3.116 \(\int \frac {(d x)^m}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=33 \[ -\frac {d (d x)^{m-1} \, _2F_1\left (2,m-1;m;-\frac {c x}{b}\right )}{b^2 (1-m)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {647, 64} \[ -\frac {d (d x)^{m-1} \, _2F_1\left (2,m-1;m;-\frac {c x}{b}\right )}{b^2 (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(b*x + c*x^2)^2,x]

[Out]

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

Rule 64

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

Rule 647

Int[((e_.)*(x_))^(m_.)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e*x)^(m + p)*(b + c*x)
^p, x], x] /; FreeQ[{b, c, e, m}, x] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.91 \[ \frac {(d x)^m \, _2F_1\left (2,m-1;m;-\frac {c x}{b}\right )}{b^2 (m-1) x} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^m/(b*x + c*x^2)^2,x]

[Out]

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

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d x\right )^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x)^m/(c^2*x^4 + 2*b*c*x^3 + b^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{{\left (c x^{2} + b x\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

integrate((d*x)^m/(c*x^2 + b*x)^2, x)

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maple [F]  time = 0.51, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{\left (c \,x^{2}+b x \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(c*x^2+b*x)^2,x)

[Out]

int((d*x)^m/(c*x^2+b*x)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{{\left (c x^{2} + b x\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^m/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

integrate((d*x)^m/(c*x^2 + b*x)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (d\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(b*x + c*x^2)^2,x)

[Out]

int((d*x)^m/(b*x + c*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x\right )^{m}}{x^{2} \left (b + c x\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**m/(c*x**2+b*x)**2,x)

[Out]

Integral((d*x)**m/(x**2*(b + c*x)**2), x)

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