### 3.959 $$\int (a+b x)^m (a^2-b^2 x^2)^p \, dx$$

Optimal. Leaf size=63 $\frac{(a+b x)^m \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1,m+2 p+2;m+p+2;\frac{a+b x}{2 a}\right )}{2 a b (m+p+1)}$

[Out]

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

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Rubi [A]  time = 0.055875, antiderivative size = 85, normalized size of antiderivative = 1.35, number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {680, 678, 69} $-\frac{2^{m+p} (a+b x)^m \left (a^2-b^2 x^2\right )^{p+1} \left (\frac{b x}{a}+1\right )^{-m-p-1} \, _2F_1\left (-m-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^m*(a^2 - b^2*x^2)^p,x]

[Out]

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

Rule 680

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^IntPart[m]*(d + e*x)^FracPart[m]
)/(1 + (e*x)/d)^FracPart[m], Int[(1 + (e*x)/d)^m*(a + c*x^2)^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])

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

Mathematica [A]  time = 0.0899089, size = 85, normalized size = 1.35 $\frac{2^{m+p} (b x-a) (a+b x)^m \left (a^2-b^2 x^2\right )^p \left (\frac{b x}{a}+1\right )^{-m-p} \, _2F_1\left (-m-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.625, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(-a + b*x)*(a + b*x))**p*(a + b*x)**m, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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