3.10.70 \(\int \frac {(a^2-b^2 x^2)^p}{a+b x} \, dx\) [970]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {692, 71} \begin {gather*} -\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)} \end {gather*}

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 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 692

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/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]) &&  !(IGtQ[m, 0] && (
IntegerQ[3*p] || IntegerQ[4*p]))

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*}

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Mathematica [A]
time = 0.18, size = 75, normalized size = 1.36 \begin {gather*} -\frac {2^{-1+p} (a-b x) \left (1+\frac {b x}{a}\right )^{-p} \left (a^2-b^2 x^2\right )^p \, _2F_1\left (1-p,1+p;2+p;\frac {a-b x}{2 a}\right )}{a b (1+p)} \end {gather*}

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 = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (-b^{2} x^{2}+a^{2}\right )^{p}}{b x +a}\, dx\]

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 = 0.00, 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((-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 = 0.00, 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((-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] Result contains complex when optimal does not.
time = 2.97, size = 321, normalized size = 5.84 \begin {gather*} \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}\: \left |{\frac {b^{2} x^{2}}{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 {gather*}

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/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 = 0.00, 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((-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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 - b^2*x^2)^p/(a + b*x),x)

[Out]

int((a^2 - b^2*x^2)^p/(a + b*x), x)

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