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]

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

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Rubi [A]  time = 0.18, antiderivative size = 148, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {(g x)^m (d+e x)^n}{a+c x^2} \, dx \]

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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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{c x^{2} + a}, x\right ) \]

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

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)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x \right )^{m} \left (e x +d \right )^{n}}{c \,x^{2}+a}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{c x^{2} + a}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

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]

Exception raised: HeuristicGCDFailed

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