### 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]

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

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Rubi [A]  time = 0.0413844, antiderivative size = 28, normalized size of antiderivative = 1., 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 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 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 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*}

Mathematica [A]  time = 0.0051315, size = 28, normalized size = 1. $-\frac{\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)}$

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]

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

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Maple [A]  time = 0.039, size = 36, normalized size = 1.3 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -bx+a \right ) \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}}{2\,b \left ( 1+p \right ) }} \end{align*}

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/(1+p)

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Maxima [A]  time = 1.39302, size = 57, normalized size = 2.04 \begin{align*} \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 )}} \end{align*}

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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Fricas [A]  time = 1.97114, size = 68, normalized size = 2.43 \begin{align*} \frac{{\left (b^{2} x^{2} - a^{2}\right )}{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{2 \,{\left (b p + b\right )}} \end{align*}

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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Sympy [A]  time = 3.82176, size = 49, normalized size = 1.75 \begin{align*} 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 ) \end{align*}

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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Giac [A]  time = 1.2646, size = 35, normalized size = 1.25 \begin{align*} -\frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p + 1}}{2 \, b{\left (p + 1\right )}} \end{align*}

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))