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

Optimal. Leaf size=194 \[ \frac{\sqrt{-a} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 c (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{\sqrt{-a} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 c (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{(d+e x)^{n+1}}{c e (n+1)} \]

[Out]

(d + e*x)^(1 + n)/(c*e*(1 + n)) + (Sqrt[-a]*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d +
e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*c*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - (Sqrt[-a]*(d + e*x)^(1 + n)*Hyperge
ometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*c*(Sqrt[c]*d + Sqrt[-a]*e)*(1 +
n))

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Rubi [A]  time = 0.226158, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1629, 712, 68} \[ \frac{\sqrt{-a} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 c (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{\sqrt{-a} (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 c (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{(d+e x)^{n+1}}{c e (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x)^n)/(a + c*x^2),x]

[Out]

(d + e*x)^(1 + n)/(c*e*(1 + n)) + (Sqrt[-a]*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d +
e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*c*(Sqrt[c]*d - Sqrt[-a]*e)*(1 + n)) - (Sqrt[-a]*(d + e*x)^(1 + n)*Hyperge
ometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(2*c*(Sqrt[c]*d + Sqrt[-a]*e)*(1 +
n))

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 712

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

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]

Rubi steps

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

Mathematica [A]  time = 0.146173, size = 170, normalized size = 0.88 \[ \frac{(d+e x)^{n+1} \left (2 \left (a e^2+c d^2\right )+e \left (\sqrt{-a} \sqrt{c} d-a e\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )-e \left (\sqrt{-a} \sqrt{c} d+a e\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )\right )}{2 c e (n+1) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.737, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{n}{x}^{2}}{c{x}^{2}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2*(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} x^{2}}{c x^{2} + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^n*x^2/(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} x^{2}}{c x^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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