### 3.971 $$\int \frac{(a^2-b^2 x^2)^p}{(a+b x)^2} \, dx$$

Optimal. Leaf size=58 $-\frac{\left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1,2 p;p;\frac{a+b x}{2 a}\right )}{2 a b (1-p) (a+b x)^2}$

[Out]

-((a^2 - b^2*x^2)^(1 + p)*Hypergeometric2F1[1, 2*p, p, (a + b*x)/(2*a)])/(2*a*b*(1 - p)*(a + b*x)^2)

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Rubi [A]  time = 0.0309562, antiderivative size = 73, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.091, Rules used = {678, 69} $-\frac{2^{p-2} \left (\frac{b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (2-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a^3 b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 - b^2*x^2)^p/(a + b*x)^2,x]

[Out]

-((2^(-2 + p)*(1 + (b*x)/a)^(-1 - p)*(a^2 - b^2*x^2)^(1 + p)*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (a - b*x)/
(2*a)])/(a^3*b*(1 + p)))

Rule 678

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^(m - 1)*(a + c*x^2)^(p + 1))/((1
+ (e*x)/d)^(p + 1)*(a/d + (c*x)/e)^(p + 1)), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a,
c, d, e, m}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.0531989, size = 75, normalized size = 1.29 $-\frac{2^{p-2} (a-b x) \left (\frac{b x}{a}+1\right )^{-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (2-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a^2 b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((2^(-2 + p)*(a - b*x)*(a^2 - b^2*x^2)^p*Hypergeometric2F1[2 - p, 1 + p, 2 + p, (a - b*x)/(2*a)])/(a^2*b*(1 +
p)*(1 + (b*x)/a)^p))

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Maple [F]  time = 0.541, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}}{ \left ( bx+a \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{b^{2} x^{2} + 2 \, a b x + a^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (- a + b x\right ) \left (a + b x\right )\right )^{p}}{\left (a + b x\right )^{2}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(-a + b*x)*(a + b*x))**p/(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-b^{2} x^{2} + a^{2}\right )}^{p}}{{\left (b x + a\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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