\(\int (-a (a^2-b^2 x^2)^p+(a+b x) (a^2-b^2 x^2)^p) \, dx\) [977]

Optimal result

Integrand size = 38, antiderivative size = 28 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {252, 251, 655} \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=-\frac {\left (a^2-b^2 x^2\right )^{p+1}}{2 b (p+1)} \]

[In]

Int[-(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))

Rule 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 655

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*} \text {integral}& = -\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] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=-\frac {\left (a^2-b^2 x^2\right )^{1+p}}{2 b (1+p)} \]

[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))

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29

method	result	size
gosper	\(-\frac {\left (b x +a \right ) \left (-b x +a \right ) \left (-b^{2} x^{2}+a^{2}\right )^{p}}{2 b \left (1+p \right )}\)	\(36\)
risch	\(-\frac {\left (-b^{2} x^{2}+a^{2}\right ) \left (-b^{2} x^{2}+a^{2}\right )^{p}}{2 b \left (1+p \right )}\)	\(37\)
parallelrisch	\(\frac {x^{2} \left (-b^{2} x^{2}+a^{2}\right )^{p} b^{3}-\left (-b^{2} x^{2}+a^{2}\right )^{p} a^{2} b}{2 b^{2} \left (1+p \right )}\)	\(53\)
norman	\(-\frac {a^{2} {\mathrm e}^{p \ln \left (-b^{2} x^{2}+a^{2}\right )}}{2 b \left (1+p \right )}+\frac {b \,x^{2} {\mathrm e}^{p \ln \left (-b^{2} x^{2}+a^{2}\right )}}{2+2 p}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=\frac {{\left (b^{2} x^{2} - a^{2}\right )} {\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{2 \, {\left (b p + b\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=b \left (\begin {cases} \frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\begin {cases} \frac {\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a^{2} - b^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=\frac {{\left (b^{2} x^{2} - a^{2}\right )} e^{\left (p \log \left (b x + a\right ) + p \log \left (-b x + a\right )\right )}}{2 \, b {\left (p + 1\right )}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=-\frac {{\left (-b^{2} x^{2} + a^{2}\right )}^{p + 1}}{2 \, b {\left (p + 1\right )}} \]

[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))

Mupad [B] (verification not implemented)

Time = 10.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \left (-a \left (a^2-b^2 x^2\right )^p+(a+b x) \left (a^2-b^2 x^2\right )^p\right ) \, dx=-\frac {{\left (a^2-b^2\,x^2\right )}^{p+1}}{2\,b\,\left (p+1\right )} \]

[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))