### 3.948 $$\int \frac{(a+b x)^m}{a^2-b^2 x^2} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0133872, antiderivative size = 38, 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 = {627, 68} $\frac{(a+b x)^m \, _2F_1\left (1,m;m+1;\frac{a+b x}{2 a}\right )}{2 a b m}$

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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]))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

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*}

Mathematica [A]  time = 0.0421651, size = 59, normalized size = 1.55 $\frac{(a+b x)^m \left (m (a+b x) \, _2F_1\left (1,m+1;m+2;\frac{a+b x}{2 a}\right )+2 a (m+1)\right )}{4 a^2 b m (m+1)}$

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 = 0.522, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{-{b}^{2}{x}^{2}+{a}^{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (b x + a\right )}^{m}}{b^{2} x^{2} - a^{2}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b x + a\right )}^{m}}{b^{2} x^{2} - a^{2}}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} - \int \frac{\left (a + b x\right )^{m}}{- a^{2} + b^{2} x^{2}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b x + a\right )}^{m}}{b^{2} x^{2} - a^{2}}\,{d x} \end{align*}

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)