3.423 \(\int \frac {x^4 (a+b x^2)^p}{(d+e x)^3} \, dx\)

Optimal. Leaf size=449 \[ \frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^4 \left (a e^2+b d^2\right )^2}-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^4 \left (a e^2+b d^2\right )^2}-\frac {d^2 \left (a+b x^2\right )^{p+1} \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^3 (p+1) \left (a e^2+b d^2\right )^3}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}+\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )^2}+\frac {\left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)} \]

[Out]

1/2*(b*x^2+a)^(1+p)/b/e^3/(1+p)-1/2*d^4*(b*x^2+a)^(1+p)/e^3/(a*e^2+b*d^2)/(e*x+d)^2+d^3*(4*a*e^2+b*d^2*(3+p))*
(b*x^2+a)^(1+p)/e^3/(a*e^2+b*d^2)^2/(e*x+d)+d*(6*a^2*e^4+3*a*b*d^2*e^2*(4+3*p)+b^2*d^4*(2*p^2+7*p+6))*x*(b*x^2
+a)^p*AppellF1(1/2,1,-p,3/2,e^2*x^2/d^2,-b*x^2/a)/e^4/(a*e^2+b*d^2)^2/((1+b*x^2/a)^p)-d*(3*a^2*e^4+2*a*b*d^2*e
^2*(5+4*p)+b^2*d^4*(2*p^2+7*p+6))*x*(b*x^2+a)^p*hypergeom([1/2, -p],[3/2],-b*x^2/a)/e^4/(a*e^2+b*d^2)^2/((1+b*
x^2/a)^p)-1/2*d^2*(6*a^2*e^4+3*a*b*d^2*e^2*(4+3*p)+b^2*d^4*(2*p^2+7*p+6))*(b*x^2+a)^(1+p)*hypergeom([1, 1+p],[
2+p],e^2*(b*x^2+a)/(a*e^2+b*d^2))/e^3/(a*e^2+b*d^2)^3/(1+p)

________________________________________________________________________________________

Rubi [A]  time = 0.94, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1651, 1654, 844, 246, 245, 757, 430, 429, 444, 68} \[ \frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^4 \left (a e^2+b d^2\right )^2}-\frac {d x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a^2 e^4+2 a b d^2 e^2 (4 p+5)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^4 \left (a e^2+b d^2\right )^2}-\frac {d^2 \left (a+b x^2\right )^{p+1} \left (6 a^2 e^4+3 a b d^2 e^2 (3 p+4)+b^2 d^4 \left (2 p^2+7 p+6\right )\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 e^3 (p+1) \left (a e^2+b d^2\right )^3}+\frac {d^3 \left (a+b x^2\right )^{p+1} \left (4 a e^2+b d^2 (p+3)\right )}{e^3 (d+e x) \left (a e^2+b d^2\right )^2}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{2 e^3 (d+e x)^2 \left (a e^2+b d^2\right )}+\frac {\left (a+b x^2\right )^{p+1}}{2 b e^3 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^2)^p)/(d + e*x)^3,x]

[Out]

(a + b*x^2)^(1 + p)/(2*b*e^3*(1 + p)) - (d^4*(a + b*x^2)^(1 + p))/(2*e^3*(b*d^2 + a*e^2)*(d + e*x)^2) + (d^3*(
4*a*e^2 + b*d^2*(3 + p))*(a + b*x^2)^(1 + p))/(e^3*(b*d^2 + a*e^2)^2*(d + e*x)) + (d*(6*a^2*e^4 + 3*a*b*d^2*e^
2*(4 + 3*p) + b^2*d^4*(6 + 7*p + 2*p^2))*x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2
])/(e^4*(b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p) - (d*(3*a^2*e^4 + 2*a*b*d^2*e^2*(5 + 4*p) + b^2*d^4*(6 + 7*p + 2*
p^2))*x*(a + b*x^2)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(e^4*(b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p)
- (d^2*(6*a^2*e^4 + 3*a*b*d^2*e^2*(4 + 3*p) + b^2*d^4*(6 + 7*p + 2*p^2))*(a + b*x^2)^(1 + p)*Hypergeometric2F
1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(2*e^3*(b*d^2 + a*e^2)^3*(1 + p))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 757

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1651

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, d
+ e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1
)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p*ExpandToSum[(m
+ 1)*(c*d^2 + a*e^2)*Q + c*d*R*(m + 1) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e, p}, x] && Po
lyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^3} \, dx &=-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}-\frac {\int \frac {\left (a+b x^2\right )^p \left (\frac {2 a d^3}{e^2}-\frac {2 d^2 \left (a e^2+b d^2 (1+p)\right ) x}{e^3}+2 d \left (a+\frac {b d^2}{e^2}\right ) x^2-2 \left (\frac {b d^2}{e}+a e\right ) x^3\right )}{(d+e x)^2} \, dx}{2 \left (b d^2+a e^2\right )}\\ &=-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\int \frac {\left (a+b x^2\right )^p \left (2 a d^2 \left (3 a+\frac {b d^2 (2+p)}{e^2}\right )-\frac {2 d \left (2 a^2 e^4+8 a b d^2 e^2 (1+p)+b^2 d^4 \left (5+7 p+2 p^2\right )\right ) x}{e^3}+\frac {2 \left (b d^2+a e^2\right )^2 x^2}{e^2}\right )}{d+e x} \, dx}{2 \left (b d^2+a e^2\right )^2}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\int \frac {\left (4 a b d^2 (1+p) \left (3 a e^2+b d^2 (2+p)\right )-\frac {4 b d (1+p) \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x}{e}\right ) \left (a+b x^2\right )^p}{d+e x} \, dx}{4 b e^2 \left (b d^2+a e^2\right )^2 (1+p)}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d+e x} \, dx}{e^4 \left (b d^2+a e^2\right )^2}-\frac {\left (d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \int \left (a+b x^2\right )^p \, dx}{e^4 \left (b d^2+a e^2\right )^2}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {\left (d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \int \left (\frac {d \left (a+b x^2\right )^p}{d^2-e^2 x^2}+\frac {e x \left (a+b x^2\right )^p}{-d^2+e^2 x^2}\right ) \, dx}{e^4 \left (b d^2+a e^2\right )^2}-\frac {\left (d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx}{e^4 \left (b d^2+a e^2\right )^2}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}-\frac {d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^4 \left (b d^2+a e^2\right )^2}+\frac {\left (d^3 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \int \frac {\left (a+b x^2\right )^p}{d^2-e^2 x^2} \, dx}{e^4 \left (b d^2+a e^2\right )^2}+\frac {\left (d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \int \frac {x \left (a+b x^2\right )^p}{-d^2+e^2 x^2} \, dx}{e^3 \left (b d^2+a e^2\right )^2}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}-\frac {d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^4 \left (b d^2+a e^2\right )^2}+\frac {\left (d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{2 e^3 \left (b d^2+a e^2\right )^2}+\frac {\left (d^3 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p}{d^2-e^2 x^2} \, dx}{e^4 \left (b d^2+a e^2\right )^2}\\ &=\frac {\left (a+b x^2\right )^{1+p}}{2 b e^3 (1+p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{2 e^3 \left (b d^2+a e^2\right ) (d+e x)^2}+\frac {d^3 \left (4 a e^2+b d^2 (3+p)\right ) \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right )^2 (d+e x)}+\frac {d \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^4 \left (b d^2+a e^2\right )^2}-\frac {d \left (3 a^2 e^4+2 a b d^2 e^2 (5+4 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e^4 \left (b d^2+a e^2\right )^2}-\frac {d^2 \left (6 a^2 e^4+3 a b d^2 e^2 (4+3 p)+b^2 d^4 \left (6+7 p+2 p^2\right )\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{2 e^3 \left (b d^2+a e^2\right )^3 (1+p)}\\ \end {align*}

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Mathematica [A]  time = 0.93, size = 462, normalized size = 1.03 \[ \frac {\left (a+b x^2\right )^p \left (\frac {d^4 \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(p-1) (d+e x)^2}-\frac {8 d^3 \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(2 p-1) (d+e x)}+\frac {6 d^2 \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}-6 d e x \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+\frac {a e^2}{b p+b}+\frac {e^2 x^2}{p+1}\right )}{2 e^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*x^2)^p)/(d + e*x)^3,x]

[Out]

((a + b*x^2)^p*((a*e^2)/(b + b*p) + (e^2*x^2)/(1 + p) - (8*d^3*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(
a/b)]*e)/(d + e*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/((-1 + 2*p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sq
rt[-(a/b)] + x))/(d + e*x))^p*(d + e*x)) + (d^4*AppellF1[2 - 2*p, -p, -p, 3 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e
*x), (d + Sqrt[-(a/b)]*e)/(d + e*x)])/((-1 + p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/
(d + e*x))^p*(d + e*x)^2) + (6*d^2*AppellF1[-2*p, -p, -p, 1 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + Sqrt[-
(a/b)]*e)/(d + e*x)])/(p*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/b)] + x))/(d + e*x))^p) - (6*d*e*
x*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^p))/(2*e^5)

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p*x^4/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*x^4/(e*x + d)^3, x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^p/(e*x+d)^3,x)

[Out]

int(x^4*(b*x^2+a)^p/(e*x+d)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{p} x^{4}}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^p/(e*x+d)^3,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*x^4/(e*x + d)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (b\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^2)^p)/(d + e*x)^3,x)

[Out]

int((x^4*(a + b*x^2)^p)/(d + e*x)^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**p/(e*x+d)**3,x)

[Out]

Timed out

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