### 3.970 $$\int \frac{(a^2-b^2 x^2)^p}{a+b x} \, dx$$

Optimal. Leaf size=55 $\frac{(a-b x) \left (a^2-b^2 x^2\right )^p \, _2F_1\left (1,2 p+1;p+1;\frac{a+b x}{2 a}\right )}{2 a b p}$

[Out]

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

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Rubi [A]  time = 0.0299029, antiderivative size = 73, normalized size of antiderivative = 1.33, 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} $-\frac{2^{p-1} \left (\frac{b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \, _2F_1\left (1-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a^2 b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 - b^2*x^2)^p/(a + b*x),x]

[Out]

-((2^(-1 + p)*(1 + (b*x)/a)^(-1 - p)*(a^2 - b^2*x^2)^(1 + p)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (a - b*x)/
(2*a)])/(a^2*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} \, 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 )^{-1+p} \, dx}{a^2}\\ &=-\frac{2^{-1+p} \left (1+\frac{b x}{a}\right )^{-1-p} \left (a^2-b^2 x^2\right )^{1+p} \, _2F_1\left (1-p,1+p;2+p;\frac{a-b x}{2 a}\right )}{a^2 b (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0516805, size = 75, normalized size = 1.36 $-\frac{2^{p-1} (a-b x) \left (\frac{b x}{a}+1\right )^{-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{a-b x}{2 a}\right )}{a b (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [F]  time = 0.532, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{p}}{bx+a}}\, 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),x)

[Out]

int((-b^2*x^2+a^2)^p/(b*x+a),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}}{b x + a}\,{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),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C]  time = 5.05242, size = 323, normalized size = 5.87 \begin{align*} \begin{cases} \frac{0^{p} \log{\left (-1 + \frac{b^{2} x^{2}}{a^{2}} \right )}}{2 b} + \frac{0^{p} \operatorname{acoth}{\left (\frac{b x}{a} \right )}}{b} + \frac{a b^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac{1}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{1}{2} - p \\ \frac{3}{2} - p \end{matrix}\middle |{\frac{a^{2}}{b^{2} x^{2}}} \right )}}{2 b^{2} x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac{a^{2 p} b x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{3}F_{2}\left (\begin{matrix} 2, 1, 1 - p \\ 2, 2 \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )}}{2 a^{2} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \frac{\left |{b^{2} x^{2}}\right |}{\left |{a^{2}}\right |} > 1 \\\frac{0^{p} \log{\left (1 - \frac{b^{2} x^{2}}{a^{2}} \right )}}{2 b} + \frac{0^{p} \operatorname{atanh}{\left (\frac{b x}{a} \right )}}{b} + \frac{a b^{2 p} p x^{2 p} e^{i \pi p} \Gamma \left (p\right ) \Gamma \left (\frac{1}{2} - p\right ){{}_{2}F_{1}\left (\begin{matrix} 1 - p, \frac{1}{2} - p \\ \frac{3}{2} - p \end{matrix}\middle |{\frac{a^{2}}{b^{2} x^{2}}} \right )}}{2 b^{2} x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac{a^{2 p} b x^{2} \Gamma \left (p\right ) \Gamma \left (1 - p\right ){{}_{3}F_{2}\left (\begin{matrix} 2, 1, 1 - p \\ 2, 2 \end{matrix}\middle |{\frac{b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )}}{2 a^{2} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0**p*log(-1 + b**2*x**2/a**2)/(2*b) + 0**p*acoth(b*x/a)/b + a*b**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma
(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (3/2 - p,), a**2/(b**2*x**2))/(2*b**2*x*gamma(3/2 - p)*gamma(p + 1)
) + a**(2*p)*b*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), b**2*x**2*exp_polar(2*I*pi)/a**2)/(2*a*
*2*gamma(-p)*gamma(p + 1)), Abs(b**2*x**2)/Abs(a**2) > 1), (0**p*log(1 - b**2*x**2/a**2)/(2*b) + 0**p*atanh(b*
x/a)/b + a*b**(2*p)*p*x**(2*p)*exp(I*pi*p)*gamma(p)*gamma(1/2 - p)*hyper((1 - p, 1/2 - p), (3/2 - p,), a**2/(b
**2*x**2))/(2*b**2*x*gamma(3/2 - p)*gamma(p + 1)) + a**(2*p)*b*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p),
(2, 2), b**2*x**2*exp_polar(2*I*pi)/a**2)/(2*a**2*gamma(-p)*gamma(p + 1)), True))

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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}}{b x + a}\,{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),x, algorithm="giac")

[Out]

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