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

Optimal. Leaf size=191 \[ \frac {x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,2;\frac {3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2}+\frac {e^2 x^3 \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,2;\frac {5}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^4}-\frac {c d e \left (a+c x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{(p+1) \left (a e^2+c d^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {757, 430, 429, 444, 68, 511, 510} \[ \frac {x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {1}{2};-p,2;\frac {3}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{d^2}+\frac {e^2 x^3 \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} F_1\left (\frac {3}{2};-p,2;\frac {5}{2};-\frac {c x^2}{a},\frac {e^2 x^2}{d^2}\right )}{3 d^4}-\frac {c d e \left (a+c x^2\right )^{p+1} \, _2F_1\left (2,p+1;p+2;\frac {e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{(p+1) \left (a e^2+c d^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^p/(d + e*x)^2,x]

[Out]

(x*(a + c*x^2)^p*AppellF1[1/2, -p, 2, 3/2, -((c*x^2)/a), (e^2*x^2)/d^2])/(d^2*(1 + (c*x^2)/a)^p) + (e^2*x^3*(a
+ c*x^2)^p*AppellF1[3/2, -p, 2, 5/2, -((c*x^2)/a), (e^2*x^2)/d^2])/(3*d^4*(1 + (c*x^2)/a)^p) - (c*d*e*(a + c*
x^2)^(1 + p)*Hypergeometric2F1[2, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/((c*d^2 + a*e^2)^2*(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 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]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 141, normalized size = 0.74 \[ \frac {\left (a+c x^2\right )^p \left (\frac {e \left (x-\sqrt {-\frac {a}{c}}\right )}{d+e x}\right )^{-p} \left (\frac {e \left (\sqrt {-\frac {a}{c}}+x\right )}{d+e x}\right )^{-p} F_1\left (1-2 p;-p,-p;2-2 p;\frac {d-\sqrt {-\frac {a}{c}} e}{d+e x},\frac {d+\sqrt {-\frac {a}{c}} e}{d+e x}\right )}{e (2 p-1) (d+e x)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (c x^{2} + a\right )}^{p}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.75, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+a \right )^{p}}{\left (e x +d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)^p/(d + e*x)^2,x)

[Out]

int((a + c*x^2)^p/(d + e*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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