3.2097 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^p}{(d+e x)^2} \, dx\)

Optimal. Leaf size=92 \[ \frac {(a e+c d x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^p \, _2F_1\left (1,2 p;p;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{(1-p) (d+e x) \left (c d^2-a e^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.30, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {677, 70, 69} \[ \frac {c d (a e+c d x) \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{-p} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^p \, _2F_1\left (2-p,p+1;p+2;-\frac {e (a e+c d x)}{c d^2-a e^2}\right )}{(p+1) \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 69

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

Rule 70

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

Rule 677

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d^m*(a + b*x + c*x^2
)^FracPart[p])/((1 + (e*x)/d)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(1 + (e*x)/d)^(m + p)*(a/d + (c*x)
/e)^p, x], 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] &&  !Int
egerQ[p] && (IntegerQ[m] || GtQ[d, 0]) &&  !(IGtQ[m, 0] && (IntegerQ[3*p] || IntegerQ[4*p]))

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 108, normalized size = 1.17 \[ \frac {c d (a e+c d x) \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{-p} ((d+e x) (a e+c d x))^p \, _2F_1\left (2-p,p+1;p+2;\frac {e (a e+c d x)}{a e^2-c d^2}\right )}{(p+1) \left (c d^2-a e^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{p}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^p/(e^2*x^2 + 2*d*e*x + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{p}}{\left (e x +d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((d + e*x)*(a*e + c*d*x))**p/(d + e*x)**2, x)

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