3.379 \(\int \frac{(g x)^m (d+e x)^n}{a+c x^2} \, dx\)

Optimal. Leaf size=148 \[ \frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{e x}{d},-\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (m+1)}+\frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{e x}{d},\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.176814, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {912, 135, 133} \[ \frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{e x}{d},-\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (m+1)}+\frac{(g x)^{m+1} (d+e x)^n \left (\frac{e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac{e x}{d},\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 912

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
 0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
 x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{(g x)^m (d+e x)^n}{a+c x^2} \, dx &=\int \left (\frac{\sqrt{-a} (g x)^m (d+e x)^n}{2 a \left (\sqrt{-a}-\sqrt{c} x\right )}+\frac{\sqrt{-a} (g x)^m (d+e x)^n}{2 a \left (\sqrt{-a}+\sqrt{c} x\right )}\right ) \, dx\\ &=-\frac{\int \frac{(g x)^m (d+e x)^n}{\sqrt{-a}-\sqrt{c} x} \, dx}{2 \sqrt{-a}}-\frac{\int \frac{(g x)^m (d+e x)^n}{\sqrt{-a}+\sqrt{c} x} \, dx}{2 \sqrt{-a}}\\ &=-\frac{\left ((d+e x)^n \left (1+\frac{e x}{d}\right )^{-n}\right ) \int \frac{(g x)^m \left (1+\frac{e x}{d}\right )^n}{\sqrt{-a}-\sqrt{c} x} \, dx}{2 \sqrt{-a}}-\frac{\left ((d+e x)^n \left (1+\frac{e x}{d}\right )^{-n}\right ) \int \frac{(g x)^m \left (1+\frac{e x}{d}\right )^n}{\sqrt{-a}+\sqrt{c} x} \, dx}{2 \sqrt{-a}}\\ &=\frac{(g x)^{1+m} (d+e x)^n \left (1+\frac{e x}{d}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{e x}{d},-\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (1+m)}+\frac{(g x)^{1+m} (d+e x)^n \left (1+\frac{e x}{d}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac{e x}{d},\frac{\sqrt{c} x}{\sqrt{-a}}\right )}{2 a g (1+m)}\\ \end{align*}

Mathematica [F]  time = 0.116095, size = 0, normalized size = 0. \[ \int \frac{(g x)^m (d+e x)^n}{a+c x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.752, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( gx \right ) ^{m} \left ( ex+d \right ) ^{n}}{c{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^m*(e*x+d)^n/(c*x^2+a),x)

[Out]

int((g*x)^m*(e*x+d)^n/(c*x^2+a),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^n*(g*x)^m/(c*x^2 + a), 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*(e*x+d)**n/(c*x**2+a),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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