{315 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(b*x^2 + a)^(5/2)*x^4/b - 4/63*(b*x^2 + a)^(5/2)*a*x^2/b^2 + 8/315*(b*x^2 + a)^(5/2)*a^2/b^3

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Fricas [A]
time = 0.86, size = 57, normalized size = 0.97 \begin {gather*} \frac {{\left (35 \, b^{4} x^{8} + 50 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} - 4 \, a^{3} b x^{2} + 8 \, a^{4}\right )} \sqrt {b x^{2} + a}}{315 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*b^4*x^8 + 50*a*b^3*x^6 + 3*a^2*b^2*x^4 - 4*a^3*b*x^2 + 8*a^4)*sqrt(b*x^2 + a)/b^3

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (51) = 102\).
time = 0.27, size = 109, normalized size = 1.85 \begin {gather*} \begin {cases} \frac {8 a^{4} \sqrt {a + b x^{2}}}{315 b^{3}} - \frac {4 a^{3} x^{2} \sqrt {a + b x^{2}}}{315 b^{2}} + \frac {a^{2} x^{4} \sqrt {a + b x^{2}}}{105 b} + \frac {10 a x^{6} \sqrt {a + b x^{2}}}{63} + \frac {b x^{8} \sqrt {a + b x^{2}}}{9} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{2}} x^{6}}{6} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*a**4*sqrt(a + b*x**2)/(315*b**3) - 4*a**3*x**2*sqrt(a + b*x**2)/(315*b**2) + a**2*x**4*sqrt(a + b
*x**2)/(105*b) + 10*a*x**6*sqrt(a + b*x**2)/63 + b*x**8*sqrt(a + b*x**2)/9, Ne(b, 0)), (a**(3/2)*x**6/6, True)
)

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Giac [A]
time = 0.92, size = 43, normalized size = 0.73 \begin {gather*} \frac {35 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} - 90 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a + 63 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2}}{315 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/315*(35*(b*x^2 + a)^(9/2) - 90*(b*x^2 + a)^(7/2)*a + 63*(b*x^2 + a)^(5/2)*a^2)/b^3

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Mupad [B]
time = 4.66, size = 53, normalized size = 0.90 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {10\,a\,x^6}{63}+\frac {b\,x^8}{9}+\frac {8\,a^4}{315\,b^3}+\frac {a^2\,x^4}{105\,b}-\frac {4\,a^3\,x^2}{315\,b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2)^(3/2),x)

[Out]

(a + b*x^2)^(1/2)*((10*a*x^6)/63 + (b*x^8)/9 + (8*a^4)/(315*b^3) + (a^2*x^4)/(105*b) - (4*a^3*x^2)/(315*b^2))

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