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

Optimal. Leaf size=295 \[ \frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)} \]

[Out]

1/4*(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^2/g/(1+m)/((1+e*x/d)^n)+1/4*(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^2/g/(1+m)/((1+e*x/d)^n)+1/4*(g*x)^(1+
m)*(e*x+d)^n*AppellF1(1+m,-n,2,2+m,-e*x/d,-x*c^(1/2)/(-a)^(1/2))/a^2/g/(1+m)/((1+e*x/d)^n)+1/4*(g*x)^(1+m)*(e*
x+d)^n*AppellF1(1+m,2,-n,2+m,x*c^(1/2)/(-a)^(1/2),-e*x/d)/a^2/g/(1+m)/((1+e*x/d)^n)

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Rubi [A]  time = 0.42, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {961, 135, 133, 912} \[ \frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,1;m+2;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[((g*x)^m*(d + e*x)^n)/(a + c*x^2)^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])])/(4*a^2*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]
])/(4*a^2*g*(1 + m)*(1 + (e*x)/d)^n) + ((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 2, 2 + m, -((e*x)/d), -(
(Sqrt[c]*x)/Sqrt[-a])])/(4*a^2*g*(1 + m)*(1 + (e*x)/d)^n) + ((g*x)^(1 + m)*(d + e*x)^n*AppellF1[1 + m, -n, 2,
2 + m, -((e*x)/d), (Sqrt[c]*x)/Sqrt[-a]])/(4*a^2*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]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rubi steps

\begin {align*} \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx &=\int \left (-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}-c x\right )^2}-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}+c x\right )^2}-\frac {c (g x)^m (d+e x)^n}{2 a \left (-a c-c^2 x^2\right )}\right ) \, dx\\ &=-\frac {c \int \frac {(g x)^m (d+e x)^n}{\left (\sqrt {-a} \sqrt {c}-c x\right )^2} \, dx}{4 a}-\frac {c \int \frac {(g x)^m (d+e x)^n}{\left (\sqrt {-a} \sqrt {c}+c x\right )^2} \, dx}{4 a}-\frac {c \int \frac {(g x)^m (d+e x)^n}{-a c-c^2 x^2} \, dx}{2 a}\\ &=-\frac {c \int \left (-\frac {\sqrt {-a} (g x)^m (d+e x)^n}{2 a c \left (\sqrt {-a}-\sqrt {c} x\right )}-\frac {\sqrt {-a} (g x)^m (d+e x)^n}{2 a c \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a}-\frac {\left (c (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}{\left (\sqrt {-a} \sqrt {c}-c x\right )^2} \, dx}{4 a}-\frac {\left (c (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}{\left (\sqrt {-a} \sqrt {c}+c x\right )^2} \, dx}{4 a}\\ &=\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 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,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {\int \frac {(g x)^m (d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 (-a)^{3/2}}+\frac {\int \frac {(g x)^m (d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 (-a)^{3/2}}\\ &=\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 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,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\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}{4 (-a)^{3/2}}+\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}{4 (-a)^{3/2}}\\ &=\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 )}{4 a^2 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 )}{4 a^2 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,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 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,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}\\ \end {align*}

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Mathematica [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^n*(g*x)^m/(c^2*x^4 + 2*a*c*x^2 + a^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (g x \right )^{m} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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