3.10.49 \(\int \frac {(a+b x)^m}{(a^2-b^2 x^2)^2} \, dx\) [949]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 44, 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-1} \, _2F_1\left (2,m-1;m;\frac {a+b x}{2 a}\right )}{4 a^2 b (1-m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m/(a^2 - b^2*x^2)^2,x]

[Out]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(102\) vs. \(2(44)=88\).
time = 0.19, size = 102, normalized size = 2.32 \begin {gather*} \frac {(a+b x)^m \left (4 a \left (\frac {1}{m}+\frac {a}{(-1+m) (a+b x)}\right )+\frac {2 (a+b x) \, _2F_1\left (1,1+m;2+m;\frac {a+b x}{2 a}\right )}{1+m}+\frac {(a+b x) \, _2F_1\left (2,1+m;2+m;\frac {a+b x}{2 a}\right )}{1+m}\right )}{16 a^4 b} \end {gather*}

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)

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {\left (b x +a \right )^{m}}{\left (-b^{2} x^{2}+a^{2}\right )^{2}}\, dx\]

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)

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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)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral((b*x + a)^m/(b^4*x^4 - 2*a^2*b^2*x^2 + a^4), 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}}{\left (- a + b x\right )^{2} \left (a + b x\right )^{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)**2,x)

[Out]

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

[Out]

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

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

[Out]

int((a + b*x)^m/(a^2 - b^2*x^2)^2, x)

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