3.271 \(\int \frac{(d^2-e^2 x^2)^p}{d+e x} \, dx\)

Optimal. Leaf size=73 \[ -\frac{2^{p-1} \left (\frac{e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^2 e (p+1)} \]

[Out]

-((2^(-1 + p)*(1 + (e*x)/d)^(-1 - p)*(d^2 - e^2*x^2)^(1 + p)*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/
(2*d)])/(d^2*e*(1 + p)))

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Rubi [A]  time = 0.0328738, antiderivative size = 73, normalized size of antiderivative = 1., 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} \[ -\frac{2^{p-1} \left (\frac{e x}{d}+1\right )^{-p-1} \left (d^2-e^2 x^2\right )^{p+1} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d^2 e (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d^2 - e^2*x^2)^p/(d + e*x),x]

[Out]

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

Mathematica [A]  time = 0.0391966, size = 75, normalized size = 1.03 \[ -\frac{2^{p-1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \left (d^2-e^2 x^2\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d e (p+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d^2 - e^2*x^2)^p/(d + e*x),x]

[Out]

-((2^(-1 + p)*(d - e*x)*(d^2 - e^2*x^2)^p*Hypergeometric2F1[1 - p, 1 + p, 2 + p, (d - e*x)/(2*d)])/(d*e*(1 + p
)*(1 + (e*x)/d)^p))

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Maple [F]  time = 0.702, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{p}}{ex+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-e^2*x^2+d^2)^p/(e*x+d),x)

[Out]

int((-e^2*x^2+d^2)^p/(e*x+d),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-e^2*x^2 + d^2)^p/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-e^2*x^2 + d^2)^p/(e*x + d), x)

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Sympy [C]  time = 6.79971, size = 323, normalized size = 4.42 \begin{align*} \begin{cases} \frac{0^{p} \log{\left (-1 + \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 e} + \frac{0^{p} \operatorname{acoth}{\left (\frac{e x}{d} \right )}}{e} + \frac{d e^{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{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac{d^{2 p} e 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{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 d^{2} \Gamma \left (- p\right ) \Gamma \left (p + 1\right )} & \text{for}\: \frac{\left |{e^{2} x^{2}}\right |}{\left |{d^{2}}\right |} > 1 \\\frac{0^{p} \log{\left (1 - \frac{e^{2} x^{2}}{d^{2}} \right )}}{2 e} + \frac{0^{p} \operatorname{atanh}{\left (\frac{e x}{d} \right )}}{e} + \frac{d e^{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{d^{2}}{e^{2} x^{2}}} \right )}}{2 e^{2} x \Gamma \left (\frac{3}{2} - p\right ) \Gamma \left (p + 1\right )} + \frac{d^{2 p} e 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{e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )}}{2 d^{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((-e**2*x**2+d**2)**p/(e*x+d),x)

[Out]

Piecewise((0**p*log(-1 + e**2*x**2/d**2)/(2*e) + 0**p*acoth(e*x/d)/e + d*e**(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,), d**2/(e**2*x**2))/(2*e**2*x*gamma(3/2 - p)*gamma(p + 1)
) + d**(2*p)*e*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p), (2, 2), e**2*x**2*exp_polar(2*I*pi)/d**2)/(2*d*
*2*gamma(-p)*gamma(p + 1)), Abs(e**2*x**2)/Abs(d**2) > 1), (0**p*log(1 - e**2*x**2/d**2)/(2*e) + 0**p*atanh(e*
x/d)/e + d*e**(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,), d**2/(e
**2*x**2))/(2*e**2*x*gamma(3/2 - p)*gamma(p + 1)) + d**(2*p)*e*x**2*gamma(p)*gamma(1 - p)*hyper((2, 1, 1 - p),
(2, 2), e**2*x**2*exp_polar(2*I*pi)/d**2)/(2*d**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 (-e^{2} x^{2} + d^{2}\right )}^{p}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-e^2*x^2 + d^2)^p/(e*x + d), x)