### 3.949 $$\int \frac{(a+b x)^m}{(a^2-b^2 x^2)^2} \, dx$$

Optimal. Leaf size=44 $-\frac{(a+b x)^{m-1} \, _2F_1\left (2,m-1;m;\frac{a+b x}{2 a}\right )}{4 a^2 b (1-m)}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0154593, antiderivative size = 44, 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} $-\frac{(a+b x)^{m-1} \, _2F_1\left (2,m-1;m;\frac{a+b x}{2 a}\right )}{4 a^2 b (1-m)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*x)^m/(a^2 - b^2*x^2)^2,x]

[Out]

-((a + b*x)^(-1 + m)*Hypergeometric2F1[2, -1 + m, m, (a + b*x)/(2*a)])/(4*a^2*b*(1 - 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}{\left (a^2-b^2 x^2\right )^2} \, dx &=\int \frac{(a+b x)^{-2+m}}{(a-b x)^2} \, dx\\ &=-\frac{(a+b x)^{-1+m} \, _2F_1\left (2,-1+m;m;\frac{a+b x}{2 a}\right )}{4 a^2 b (1-m)}\\ \end{align*}

Mathematica [B]  time = 0.150645, size = 102, normalized size = 2.32 $\frac{(a+b x)^m \left (\frac{2 (a+b x) \, _2F_1\left (1,m+1;m+2;\frac{a+b x}{2 a}\right )}{m+1}+\frac{(a+b x) \, _2F_1\left (2,m+1;m+2;\frac{a+b x}{2 a}\right )}{m+1}+4 a \left (\frac{a}{(m-1) (a+b x)}+\frac{1}{m}\right )\right )}{16 a^4 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((a + b*x)^m*(4*a*(m^(-1) + a/((-1 + m)*(a + b*x))) + (2*(a + b*x)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*x
)/(2*a)])/(1 + m) + ((a + b*x)*Hypergeometric2F1[2, 1 + m, 2 + m, (a + b*x)/(2*a)])/(1 + m)))/(16*a^4*b)

________________________________________________________________________________________

Maple [F]  time = 0.505, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( -{b}^{2}{x}^{2}+{a}^{2} \right ) ^{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)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{{\left (b^{2} x^{2} - a^{2}\right )}^{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)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}}{b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}}, 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)^2,x, algorithm="fricas")

[Out]

integral((b*x + a)^m/(b^4*x^4 - 2*a^2*b^2*x^2 + a^4), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{{\left (b^{2} x^{2} - a^{2}\right )}^{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)^2,x, algorithm="giac")

[Out]

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