3.5.16 \(\int \frac {x^4 (a+b x^2)^p}{(d+e x)^2} \, dx\) [416]

Optimal. Leaf size=392 \[ -\frac {d (4+3 p) \left (a+b x^2\right )^{1+p}}{b e^3 (1+p) (3+2 p)}-\frac {d^4 \left (a+b x^2\right )^{1+p}}{e^3 \left (b d^2+a e^2\right ) (d+e x)}+\frac {(d+e x) \left (a+b x^2\right )^{1+p}}{b e^3 (3+2 p)}-\frac {2 d^2 \left (2 a e^2+b d^2 (2+p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e^4 \left (b d^2+a e^2\right )}-\frac {\left (a^2 e^4-2 a b d^2 e^2 (4+3 p)-2 b^2 d^4 \left (6+7 p+2 p^2\right )\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b e^4 \left (b d^2+a e^2\right ) (3+2 p)}+\frac {d^3 \left (2 a e^2+b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p} \, _2F_1\left (1,1+p;2+p;\frac {e^2 \left (a+b x^2\right )}{b d^2+a e^2}\right )}{e^3 \left (b d^2+a e^2\right )^2 (1+p)} \]

[Out]

-d*(4+3*p)*(b*x^2+a)^(1+p)/b/e^3/(1+p)/(3+2*p)-d^4*(b*x^2+a)^(1+p)/e^3/(a*e^2+b*d^2)/(e*x+d)+(e*x+d)*(b*x^2+a)
^(1+p)/b/e^3/(3+2*p)-2*d^2*(2*a*e^2+b*d^2*(2+p))*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)/((1+b*x^2/a)^p)-(a^2*e^4-2*a*b*d^2*e^2*(4+3*p)-2*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)/b/e^4/(a*e^2+b*d^2)/(3+2*p)/((1+b*x^2/a)^p)+d^3*(2*a*e^2+b*d^2*(2+p))*(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)^2/(1+p)

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Rubi [A]
time = 0.58, antiderivative size = 392, 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 = {1665, 1668, 858, 252, 251, 771, 441, 440, 455, 70} \begin {gather*} -\frac {x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 e^4-2 a b d^2 e^2 (3 p+4)-2 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 )}{b e^4 (2 p+3) \left (a e^2+b d^2\right )}-\frac {2 d^2 x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (2 a e^2+b d^2 (p+2)\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 )}-\frac {d^4 \left (a+b x^2\right )^{p+1}}{e^3 (d+e x) \left (a e^2+b d^2\right )}+\frac {d^3 \left (a+b x^2\right )^{p+1} \left (2 a e^2+b d^2 (p+2)\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 )}{e^3 (p+1) \left (a e^2+b d^2\right )^2}-\frac {d (3 p+4) \left (a+b x^2\right )^{p+1}}{b e^3 (p+1) (2 p+3)}+\frac {(d+e x) \left (a+b x^2\right )^{p+1}}{b e^3 (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 251

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 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[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 440

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 441

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 455

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 771

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 858

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 1665

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 1668

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

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Mathematica [F]
time = 0.71, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 \left (a+b x^2\right )^p}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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