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

Optimal. Leaf size=336 \[ -\frac {b p x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 p\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b d p \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}-\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right )}{(d+e x) \left (a e^2+b d^2\right )^2}+\frac {d \left (a+b x^2\right )^{p+1}}{2 (d+e x)^2 \left (a e^2+b d^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {835, 844, 246, 245, 757, 430, 429, 444, 68} \[ -\frac {b p x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 a e^2+b d^2 (2 p+1)\right ) F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b (2 p+1) x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a e^2+b d^2 p\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{e \left (a e^2+b d^2\right )^2}+\frac {b d p \left (a+b x^2\right )^{p+1} \left (3 a e^2+b d^2 (2 p+1)\right ) \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 (p+1) \left (a e^2+b d^2\right )^3}-\frac {\left (a+b x^2\right )^{p+1} \left (a e^2+b d^2 p\right )}{(d+e x) \left (a e^2+b d^2\right )^2}+\frac {d \left (a+b x^2\right )^{p+1}}{2 (d+e x)^2 \left (a e^2+b d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*x^2)^p)/(d + e*x)^3,x]

[Out]

(d*(a + b*x^2)^(1 + p))/(2*(b*d^2 + a*e^2)*(d + e*x)^2) - ((a*e^2 + b*d^2*p)*(a + b*x^2)^(1 + p))/((b*d^2 + a*
e^2)^2*(d + e*x)) - (b*p*(3*a*e^2 + b*d^2*(1 + 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*(b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p) + (b*(1 + 2*p)*(a*e^2 + b*d^2*p)*x*(a + b*x^2)^p*Hyperg
eometric2F1[1/2, -p, 3/2, -((b*x^2)/a)])/(e*(b*d^2 + a*e^2)^2*(1 + (b*x^2)/a)^p) + (b*d*p*(3*a*e^2 + b*d^2*(1
+ 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)])/(2*(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 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(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*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 229, normalized size = 0.68 \[ \frac {\left (a+b x^2\right )^p \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} \left (2 (p-1) (d+e x) F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )+d (1-2 p) F_1\left (2-2 p;-p,-p;3-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )\right )}{2 e^2 (p-1) (2 p-1) (d+e x)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + b*x^2)^p*(2*(-1 + p)*(d + e*x)*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (d - Sqrt[-(a/b)]*e)/(d + e*x), (d + S
qrt[-(a/b)]*e)/(d + e*x)] + d*(1 - 2*p)*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)]))/(2*e^2*(-1 + p)*(-1 + 2*p)*((e*(-Sqrt[-(a/b)] + x))/(d + e*x))^p*((e*(Sqrt[-(a/
b)] + x))/(d + e*x))^p*(d + e*x)^2)

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p} x}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^2 + a)^p*x/(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 \[ \int \frac {{\left (b x^{2} + a\right )}^{p} x}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {x \left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \[ \int \frac {{\left (b x^{2} + a\right )}^{p} x}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*x^2)^p)/(d + e*x)^3,x)

[Out]

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

[Out]

Timed out

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