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

Optimal. Leaf size=421 \[ \frac{2 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4}+\frac{e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^6}-\frac{e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d^3 (p+1) \left (a e^2+b d^2\right )}-\frac{b e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d (p+1) \left (a e^2+b d^2\right )^2}+\frac{e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{a d^3 (p+1)}-\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^2 x} \]

[Out]

(2*e^2*x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^4*(1 + (b*x^2)/a)^p) + (e^2*
x*(a + b*x^2)^p*AppellF1[1/2, -p, 2, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^4*(1 + (b*x^2)/a)^p) + (e^4*x^3*(a
+ b*x^2)^p*AppellF1[3/2, -p, 2, 5/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(3*d^6*(1 + (b*x^2)/a)^p) - ((a + b*x^2)^p*
Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(d^2*x*(1 + (b*x^2)/a)^p) - (e^3*(a + b*x^2)^(1 + p)*Hypergeom
etric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(d^3*(b*d^2 + a*e^2)*(1 + p)) + (e*(a + b*x^2)^(
1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(a*d^3*(1 + p)) - (b*e^3*(a + b*x^2)^(1 + p)*Hyperge
ometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(d*(b*d^2 + a*e^2)^2*(1 + p))

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Rubi [A]  time = 0.446496, antiderivative size = 421, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 12, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {961, 365, 364, 266, 65, 757, 430, 429, 444, 68, 511, 510} \[ \frac{2 e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,1;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4}+\frac{e^2 x \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,2;\frac{3}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4}+\frac{e^4 x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,2;\frac{5}{2};-\frac{b x^2}{a},\frac{e^2 x^2}{d^2}\right )}{3 d^6}-\frac{e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d^3 (p+1) \left (a e^2+b d^2\right )}-\frac{b e^3 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{d (p+1) \left (a e^2+b d^2\right )^2}+\frac{e \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac{b x^2}{a}+1\right )}{a d^3 (p+1)}-\frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b x^2}{a}\right )}{d^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*x*(a + b*x^2)^p*AppellF1[1/2, -p, 1, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^4*(1 + (b*x^2)/a)^p) + (e^2*
x*(a + b*x^2)^p*AppellF1[1/2, -p, 2, 3/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(d^4*(1 + (b*x^2)/a)^p) + (e^4*x^3*(a
+ b*x^2)^p*AppellF1[3/2, -p, 2, 5/2, -((b*x^2)/a), (e^2*x^2)/d^2])/(3*d^6*(1 + (b*x^2)/a)^p) - ((a + b*x^2)^p*
Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^2)/a)])/(d^2*x*(1 + (b*x^2)/a)^p) - (e^3*(a + b*x^2)^(1 + p)*Hypergeom
etric2F1[1, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(d^3*(b*d^2 + a*e^2)*(1 + p)) + (e*(a + b*x^2)^(
1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^2)/a])/(a*d^3*(1 + p)) - (b*e^3*(a + b*x^2)^(1 + p)*Hyperge
ometric2F1[2, 1 + p, 2 + p, (e^2*(a + b*x^2))/(b*d^2 + a*e^2)])/(d*(b*d^2 + a*e^2)^2*(1 + p))

Rule 961

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

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
 1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 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 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 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 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 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 511

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

Rule 510

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

Rubi steps

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

Mathematica [F]  time = 0.13013, size = 0, normalized size = 0. \[ \int \frac{\left (a+b x^2\right )^p}{x^2 (d+e x)^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.64, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p}}{{x}^{2} \left ( ex+d \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{p}}{e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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