3.22 \(\int x (a+b x)^p \, dx\)

Optimal. Leaf size=39 \[ \frac {(a+b x)^{p+2}}{b^2 (p+2)}-\frac {a (a+b x)^{p+1}}{b^2 (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {43} \[ \frac {(a+b x)^{p+2}}{b^2 (p+2)}-\frac {a (a+b x)^{p+1}}{b^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x)^p,x]

[Out]

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

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])

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.85 \[ \frac {(a+b x)^{p+1} (b (p+1) x-a)}{b^2 (p+1) (p+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x)^p,x]

[Out]

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

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fricas [A]  time = 0.43, size = 53, normalized size = 1.36 \[ \frac {{\left (a b p x + {\left (b^{2} p + b^{2}\right )} x^{2} - a^{2}\right )} {\left (b x + a\right )}^{p}}{b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.24, size = 76, normalized size = 1.95 \[ \frac {{\left (b x + a\right )}^{p} b^{2} p x^{2} + {\left (b x + a\right )}^{p} a b p x + {\left (b x + a\right )}^{p} b^{2} x^{2} - {\left (b x + a\right )}^{p} a^{2}}{b^{2} p^{2} + 3 \, b^{2} p + 2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 36, normalized size = 0.92 \[ -\frac {\left (-x p b -b x +a \right ) \left (b x +a \right )^{p +1}}{\left (p^{2}+3 p +2\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x+a)^p,x)

[Out]

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

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maxima [A]  time = 0.43, size = 42, normalized size = 1.08 \[ \frac {{\left (b^{2} {\left (p + 1\right )} x^{2} + a b p x - a^{2}\right )} {\left (b x + a\right )}^{p}}{{\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 94, normalized size = 2.41 \[ \left \{\begin {array}{cl} -\frac {a\,\ln \left (a+b\,x\right )-b\,x}{b^2} & \text {\ if\ \ }p=-1\\ \frac {\ln \left (a+b\,x\right )+\frac {a}{a+b\,x}}{b^2} & \text {\ if\ \ }p=-2\\ \frac {2\,\left (\frac {{\left (a+b\,x\right )}^{p+2}}{2\,p+4}-\frac {a\,{\left (a+b\,x\right )}^{p+1}}{2\,p+2}\right )}{b^2} & \text {\ if\ \ }p\neq -1\wedge p\neq -2 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x)^p,x)

[Out]

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

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sympy [A]  time = 0.65, size = 201, normalized size = 5.15 \[ \begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {a \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} + \frac {a}{a b^{2} + b^{3} x} + \frac {b x \log {\left (\frac {a}{b} + x \right )}}{a b^{2} + b^{3} x} & \text {for}\: p = -2 \\- \frac {a \log {\left (\frac {a}{b} + x \right )}}{b^{2}} + \frac {x}{b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x\right )^{p}}{b^{2} p^{2} + 3 b^{2} p + 2 b^{2}} + \frac {a b p x \left (a + b x\right )^{p}}{b^{2} p^{2} + 3 b^{2} p + 2 b^{2}} + \frac {b^{2} p x^{2} \left (a + b x\right )^{p}}{b^{2} p^{2} + 3 b^{2} p + 2 b^{2}} + \frac {b^{2} x^{2} \left (a + b x\right )^{p}}{b^{2} p^{2} + 3 b^{2} p + 2 b^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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