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

Optimal. Leaf size=368 \[ -\frac {e^3 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,2;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^5}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,2;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}+\frac {e^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^2 (p+1) \left (a e^2+b d^2\right )}+\frac {b e^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{(p+1) \left (a e^2+b d^2\right )^2}-\frac {\left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a d^2 (p+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.43, antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {961, 266, 65, 757, 430, 429, 444, 68, 511, 510} \[ -\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,1;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e x \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,2;\frac {3}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^3}-\frac {e^3 x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,2;\frac {5}{2};-\frac {b x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^5}+\frac {e^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{2 d^2 (p+1) \left (a e^2+b d^2\right )}+\frac {b e^2 \left (a+b x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (b x^2+a\right )}{b d^2+a e^2}\right )}{(p+1) \left (a e^2+b d^2\right )^2}-\frac {\left (a+b x^2\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b x^2}{a}+1\right )}{2 a d^2 (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^p/(x*(d + e*x)^2),x]

[Out]

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

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 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 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 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 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])

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 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 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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.35, size = 303, normalized size = 0.82 \[ \frac {\left (a+b x^2\right )^p \left (\frac {\left (\frac {a}{b x^2}+1\right )^{-p} \, _2F_1\left (-p,-p;1-p;-\frac {a}{b x^2}\right )-\left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (-2 p;-p,-p;1-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{p}-\frac {2 d \left (\frac {e \left (x-\sqrt {-\frac {a}{b}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{b}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{b}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{b}} e}{d+e x}\right )}{(2 p-1) (d+e x)}\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{p}}{e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2} x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^p/(x*(d + e*x)^2),x)

[Out]

int((a + b*x^2)^p/(x*(d + e*x)^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________