3.371 \(\int \frac{x^3 (d+e x)^n}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=297 \[ -\frac{(d+e x)^{n+1} \left (\sqrt{-a} \sqrt{c} d e n+a e^2 (n+2)+2 c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 c^{3/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{n+1} \left (\sqrt{-a} d e n-\frac{a e^2 (n+2)+2 c d^2}{\sqrt{c}}\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 c (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{a (d-e x) (d+e x)^{n+1}}{2 c \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

[Out]

(a*(d - e*x)*(d + e*x)^(1 + n))/(2*c*(c*d^2 + a*e^2)*(a + c*x^2)) + ((Sqrt[-a]*d*e*n - (2*c*d^2 + a*e^2*(2 + n
))/Sqrt[c])*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)]
)/(4*c*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n)) - ((2*c*d^2 + Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))
*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*c^(3/2
)*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n))

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Rubi [A]  time = 0.410105, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {1649, 831, 68} \[ -\frac{(d+e x)^{n+1} \left (\sqrt{-a} \sqrt{c} d e n+a e^2 (n+2)+2 c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 c^{3/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}+\frac{(d+e x)^{n+1} \left (\sqrt{-a} d e n-\frac{a e^2 (n+2)+2 c d^2}{\sqrt{c}}\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 c (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{a (d-e x) (d+e x)^{n+1}}{2 c \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(d + e*x)^n)/(a + c*x^2)^2,x]

[Out]

(a*(d - e*x)*(d + e*x)^(1 + n))/(2*c*(c*d^2 + a*e^2)*(a + c*x^2)) + ((Sqrt[-a]*d*e*n - (2*c*d^2 + a*e^2*(2 + n
))/Sqrt[c])*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)]
)/(4*c*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n)) - ((2*c*d^2 + Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))
*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*c^(3/2
)*(Sqrt[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n))

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, -Simp[((d + e*x)^(m + 1)*(a + c*x^2)^(p + 1)*(a*(e*f - d*g) + (c*d*f + a*e*g)*x))/(2*a*(p +
1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSu
m[2*a*(p + 1)*(c*d^2 + a*e^2)*Q + c*d^2*f*(2*p + 3) - a*e*(d*g*m - e*f*(m + 2*p + 3)) + e*(c*d*f + a*e*g)*(m +
2*p + 4)*x, x], x], x]] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] &
&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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]

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

Mathematica [A]  time = 0.845067, size = 247, normalized size = 0.83 \[ -\frac{(d+e x)^{n+1} \left (\frac{\left (-\sqrt{-a} \sqrt{c} d e n+a e^2 (n+2)+2 c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{(n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}+\frac{\left (\sqrt{-a} \sqrt{c} d e n+a e^2 (n+2)+2 c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{(n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{2 a \sqrt{c} (e x-d)}{a+c x^2}\right )}{4 c^{3/2} \left (a e^2+c d^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(d + e*x)^n)/(a + c*x^2)^2,x]

[Out]

-((d + e*x)^(1 + n)*((2*a*Sqrt[c]*(-d + e*x))/(a + c*x^2) + ((2*c*d^2 - Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n)
)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((Sqrt[c]*d - Sqrt[-a]*e)*
(1 + n)) + ((2*c*d^2 + Sqrt[-a]*Sqrt[c]*d*e*n + a*e^2*(2 + n))*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d
+ e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/((Sqrt[c]*d + Sqrt[-a]*e)*(1 + n))))/(4*c^(3/2)*(c*d^2 + a*e^2))

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Maple [F]  time = 0.769, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{n}{x}^{3}}{ \left ( c{x}^{2}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(e*x+d)^n/(c*x^2+a)^2,x)

[Out]

int(x^3*(e*x+d)^n/(c*x^2+a)^2,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^{3}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((e*x + d)^n*x^3/(c*x^2 + a)^2, 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^{3}}{c^{2} x^{4} + 2 \, a c x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

integral((e*x + d)^n*x^3/(c^2*x^4 + 2*a*c*x^2 + a^2), 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(x**3*(e*x+d)**n/(c*x**2+a)**2,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} x^{3}}{{\left (c x^{2} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*x+d)^n/(c*x^2+a)^2,x, algorithm="giac")

[Out]

integrate((e*x + d)^n*x^3/(c*x^2 + a)^2, x)