3.21.88 \(\int \frac {(d+e x)^m}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [2088]

Optimal. Leaf size=61 \[ -\frac {e (d+e x)^{-1+m} \, _2F_1\left (2,-1+m;m;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{\left (c d^2-a e^2\right )^2 (1-m)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {640, 70} \begin {gather*} -\frac {e (d+e x)^{m-1} \, _2F_1\left (2,m-1;m;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{(1-m) \left (c d^2-a e^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

-((e*(d + e*x)^(-1 + m)*Hypergeometric2F1[2, -1 + m, m, (c*d*(d + e*x))/(c*d^2 - a*e^2)])/((c*d^2 - a*e^2)^2*(
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 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
 IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 59, normalized size = 0.97 \begin {gather*} \frac {e (d+e x)^{-1+m} \, _2F_1\left (2,-1+m;m;-\frac {c d (d+e x)}{-c d^2+a e^2}\right )}{\left (-c d^2+a e^2\right )^2 (-1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^m/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]

[Out]

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

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Maple [F]
time = 0.48, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="maxima")

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m/((d + e*x)**2*(a*e + c*d*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((e*x+d)^m/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm="giac")

[Out]

integrate((x*e + d)^m/(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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