3.5.22 \(\int \frac {(a+b x^2)^p}{x^2 (d+e x)^2} \, dx\) [422]

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

[Out]

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

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Rubi [A]
time = 0.30, antiderivative size = 421, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {975, 372, 371, 272, 67, 771, 441, 440, 455, 70, 525, 524} \begin {gather*} \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 {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 \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 {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 {\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}-\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 )} \end {gather*}

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 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

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 272

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 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 975

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])

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 \text {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 \text {Subst}\left (\int \frac {(a+b x)^p}{-d^2+e^2 x} \, dx,x,x^2\right )}{d^3}-\frac {e^3 \text {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*}

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Mathematica [A]
time = 0.53, size = 342, normalized size = 0.81 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (\frac {d e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\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 )}{(-1+2 p) (d+e x)}+\frac {e \left (\frac {e \left (-\sqrt {-\frac {a}{b}}+x\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}-\frac {d \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 )}{x}-\frac {e \left (1+\frac {a}{b x^2}\right )^{-p} \, _2F_1\left (-p,-p;1-p;-\frac {a}{b x^2}\right )}{p}\right )}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{2}+a \right )^{p}}{x^{2} \left (e x +d \right )^{2}}\, dx\]

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 = 0.00, 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((b*x^2+a)^p/x^2/(e*x+d)^2,x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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