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

Optimal. Leaf size=163 \[ -\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )} \]

[Out]

-1/2*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(-e*(-a)^(1/2)+d*c^(1/2)))/(1+n)/c^(1/2)/(-e*(-a)^
(1/2)+d*c^(1/2))-1/2*(e*x+d)^(1+n)*hypergeom([1, 1+n],[2+n],(e*x+d)*c^(1/2)/(e*(-a)^(1/2)+d*c^(1/2)))/(1+n)/c^
(1/2)/(e*(-a)^(1/2)+d*c^(1/2))

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Rubi [A]  time = 0.09, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {831, 68} \[ -\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}-\frac {(d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {c} (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x)^n)/(a + c*x^2),x]

[Out]

-((d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*Sqrt[
c]*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - ((d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x
))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*Sqrt[c]*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))

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 831

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

Rubi steps

\begin {align*} \int \frac {x (d+e x)^n}{a+c x^2} \, dx &=\int \left (-\frac {(d+e x)^n}{2 \sqrt {c} \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {(d+e x)^n}{2 \sqrt {c} \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx\\ &=-\frac {\int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{2 \sqrt {c}}+\frac {\int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{2 \sqrt {c}}\\ &=-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 \sqrt {c} \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 \sqrt {c} \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 151, normalized size = 0.93 \[ -\frac {(d+e x)^{n+1} \left (\left (\sqrt {-a} e+\sqrt {c} d\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )+\left (\sqrt {c} d-\sqrt {-a} e\right ) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )\right )}{2 \sqrt {c} (n+1) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((d + e*x)^(1 + n)*((Sqrt[c]*d + Sqrt[-a]*e)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt
[c]*d - Sqrt[-a]*e)] + (Sqrt[c]*d - Sqrt[-a]*e)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c
]*d + Sqrt[-a]*e)]))/(Sqrt[c]*(c*d^2 + a*e^2)*(1 + n))

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{n} x}{c x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x \left (e x +d \right )^{n}}{c \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(d + e*x)^n)/(a + c*x^2),x)

[Out]

int((x*(d + e*x)^n)/(a + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(d + e*x)**n/(a + c*x**2), x)

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