3.1.23 \(\int x^2 (a+b x)^p \, dx\) [23]

Optimal. Leaf size=60 \[ \frac {a^2 (a+b x)^{1+p}}{b^3 (1+p)}-\frac {2 a (a+b x)^{2+p}}{b^3 (2+p)}+\frac {(a+b x)^{3+p}}{b^3 (3+p)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \begin {gather*} \frac {a^2 (a+b x)^{p+1}}{b^3 (p+1)}-\frac {2 a (a+b x)^{p+2}}{b^3 (p+2)}+\frac {(a+b x)^{p+3}}{b^3 (p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x)^p,x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 57, normalized size = 0.95 \begin {gather*} \frac {(a+b x)^{1+p} \left (2 a^2-2 a b (1+p) x+b^2 \left (2+3 p+p^2\right ) x^2\right )}{b^3 (1+p) (2+p) (3+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x)^p,x]

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 4.46, size = 496, normalized size = 8.27 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {x^3 a^p}{3},b\text {==}0\right \},\left \{\frac {\frac {a^2 \left (3+2 \text {Log}\left [\frac {a+b x}{b}\right ]\right )}{2}+2 a b x \left (1+\text {Log}\left [\frac {a+b x}{b}\right ]\right )+b^2 x^2 \text {Log}\left [\frac {a+b x}{b}\right ]}{b^3 \left (a^2+2 a b x+b^2 x^2\right )},p\text {==}-3\right \},\left \{\frac {-2 a^2 \left (1+\text {Log}\left [\frac {a+b x}{b}\right ]\right )-2 a b x \text {Log}\left [\frac {a+b x}{b}\right ]+b^2 x^2}{b^3 \left (a+b x\right )},p\text {==}-2\right \},\left \{\frac {a^2 \text {Log}\left [\frac {a}{b}+x\right ]-a b x+\frac {b^2 x^2}{2}}{b^3},p\text {==}-1\right \}\right \},\frac {2 a^3 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}-\frac {2 a^2 b p x \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}+\frac {a b^2 p x^2 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}+\frac {a b^2 p^2 x^2 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}+\frac {2 b^3 x^3 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}+\frac {3 b^3 p x^3 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}+\frac {b^3 p^2 x^3 \left (a+b x\right )^p}{6 b^3+11 b^3 p+6 b^3 p^2+b^3 p^3}\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[x^2*(a+b*x)^p,x]')

[Out]

Piecewise[{{x ^ 3 a ^ p / 3, b == 0}, {(a ^ 2 (3 + 2 Log[(a + b x) / b]) / 2 + 2 a b x (1 + Log[(a + b x) / b]
) + b ^ 2 x ^ 2 Log[(a + b x) / b]) / (b ^ 3 (a ^ 2 + 2 a b x + b ^ 2 x ^ 2)), p == -3}, {(-2 a ^ 2 (1 + Log[(
a + b x) / b]) - 2 a b x Log[(a + b x) / b] + b ^ 2 x ^ 2) / (b ^ 3 (a + b x)), p == -2}, {(a ^ 2 Log[a / b +
x] - a b x + b ^ 2 x ^ 2 / 2) / b ^ 3, p == -1}}, 2 a ^ 3 (a + b x) ^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^
2 + b ^ 3 p ^ 3) - 2 a ^ 2 b p x (a + b x) ^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^ 2 + b ^ 3 p ^ 3) + a b ^
2 p x ^ 2 (a + b x) ^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^ 2 + b ^ 3 p ^ 3) + a b ^ 2 p ^ 2 x ^ 2 (a + b x)
^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^ 2 + b ^ 3 p ^ 3) + 2 b ^ 3 x ^ 3 (a + b x) ^ p / (6 b ^ 3 + 11 b ^
3 p + 6 b ^ 3 p ^ 2 + b ^ 3 p ^ 3) + 3 b ^ 3 p x ^ 3 (a + b x) ^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^ 2 + b
^ 3 p ^ 3) + b ^ 3 p ^ 2 x ^ 3 (a + b x) ^ p / (6 b ^ 3 + 11 b ^ 3 p + 6 b ^ 3 p ^ 2 + b ^ 3 p ^ 3)]

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Maple [A]
time = 0.03, size = 73, normalized size = 1.22

method	result	size
gosper	\(\frac {\left (b x +a \right )^{1+p} \left (b^{2} p^{2} x^{2}+3 x^{2} b^{2} p -2 x a p b +2 x^{2} b^{2}-2 a x b +2 a^{2}\right )}{b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}\)	\(73\)
risch	\(\frac {\left (b^{3} p^{2} x^{3}+a \,b^{2} p^{2} x^{2}+3 b^{3} p \,x^{3}+x^{2} a p \,b^{2}+2 x^{3} b^{3}-2 a^{2} p x b +2 a^{3}\right ) \left (b x +a \right )^{p}}{\left (2+p \right ) \left (3+p \right ) \left (1+p \right ) b^{3}}\)	\(88\)
norman	\(\frac {x^{3} {\mathrm e}^{p \ln \left (b x +a \right )}}{3+p}+\frac {a p \,x^{2} {\mathrm e}^{p \ln \left (b x +a \right )}}{b \left (p^{2}+5 p +6\right )}+\frac {2 a^{3} {\mathrm e}^{p \ln \left (b x +a \right )}}{b^{3} \left (p^{3}+6 p^{2}+11 p +6\right )}-\frac {2 p \,a^{2} x \,{\mathrm e}^{p \ln \left (b x +a \right )}}{b^{2} \left (p^{3}+6 p^{2}+11 p +6\right )}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 68, normalized size = 1.13 \begin {gather*} \frac {{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{3} + {\left (p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + 2 \, a^{3}\right )} {\left (b x + a\right )}^{p}}{{\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^2*(b*x+a)^p,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.33, size = 96, normalized size = 1.60 \begin {gather*} -\frac {{\left (2 \, a^{2} b p x - {\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{3} - 2 \, a^{3} - {\left (a b^{2} p^{2} + a b^{2} p\right )} x^{2}\right )} {\left (b x + a\right )}^{p}}{b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.40, size = 597, normalized size = 9.95 \begin {gather*} \begin {cases} \frac {a^{p} x^{3}}{3} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {3 a^{2}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {4 a b x}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} + \frac {2 b^{2} x^{2} \log {\left (\frac {a}{b} + x \right )}}{2 a^{2} b^{3} + 4 a b^{4} x + 2 b^{5} x^{2}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} - \frac {2 a^{2}}{a b^{3} + b^{4} x} - \frac {2 a b x \log {\left (\frac {a}{b} + x \right )}}{a b^{3} + b^{4} x} + \frac {b^{2} x^{2}}{a b^{3} + b^{4} x} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{b^{3}} - \frac {a x}{b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} - \frac {2 a^{2} b p x \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac {a b^{2} p^{2} x^{2} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac {a b^{2} p x^{2} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac {b^{3} p^{2} x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac {3 b^{3} p x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} + \frac {2 b^{3} x^{3} \left (a + b x\right )^{p}}{b^{3} p^{3} + 6 b^{3} p^{2} + 11 b^{3} p + 6 b^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**p*x**3/3, Eq(b, 0)), (2*a**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 3*a**2/(2*a
**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 4*a*b*x
/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2) + 2*b**2*x**2*log(a/b + x)/(2*a**2*b**3 + 4*a*b**4*x + 2*b**5*x**2),
Eq(p, -3)), (-2*a**2*log(a/b + x)/(a*b**3 + b**4*x) - 2*a**2/(a*b**3 + b**4*x) - 2*a*b*x*log(a/b + x)/(a*b**3
+ b**4*x) + b**2*x**2/(a*b**3 + b**4*x), Eq(p, -2)), (a**2*log(a/b + x)/b**3 - a*x/b**2 + x**2/(2*b), Eq(p, -
1)), (2*a**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) - 2*a**2*b*p*x*(a + b*x)**p/(b**3*p**
3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + a*b**2*p**2*x**2*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6
*b**3) + a*b**2*p*x**2*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + b**3*p**2*x**3*(a + b*x)*
*p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3) + 3*b**3*p*x**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b
**3*p + 6*b**3) + 2*b**3*x**3*(a + b*x)**p/(b**3*p**3 + 6*b**3*p**2 + 11*b**3*p + 6*b**3), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (60) = 120\).
time = 0.00, size = 154, normalized size = 2.57 \begin {gather*} \frac {2 a^{3} \mathrm {e}^{p \ln \left (a+b x\right )}-2 a^{2} b p x \mathrm {e}^{p \ln \left (a+b x\right )}+a b^{2} p^{2} x^{2} \mathrm {e}^{p \ln \left (a+b x\right )}+a b^{2} p x^{2} \mathrm {e}^{p \ln \left (a+b x\right )}+b^{3} p^{2} x^{3} \mathrm {e}^{p \ln \left (a+b x\right )}+3 b^{3} p x^{3} \mathrm {e}^{p \ln \left (a+b x\right )}+2 b^{3} x^{3} \mathrm {e}^{p \ln \left (a+b x\right )}}{b^{3} p^{3}+6 b^{3} p^{2}+11 b^{3} p+6 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.60, size = 192, normalized size = 3.20 \begin {gather*} \left \{\begin {array}{cl} \frac {2\,a^2\,\ln \left (a+b\,x\right )+b^2\,x^2-2\,a\,b\,x}{2\,b^3} & \text {\ if\ \ }p=-1\\ \frac {x}{b^2}-\frac {a^2}{b^3\,\left (a+b\,x\right )}-\frac {2\,a\,\ln \left (a+b\,x\right )}{b^3} & \text {\ if\ \ }p=-2\\ \frac {\ln \left (a+b\,x\right )+\frac {2\,a}{a+b\,x}-\frac {a^2}{2\,{\left (a+b\,x\right )}^2}}{b^3} & \text {\ if\ \ }p=-3\\ \frac {2\,{\left (a+b\,x\right )}^{p+1}\,\left (8\,a^2-8\,a\,b\,p\,x-8\,a\,b\,x+4\,b^2\,p^2\,x^2+12\,b^2\,p\,x^2+8\,b^2\,x^2\right )}{b^3\,\left (8\,p^3+48\,p^2+88\,p+48\right )} & \text {\ if\ \ }p\neq -1\wedge p\neq -2\wedge p\neq -3 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x)^p,x)

[Out]

piecewise(p == -1, (2*a^2*log(a + b*x) + b^2*x^2 - 2*a*b*x)/(2*b^3), p == -2, x/b^2 - a^2/(b^3*(a + b*x)) - (2
*a*log(a + b*x))/b^3, p == -3, (log(a + b*x) + (2*a)/(a + b*x) - a^2/(2*(a + b*x)^2))/b^3, p ~= -1 & p ~= -2 &
p ~= -3, (2*(a + b*x)^(p + 1)*(8*a^2 + 8*b^2*x^2 + 12*b^2*p*x^2 - 8*a*b*x + 4*b^2*p^2*x^2 - 8*a*b*p*x))/(b^3*
(88*p + 48*p^2 + 8*p^3 + 48)))

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