3.11.40 \(\int x^5 (a+b x^2)^p \, dx\) [1040]

Optimal. Leaf size=72 \[ \frac {a^2 \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac {a \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {\left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 45} \begin {gather*} \frac {a^2 \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac {a \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {\left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^p,x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 64, normalized size = 0.89 \begin {gather*} \frac {\left (a+b x^2\right )^{1+p} \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{2 b^3 (1+p) (2+p) (3+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(2*b^3*(1 + p)*(2 + p)*(3 + p))

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Maple [A]
time = 0.05, size = 80, normalized size = 1.11

method	result	size
gosper	\(\frac {\left (b \,x^{2}+a \right )^{1+p} \left (b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+2 b^{2} x^{4}-2 a b p \,x^{2}-2 a b \,x^{2}+2 a^{2}\right )}{2 b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\)	\(80\)
risch	\(\frac {\left (b^{3} p^{2} x^{6}+3 b^{3} p \,x^{6}+a \,b^{2} p^{2} x^{4}+2 b^{3} x^{6}+a \,b^{2} p \,x^{4}-2 a^{2} b p \,x^{2}+2 a^{3}\right ) \left (b \,x^{2}+a \right )^{p}}{2 \left (2+p \right ) \left (3+p \right ) \left (1+p \right ) b^{3}}\)	\(93\)
norman	\(\frac {a^{3} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}+\frac {x^{6} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{6+2 p}+\frac {a p \,x^{4} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{2 b \left (p^{2}+5 p +6\right )}-\frac {p \,a^{2} x^{2} {\mathrm e}^{p \ln \left (b \,x^{2}+a \right )}}{b^{2} \left (p^{3}+6 p^{2}+11 p +6\right )}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2+a)^(1+p)*(b^2*p^2*x^4+3*b^2*p*x^4+2*b^2*x^4-2*a*b*p*x^2-2*a*b*x^2+2*a^2)/b^3/(p^3+6*p^2+11*p+6)

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Maxima [A]
time = 0.29, size = 73, normalized size = 1.01 \begin {gather*} \frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} + {\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3)*(b*x^2 + a)^p/((p^3 + 6*p^2 + 11*p
+ 6)*b^3)

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Fricas [A]
time = 1.47, size = 98, normalized size = 1.36 \begin {gather*} \frac {{\left ({\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} + {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{4} + 2 \, a^{3}\right )} {\left (b x^{2} + a\right )}^{p}}{2 \, {\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b^3*p^2 + 3*b^3*p + 2*b^3)*x^6 - 2*a^2*b*p*x^2 + (a*b^2*p^2 + a*b^2*p)*x^4 + 2*a^3)*(b*x^2 + a)^p/(b^3*p
^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 920 vs. \(2 (58) = 116\).
time = 1.37, size = 920, normalized size = 12.78 \begin {gather*} \begin {cases} \frac {a^{p} x^{6}}{6} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {3 a^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} + \frac {a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {b^{3} p^{2} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {3 b^{3} p x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {2 b^{3} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**p,x)

[Out]

Piecewise((a**p*x**6/6, Eq(b, 0)), (2*a**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2
*a**2*log(x + sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 3*a**2/(4*a**2*b**3 + 8*a*b**4*x**2 +
4*b**5*x**4) + 4*a*b*x**2*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(x +
sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x*
*4) + 2*b**2*x**4*log(x - sqrt(-a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(x + sqrt(-
a/b))/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(p, -3)), (-2*a**2*log(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4
*x**2) - 2*a**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*lo
g(x - sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(x + sqrt(-a/b))/(2*a*b**3 + 2*b**4*x**2) + b**2*x*
*4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -2)), (a**2*log(x - sqrt(-a/b))/(2*b**3) + a**2*log(x + sqrt(-a/b))/(2*b**3
) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p
+ 12*b**3) - 2*a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p**2*
x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*b**
3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22
*b**3*p + 12*b**3) + 3*b**3*p*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 2*b**3
*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3), True))

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Giac [A]
time = 1.22, size = 132, normalized size = 1.83 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} p - 2 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a p + 2 \, {\left (b x^{2} + a\right )}^{3} {\left (b x^{2} + a\right )}^{p} - 6 \, {\left (b x^{2} + a\right )}^{2} {\left (b x^{2} + a\right )}^{p} a}{2 \, {\left (b^{3} p^{2} + 5 \, b^{3} p + 6 \, b^{3}\right )}} + \frac {{\left (b x^{2} + a\right )}^{p + 1} a^{2}}{2 \, b^{3} {\left (p + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x^2 + a)^3*(b*x^2 + a)^p*p - 2*(b*x^2 + a)^2*(b*x^2 + a)^p*a*p + 2*(b*x^2 + a)^3*(b*x^2 + a)^p - 6*(b*
x^2 + a)^2*(b*x^2 + a)^p*a)/(b^3*p^2 + 5*b^3*p + 6*b^3) + 1/2*(b*x^2 + a)^(p + 1)*a^2/(b^3*(p + 1))

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Mupad [B]
time = 4.90, size = 117, normalized size = 1.62 \begin {gather*} {\left (b\,x^2+a\right )}^p\,\left (\frac {a^3}{b^3\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {x^6\,\left (p^2+3\,p+2\right )}{2\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {a^2\,p\,x^2}{b^2\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {a\,p\,x^4\,\left (p+1\right )}{2\,b\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*x^2)^p,x)

[Out]

(a + b*x^2)^p*(a^3/(b^3*(11*p + 6*p^2 + p^3 + 6)) + (x^6*(3*p + p^2 + 2))/(2*(11*p + 6*p^2 + p^3 + 6)) - (a^2*
p*x^2)/(b^2*(11*p + 6*p^2 + p^3 + 6)) + (a*p*x^4*(p + 1))/(2*b*(11*p + 6*p^2 + p^3 + 6)))

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