3.10.48 \(\int \frac {(a+b x)^m}{a^2-b^2 x^2} \, dx\) [948]

Optimal. Leaf size=38 \[ \frac {(a+b x)^m \, _2F_1\left (1,m;1+m;\frac {a+b x}{2 a}\right )}{2 a b m} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 = {641, 70} \begin {gather*} \frac {(a+b x)^m \, _2F_1\left (1,m;m+1;\frac {a+b x}{2 a}\right )}{2 a b m} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m/(a^2 - b^2*x^2),x]

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

\begin {align*} \int \frac {(a+b x)^m}{a^2-b^2 x^2} \, dx &=\int \frac {(a+b x)^{-1+m}}{a-b x} \, dx\\ &=\frac {(a+b x)^m \, _2F_1\left (1,m;1+m;\frac {a+b x}{2 a}\right )}{2 a b m}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 59, normalized size = 1.55 \begin {gather*} \frac {(a+b x)^m \left (2 a (1+m)+m (a+b x) \, _2F_1\left (1,1+m;2+m;\frac {a+b x}{2 a}\right )\right )}{4 a^2 b m (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^m*(2*a*(1 + m) + m*(a + b*x)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*x)/(2*a)]))/(4*a^2*b*m*(1 +
m))

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m}}{-b^{2} x^{2}+a^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m/(-b^2*x^2+a^2),x)

[Out]

int((b*x+a)^m/(-b^2*x^2+a^2),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*x+a)^m/(-b^2*x^2+a^2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {\left (a + b x\right )^{m}}{- a^{2} + b^{2} x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral((a + b*x)**m/(-a**2 + b**2*x**2), x)

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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*x+a)^m/(-b^2*x^2+a^2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^m}{a^2-b^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^m/(a^2 - b^2*x^2),x)

[Out]

int((a + b*x)^m/(a^2 - b^2*x^2), x)

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