### 3.964 $$\int \frac{(1-\frac{e^2 x^2}{d^2})^p}{d+e x} \, dx$$

Optimal. Leaf size=41 $\frac{2^p \left (\frac{d+e x}{d}\right )^p \, _2F_1\left (-p,p;p+1;\frac{d+e x}{2 d}\right )}{e p}$

[Out]

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

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Rubi [A]  time = 0.03575, antiderivative size = 54, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {676, 69} $-\frac{2^{p-1} \left (\frac{d-e x}{d}\right )^{p+1} \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{e (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 676

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a^(p + 1)*d^(m - 1)*((d - 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]) && GtQ[a, 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 (1-\frac{e^2 x^2}{d^2}\right )^p}{d+e x} \, dx &=\frac{\left (\left (\frac{d-e x}{d}\right )^{1+p} \left (\frac{1}{d}-\frac{e x}{d^2}\right )^{-1-p}\right ) \int \left (\frac{1}{d}-\frac{e x}{d^2}\right )^p \left (1+\frac{e x}{d}\right )^{-1+p} \, dx}{d^2}\\ &=-\frac{2^{-1+p} \left (\frac{d-e x}{d}\right )^{1+p} \, _2F_1\left (1-p,1+p;2+p;\frac{d-e x}{2 d}\right )}{e (1+p)}\\ \end{align*}

Mathematica [A]  time = 0.0859912, size = 76, normalized size = 1.85 $-\frac{2^{p-1} (d-e x) \left (\frac{e x}{d}+1\right )^{-p} \left (1-\frac{e^2 x^2}{d^2}\right )^p \, _2F_1\left (1-p,p+1;p+2;\frac{d-e x}{2 d}\right )}{d e (p+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((2^(-1 + p)*(d - e*x)*(1 - (e^2*x^2)/d^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 = 0.529, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ex+d} \left ( 1-{\frac{{e}^{2}{x}^{2}}{{d}^{2}}} \right ) ^{p}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1-e^2*x^2/d^2)^p/(e*x+d),x)

[Out]

int((1-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 (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d),x, algorithm="maxima")

[Out]

integrate((-e^2*x^2/d^2 + 1)^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 (-\frac{e^{2} x^{2} - d^{2}}{d^{2}}\right )^{p}}{e x + d}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 5.18054, size = 323, normalized size = 7.88 \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 d^{- 2 p} 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{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 d^{- 2 p} 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{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((1-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*d**(-2*p)*e**(2*p)*p*x**(2*p)*exp(I*p
i*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)*ga
mma(p + 1)) + 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*d**(-2*p)*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)) + 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 (-\frac{e^{2} x^{2}}{d^{2}} + 1\right )}^{p}}{e x + d}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-e^2*x^2/d^2)^p/(e*x+d),x, algorithm="giac")

[Out]

integrate((-e^2*x^2/d^2 + 1)^p/(e*x + d), x)