3.10.64 \(\int \frac {(1-\frac {e^2 x^2}{d^2})^p}{d+e x} \, dx\) [964]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {690, 71} \begin {gather*} -\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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 690

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/e)*x)^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]))

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*}

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Mathematica [A]
time = 0.22, size = 76, normalized size = 1.85 \begin {gather*} -\frac {2^{-1+p} (d-e x) \left (1+\frac {e x}{d}\right )^{-p} \left (1-\frac {e^2 x^2}{d^2}\right )^p \, _2F_1\left (1-p,1+p;2+p;\frac {d-e x}{2 d}\right )}{d e (1+p)} \end {gather*}

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 = 0.13, size = 0, normalized size = 0.00 \[\int \frac {\left (1-\frac {e^{2} x^{2}}{d^{2}}\right )^{p}}{e x +d}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((-x^2*e^2/d^2 + 1)^p/(x*e + d), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Sympy [C] Result contains complex when optimal does not.
time = 2.91, size = 321, normalized size = 7.83 \begin {gather*} \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}\: \left |{\frac {e^{2} x^{2}}{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 {gather*}

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*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*d**(2*p)*e**2*x*gamma(3/2 - p)*gam
ma(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/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*d**(2*p)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((-x^2*e^2/d^2 + 1)^p/(x*e + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (1-\frac {e^2\,x^2}{d^2}\right )}^p}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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