3.21.92 \(\int (d+e x)^3 (a d e+(c d^2+a e^2) x+c d e x^2)^p \, dx\) [2092]

Optimal. Leaf size=95 \[ -\frac {(a e+c d x) (d+e x)^4 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^p \, _2F_1\left (1,5+2 p;5+p;\frac {c d (d+e x)}{c d^2-a e^2}\right )}{\left (c d^2-a e^2\right ) (4+p)} \]

[Out]

-(c*d*x+a*e)*(e*x+d)^4*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p*hypergeom([1, 5+2*p],[5+p],c*d*(e*x+d)/(-a*e^2+c*d^
2))/(-a*e^2+c*d^2)/(4+p)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 124, normalized size of antiderivative = 1.31, 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 (c d^2-a e^2\right )^3 (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 (-p-3,p+1;p+2;-\frac {e (a e+c d x)}{c d^2-a e^2}\right )}{c^4 d^4 (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^p,x]

[Out]

((c*d^2 - a*e^2)^3*(a*e + c*d*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^p*Hypergeometric2F1[-3 - p, 1 + p, 2
+ p, -((e*(a*e + c*d*x))/(c*d^2 - a*e^2))])/(c^4*d^4*(1 + p)*((c*d*(d + e*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 (d+e x)^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^p \, dx &=\left (d^3 (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 )^{3+p} \, dx\\ &=\frac {\left (\left (c d-\frac {a e^2}{d}\right )^3 (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 )^{3+p} \, dx}{c^3}\\ &=\frac {\left (c d^2-a e^2\right )^3 (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 (-3-p,1+p;2+p;-\frac {e (a e+c d x)}{c d^2-a e^2}\right )}{c^4 d^4 (1+p)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.07, size = 112, normalized size = 1.18 \begin {gather*} \frac {\left (c d^2-a e^2\right )^3 (a e+c d x) \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{-p} ((a e+c d x) (d+e x))^p \, _2F_1\left (-3-p,1+p;2+p;\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{c^4 d^4 (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^p,x]

[Out]

((c*d^2 - a*e^2)^3*(a*e + c*d*x)*((a*e + c*d*x)*(d + e*x))^p*Hypergeometric2F1[-3 - p, 1 + p, 2 + p, (e*(a*e +
c*d*x))/(-(c*d^2) + a*e^2)])/(c^4*d^4*(1 + p)*((c*d*(d + e*x))/(c*d^2 - a*e^2))^p)

________________________________________________________________________________________

Maple [F]
time = 0.45, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{3} \left (a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p,x)

[Out]

int((e*x+d)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p,x)

________________________________________________________________________________________

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)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p,x, algorithm="maxima")

[Out]

integrate((x*e + d)^3*(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^p, x)

________________________________________________________________________________________

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)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p,x, algorithm="fricas")

[Out]

integral((x^3*e^3 + 3*d*x^2*e^2 + 3*d^2*x*e + d^3)*(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)^p, x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**p,x)

[Out]

Timed out

________________________________________________________________________________________

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)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^p,x, algorithm="giac")

[Out]

integrate((x*e + d)^3*(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x)^p, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d+e\,x\right )}^3\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^p,x)

[Out]

int((d + e*x)^3*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^p, x)

________________________________________________________________________________________