### 3.738 $$\int \frac{(a+c x^2)^p}{(d+e x)^3} \, dx$$

Optimal. Leaf size=322 $\frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )^3}+\frac{c e \left (a+c x^2\right )^{p+1} \left (2 a e^2+c d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{4 (p+1) \left (a e^2+c d^2\right )^3}-\frac{d^2 e \left (a+c x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+c d^2\right )}$

[Out]

-(d^2*e*(a + c*x^2)^(1 + p))/(4*(c*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x*(a + c*x^2)^p*AppellF1[1/2, -p, 3, 3/2
, -((c*x^2)/a), (e^2*x^2)/d^2])/(d^3*(1 + (c*x^2)/a)^p) + (e^2*x^3*(a + c*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -((
c*x^2)/a), (e^2*x^2)/d^2])/(d^5*(1 + (c*x^2)/a)^p) + (c*e*(2*a*e^2 + c*d^2*(1 + p))*(a + c*x^2)^(1 + p)*Hyperg
eometric2F1[2, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(4*(c*d^2 + a*e^2)^3*(1 + p)) - (3*c^2*d^2*e*
(a + c*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(2*(c*d^2 + a*e^2)^
3*(1 + p))

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Rubi [A]  time = 0.315816, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.529, Rules used = {757, 430, 429, 444, 68, 511, 510, 446, 78} $\frac{x \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{1}{2};-p,3;\frac{3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3}+\frac{e^2 x^3 \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{3}{2};-p,3;\frac{5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5}-\frac{3 c^2 d^2 e \left (a+c x^2\right )^{p+1} \, _2F_1\left (3,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{2 (p+1) \left (a e^2+c d^2\right )^3}+\frac{c e \left (a+c x^2\right )^{p+1} \left (2 a e^2+c d^2 (p+1)\right ) \, _2F_1\left (2,p+1;p+2;\frac{e^2 \left (c x^2+a\right )}{c d^2+a e^2}\right )}{4 (p+1) \left (a e^2+c d^2\right )^3}-\frac{d^2 e \left (a+c x^2\right )^{p+1}}{4 \left (d^2-e^2 x^2\right )^2 \left (a e^2+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + c*x^2)^p/(d + e*x)^3,x]

[Out]

-(d^2*e*(a + c*x^2)^(1 + p))/(4*(c*d^2 + a*e^2)*(d^2 - e^2*x^2)^2) + (x*(a + c*x^2)^p*AppellF1[1/2, -p, 3, 3/2
, -((c*x^2)/a), (e^2*x^2)/d^2])/(d^3*(1 + (c*x^2)/a)^p) + (e^2*x^3*(a + c*x^2)^p*AppellF1[3/2, -p, 3, 5/2, -((
c*x^2)/a), (e^2*x^2)/d^2])/(d^5*(1 + (c*x^2)/a)^p) + (c*e*(2*a*e^2 + c*d^2*(1 + p))*(a + c*x^2)^(1 + p)*Hyperg
eometric2F1[2, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(4*(c*d^2 + a*e^2)^3*(1 + p)) - (3*c^2*d^2*e*
(a + c*x^2)^(1 + p)*Hypergeometric2F1[3, 1 + p, 2 + p, (e^2*(a + c*x^2))/(c*d^2 + a*e^2)])/(2*(c*d^2 + a*e^2)^
3*(1 + p))

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])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

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

Mathematica [A]  time = 0.121307, size = 142, normalized size = 0.44 $\frac{\left (a+c x^2\right )^p \left (\frac{e \left (x-\sqrt{-\frac{a}{c}}\right )}{d+e x}\right )^{-p} \left (\frac{e \left (\sqrt{-\frac{a}{c}}+x\right )}{d+e x}\right )^{-p} F_1\left (2-2 p;-p,-p;3-2 p;\frac{d-\sqrt{-\frac{a}{c}} e}{d+e x},\frac{d+\sqrt{-\frac{a}{c}} e}{d+e x}\right )}{2 e (p-1) (d+e x)^2}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.58, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c{x}^{2}+a \right ) ^{p}}{ \left ( ex+d \right ) ^{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + a\right )}^{p}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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