3.5.15 \(\int x^3 (a+b x^2)^{9/2} \, dx\) [415]

Optimal. Leaf size=38 \[ -\frac {a \left (a+b x^2\right )^{11/2}}{11 b^2}+\frac {\left (a+b x^2\right )^{13/2}}{13 b^2} \]

[Out]

-1/11*a*(b*x^2+a)^(11/2)/b^2+1/13*(b*x^2+a)^(13/2)/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 = {272, 45} \begin {gather*} \frac {\left (a+b x^2\right )^{13/2}}{13 b^2}-\frac {a \left (a+b x^2\right )^{11/2}}{11 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^(9/2),x]

[Out]

-1/11*(a*(a + b*x^2)^(11/2))/b^2 + (a + b*x^2)^(13/2)/(13*b^2)

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^3 \left (a+b x^2\right )^{9/2} \, dx &=\frac {1}{2} \text {Subst}\left (\int x (a+b x)^{9/2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (-\frac {a (a+b x)^{9/2}}{b}+\frac {(a+b x)^{11/2}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {a \left (a+b x^2\right )^{11/2}}{11 b^2}+\frac {\left (a+b x^2\right )^{13/2}}{13 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 28, normalized size = 0.74 \begin {gather*} \frac {\left (a+b x^2\right )^{11/2} \left (-2 a+11 b x^2\right )}{143 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^(9/2),x]

[Out]

((a + b*x^2)^(11/2)*(-2*a + 11*b*x^2))/(143*b^2)

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Maple [A]
time = 0.05, size = 34, normalized size = 0.89

method	result	size
gosper	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {11}{2}} \left (-11 b \,x^{2}+2 a \right )}{143 b^{2}}\)	\(25\)
default	\(\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{13 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {11}{2}}}{143 b^{2}}\)	\(34\)
trager	\(-\frac {\left (-11 b^{6} x^{12}-53 a \,b^{5} x^{10}-100 a^{2} b^{4} x^{8}-90 a^{3} x^{6} b^{3}-35 a^{4} b^{2} x^{4}-a^{5} b \,x^{2}+2 a^{6}\right ) \sqrt {b \,x^{2}+a}}{143 b^{2}}\)	\(80\)
risch	\(-\frac {\left (-11 b^{6} x^{12}-53 a \,b^{5} x^{10}-100 a^{2} b^{4} x^{8}-90 a^{3} x^{6} b^{3}-35 a^{4} b^{2} x^{4}-a^{5} b \,x^{2}+2 a^{6}\right ) \sqrt {b \,x^{2}+a}}{143 b^{2}}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/13*x^2*(b*x^2+a)^(11/2)/b-2/143*a*(b*x^2+a)^(11/2)/b^2

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Maxima [A]
time = 0.33, size = 33, normalized size = 0.87 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {11}{2}} x^{2}}{13 \, b} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a}{143 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(9/2),x, algorithm="maxima")

[Out]

1/13*(b*x^2 + a)^(11/2)*x^2/b - 2/143*(b*x^2 + a)^(11/2)*a/b^2

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (30) = 60\).
time = 0.60, size = 78, normalized size = 2.05 \begin {gather*} \frac {{\left (11 \, b^{6} x^{12} + 53 \, a b^{5} x^{10} + 100 \, a^{2} b^{4} x^{8} + 90 \, a^{3} b^{3} x^{6} + 35 \, a^{4} b^{2} x^{4} + a^{5} b x^{2} - 2 \, a^{6}\right )} \sqrt {b x^{2} + a}}{143 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(9/2),x, algorithm="fricas")

[Out]

1/143*(11*b^6*x^12 + 53*a*b^5*x^10 + 100*a^2*b^4*x^8 + 90*a^3*b^3*x^6 + 35*a^4*b^2*x^4 + a^5*b*x^2 - 2*a^6)*sq
rt(b*x^2 + a)/b^2

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (31) = 62\).
time = 0.96, size = 156, normalized size = 4.11 \begin {gather*} \begin {cases} - \frac {2 a^{6} \sqrt {a + b x^{2}}}{143 b^{2}} + \frac {a^{5} x^{2} \sqrt {a + b x^{2}}}{143 b} + \frac {35 a^{4} x^{4} \sqrt {a + b x^{2}}}{143} + \frac {90 a^{3} b x^{6} \sqrt {a + b x^{2}}}{143} + \frac {100 a^{2} b^{2} x^{8} \sqrt {a + b x^{2}}}{143} + \frac {53 a b^{3} x^{10} \sqrt {a + b x^{2}}}{143} + \frac {b^{4} x^{12} \sqrt {a + b x^{2}}}{13} & \text {for}\: b \neq 0 \\\frac {a^{\frac {9}{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)**(9/2),x)

[Out]

Piecewise((-2*a**6*sqrt(a + b*x**2)/(143*b**2) + a**5*x**2*sqrt(a + b*x**2)/(143*b) + 35*a**4*x**4*sqrt(a + b*
x**2)/143 + 90*a**3*b*x**6*sqrt(a + b*x**2)/143 + 100*a**2*b**2*x**8*sqrt(a + b*x**2)/143 + 53*a*b**3*x**10*sq
rt(a + b*x**2)/143 + b**4*x**12*sqrt(a + b*x**2)/13, Ne(b, 0)), (a**(9/2)*x**4/4, True))

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Giac [A]
time = 1.52, size = 29, normalized size = 0.76 \begin {gather*} \frac {11 \, {\left (b x^{2} + a\right )}^{\frac {13}{2}} - 13 \, {\left (b x^{2} + a\right )}^{\frac {11}{2}} a}{143 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(b*x^2+a)^(9/2),x, algorithm="giac")

[Out]

1/143*(11*(b*x^2 + a)^(13/2) - 13*(b*x^2 + a)^(11/2)*a)/b^2

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Mupad [B]
time = 4.74, size = 29, normalized size = 0.76 \begin {gather*} -\frac {13\,a\,{\left (b\,x^2+a\right )}^{11/2}-11\,{\left (b\,x^2+a\right )}^{13/2}}{143\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^(9/2),x)

[Out]

-(13*a*(a + b*x^2)^(11/2) - 11*(a + b*x^2)^(13/2))/(143*b^2)

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