3.436 \(\int \frac{(g x)^m (a+c x^2)^p}{(d+e x)^3} \, dx\)

Optimal. Leaf size=321 \[ \frac{x (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};-p,3;\frac{m+3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3 (m+1)}-\frac{3 e x^2 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+2}{2};-p,3;\frac{m+4}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4 (m+2)}+\frac{3 e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+3}{2};-p,3;\frac{m+5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5 (m+3)}-\frac{e^3 x^4 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+4}{2};-p,3;\frac{m+6}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^6 (m+4)} \]

[Out]

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

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Rubi [A]  time = 0.373623, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {962, 511, 510} \[ \frac{x (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};-p,3;\frac{m+3}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3 (m+1)}-\frac{3 e x^2 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+2}{2};-p,3;\frac{m+4}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4 (m+2)}+\frac{3 e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+3}{2};-p,3;\frac{m+5}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5 (m+3)}-\frac{e^3 x^4 (g x)^m \left (a+c x^2\right )^p \left (\frac{c x^2}{a}+1\right )^{-p} F_1\left (\frac{m+4}{2};-p,3;\frac{m+6}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^6 (m+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 962

Int[((g_.)*(x_))^(n_.)*((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(g*x)^n/x^n, Int
[ExpandIntegrand[x^n*(a + c*x^2)^p, (d/(d^2 - e^2*x^2) - (e*x)/(d^2 - e^2*x^2))^(-m), x], x], x] /; FreeQ[{a,
c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] && ILtQ[m, 0] &&  !IntegerQ[p] &&  !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{(g x)^m \left (a+c x^2\right )^p}{(d+e x)^3} \, dx &=\left (x^{-m} (g x)^m\right ) \int \left (\frac{d^3 x^m \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}-\frac{3 d^2 e x^{1+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{3 d e^2 x^{2+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3}+\frac{e^3 x^{3+m} \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3}\right ) \, dx\\ &=\left (d^3 x^{-m} (g x)^m\right ) \int \frac{x^m \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 d^2 e x^{-m} (g x)^m\right ) \int \frac{x^{1+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2 x^{-m} (g x)^m\right ) \int \frac{x^{2+m} \left (a+c x^2\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (e^3 x^{-m} (g x)^m\right ) \int \frac{x^{3+m} \left (a+c x^2\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx\\ &=\left (d^3 x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{x^m \left (1+\frac{c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx-\left (3 d^2 e x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{x^{1+m} \left (1+\frac{c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (3 d e^2 x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{x^{2+m} \left (1+\frac{c x^2}{a}\right )^p}{\left (d^2-e^2 x^2\right )^3} \, dx+\left (e^3 x^{-m} (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p}\right ) \int \frac{x^{3+m} \left (1+\frac{c x^2}{a}\right )^p}{\left (-d^2+e^2 x^2\right )^3} \, dx\\ &=\frac{x (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{1+m}{2};-p,3;\frac{3+m}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^3 (1+m)}-\frac{3 e x^2 (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{2+m}{2};-p,3;\frac{4+m}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^4 (2+m)}+\frac{3 e^2 x^3 (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{3+m}{2};-p,3;\frac{5+m}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^5 (3+m)}-\frac{e^3 x^4 (g x)^m \left (a+c x^2\right )^p \left (1+\frac{c x^2}{a}\right )^{-p} F_1\left (\frac{4+m}{2};-p,3;\frac{6+m}{2};-\frac{c x^2}{a},\frac{e^2 x^2}{d^2}\right )}{d^6 (4+m)}\\ \end{align*}

Mathematica [F]  time = 0.201937, size = 0, normalized size = 0. \[ \int \frac{(g x)^m \left (a+c x^2\right )^p}{(d+e x)^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.699, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m} \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((g*x)^m*(c*x^2+a)^p/(e*x+d)^3,x)

[Out]

int((g*x)^m*(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 (g x\right )^{m}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + a)^p*(g*x)^m/(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} \left (g x\right )^{m}}{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((g*x)^m*(c*x^2+a)^p/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((c*x^2 + a)^p*(g*x)^m/(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((g*x)**m*(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 (g x\right )^{m}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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