3.21.96 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^p}{d+e x} \, dx\) [2096]

Optimal. Leaf size=93 \[ \frac {\left (-\frac {e (a e+c d x)}{c d^2-a e^2}\right )^{-p} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^p \, _2F_1\left (-p,p;1+p;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{e p} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {691, 72, 71} \begin {gather*} \frac {\left (-\frac {e (a e+c d 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 (-p,p;p+1;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{e p} \end {gather*}

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),x]

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 72

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 691

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

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Mathematica [A]
time = 0.05, size = 81, normalized size = 0.87 \begin {gather*} \frac {\left (\frac {e (a e+c d x)}{-c d^2+a e^2}\right )^{-p} ((a e+c d x) (d+e x))^p \, _2F_1\left (-p,p;1+p;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{e p} \end {gather*}

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),x]

[Out]

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

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Maple [F]
time = 0.36, size = 0, normalized size = 0.00 \[\int \frac {\left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )^{p}}{e x +d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

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),x)

[Out]

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

[Out]

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

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

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),x)

[Out]

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

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