3.977 \(\int (-a (a^2-b^2 x^2)^p+(a+b x) (a^2-b^2 x^2)^p) \, dx\)

Optimal. Leaf size=28 \[ -\frac {\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {246, 245, 641} \[ -\frac {\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[-(a*(a^2 - b^2*x^2)^p) + (a + b*x)*(a^2 - b^2*x^2)^p,x]

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin {align*} \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx &=-\left (a \int \left (a^2-b^2 x^2\right )^p \, dx\right )+\int (a+b x) \left (a^2-b^2 x^2\right )^p \, dx\\ &=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)}+a \int \left (a^2-b^2 x^2\right )^p \, dx-\left (a \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p}\right ) \int \left (1-\frac {b^2 x^2}{a^2}\right )^p \, dx\\ &=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)}-a x \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {b^2 x^2}{a^2}\right )+\left (a \left (a^2-b^2 x^2\right )^p \left (1-\frac {b^2 x^2}{a^2}\right )^{-p}\right ) \int \left (1-\frac {b^2 x^2}{a^2}\right )^p \, dx\\ &=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ -\frac {\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 36, normalized size = 1.29 \[ \frac {{\left (b^{2} x^{2} - a^{2}\right )} {\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{2 \, {\left (b p + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 26, normalized size = 0.93 \[ -\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p + 1}}{2 \, b {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 36, normalized size = 1.29 \[ -\frac {\left (-b x +a \right ) \left (b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )^{p}}{2 \left (p +1\right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-a*(-b^2*x^2+a^2)^p+(b*x+a)*(-b^2*x^2+a^2)^p,x)

[Out]

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

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maxima [A]  time = 1.87, size = 42, normalized size = 1.50 \[ \frac {{\left (b^{2} x^{2} - a^{2}\right )} e^{\left (p \log \left (b x + a\right ) + p \log \left (-b x + a\right )\right )}}{2 \, b {\left (p + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.46, size = 26, normalized size = 0.93 \[ -\frac {{\left (a^2-b^2\,x^2\right )}^{p+1}}{2\,b\,\left (p+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^p*(a + b*x) - a*(a^2 - b^2*x^2)^p,x)

[Out]

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

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sympy [A]  time = 4.02, size = 49, normalized size = 1.75 \[ b \left (\begin {cases} \frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\begin {cases} \frac {\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a^{2} - b^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*Piecewise((x**2*(a**2)**p/2, Eq(b**2, 0)), (-Piecewise(((a**2 - b**2*x**2)**(p + 1)/(p + 1), Ne(p, -1)), (lo
g(a**2 - b**2*x**2), True))/(2*b**2), True))

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