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

Optimal. Leaf size=513 \[ -\frac{c (d+e x)^{n+1} (a e+c d x)}{2 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac{e (d+e x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{a^2 d^2 (n+1)}-\frac{c (d+e x)^{n+1} \left (\sqrt{-a} \sqrt{c} d e n+a e^2 (1-n)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 (-a)^{5/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{c (d+e x)^{n+1} \left (-\sqrt{-a} \sqrt{c} d e n+a e^2 (1-n)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 (-a)^{5/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}-\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 (-a)^{5/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}+\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 (-a)^{5/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )} \]

[Out]

-(c*(a*e + c*d*x)*(d + e*x)^(1 + n))/(2*a^2*(c*d^2 + a*e^2)*(a + c*x^2)) - (c*(d + e*x)^(1 + n)*Hypergeometric
2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*(-a)^(5/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(1
+ n)) - (c*(c*d^2 + a*e^2*(1 - n) + Sqrt[-a]*Sqrt[c]*d*e*n)*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 +
n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(4*(-a)^(5/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 +
n)) + (c*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(
2*(-a)^(5/2)*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n)) + (c*(c*d^2 + a*e^2*(1 - n) - Sqrt[-a]*Sqrt[c]*d*e*n)*(d + e*x)
^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*(-a)^(5/2)*(Sqrt
[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n)) + (e*(d + e*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (e*
x)/d])/(a^2*d^2*(1 + n))

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Rubi [A]  time = 0.698169, antiderivative size = 513, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {961, 65, 741, 831, 68, 712} \[ -\frac{c (d+e x)^{n+1} (a e+c d x)}{2 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac{e (d+e x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{a^2 d^2 (n+1)}-\frac{c (d+e x)^{n+1} \left (\sqrt{-a} \sqrt{c} d e n+a e^2 (1-n)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{4 (-a)^{5/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (a e^2+c d^2\right )}+\frac{c (d+e x)^{n+1} \left (-\sqrt{-a} \sqrt{c} d e n+a e^2 (1-n)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{4 (-a)^{5/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right ) \left (a e^2+c d^2\right )}-\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 (-a)^{5/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}+\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 (-a)^{5/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(a*e + c*d*x)*(d + e*x)^(1 + n))/(2*a^2*(c*d^2 + a*e^2)*(a + c*x^2)) - (c*(d + e*x)^(1 + n)*Hypergeometric
2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(2*(-a)^(5/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(1
+ n)) - (c*(c*d^2 + a*e^2*(1 - n) + Sqrt[-a]*Sqrt[c]*d*e*n)*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 +
n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(4*(-a)^(5/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 +
n)) + (c*(d + e*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(
2*(-a)^(5/2)*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n)) + (c*(c*d^2 + a*e^2*(1 - n) - Sqrt[-a]*Sqrt[c]*d*e*n)*(d + e*x)
^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(4*(-a)^(5/2)*(Sqrt
[c]*d + Sqrt[-a]*e)*(c*d^2 + a*e^2)*(1 + n)) + (e*(d + e*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (e*
x)/d])/(a^2*d^2*(1 + 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])

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(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*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

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]

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]

Rubi steps

\begin{align*} \int \frac{(d+e x)^n}{x^2 \left (a+c x^2\right )^2} \, dx &=\int \left (\frac{(d+e x)^n}{a^2 x^2}-\frac{c (d+e x)^n}{a \left (a+c x^2\right )^2}-\frac{c (d+e x)^n}{a^2 \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{(d+e x)^n}{x^2} \, dx}{a^2}-\frac{c \int \frac{(d+e x)^n}{a+c x^2} \, dx}{a^2}-\frac{c \int \frac{(d+e x)^n}{\left (a+c x^2\right )^2} \, dx}{a}\\ &=-\frac{c (a e+c d x) (d+e x)^{1+n}}{2 a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a^2 d^2 (1+n)}-\frac{c \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}{a^2}+\frac{c \int \frac{(d+e x)^n \left (-c d^2-a e^2 (1-n)+c d e n x\right )}{a+c x^2} \, dx}{2 a^2 \left (c d^2+a e^2\right )}\\ &=-\frac{c (a e+c d x) (d+e x)^{1+n}}{2 a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a^2 d^2 (1+n)}+\frac{c \int \frac{(d+e x)^n}{\sqrt{-a}-\sqrt{c} x} \, dx}{2 (-a)^{5/2}}+\frac{c \int \frac{(d+e x)^n}{\sqrt{-a}+\sqrt{c} x} \, dx}{2 (-a)^{5/2}}+\frac{c \int \left (\frac{\left (\sqrt{-a} \left (-c d^2-a e^2 (1-n)\right )-a \sqrt{c} d e n\right ) (d+e x)^n}{2 a \left (\sqrt{-a}-\sqrt{c} x\right )}+\frac{\left (\sqrt{-a} \left (-c d^2-a e^2 (1-n)\right )+a \sqrt{c} d e n\right ) (d+e x)^n}{2 a \left (\sqrt{-a}+\sqrt{c} x\right )}\right ) \, dx}{2 a^2 \left (c d^2+a e^2\right )}\\ &=-\frac{c (a e+c d x) (d+e x)^{1+n}}{2 a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac{c (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 (-a)^{5/2} \left (\sqrt{c} d-\sqrt{-a} e\right ) (1+n)}+\frac{c (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 (-a)^{5/2} \left (\sqrt{c} d+\sqrt{-a} e\right ) (1+n)}+\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a^2 d^2 (1+n)}+\frac{\left (c \left (c d^2+a e^2 (1-n)-\sqrt{-a} \sqrt{c} d e n\right )\right ) \int \frac{(d+e x)^n}{\sqrt{-a}-\sqrt{c} x} \, dx}{4 (-a)^{5/2} \left (c d^2+a e^2\right )}+\frac{\left (c \left (c d^2+a e^2 (1-n)+\sqrt{-a} \sqrt{c} d e n\right )\right ) \int \frac{(d+e x)^n}{\sqrt{-a}+\sqrt{c} x} \, dx}{4 (-a)^{5/2} \left (c d^2+a e^2\right )}\\ &=-\frac{c (a e+c d x) (d+e x)^{1+n}}{2 a^2 \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac{c (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 (-a)^{5/2} \left (\sqrt{c} d-\sqrt{-a} e\right ) (1+n)}-\frac{c \left (c d^2+a e^2 (1-n)+\sqrt{-a} \sqrt{c} d e 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 (-a)^{5/2} \left (\sqrt{c} d-\sqrt{-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}+\frac{c (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 (-a)^{5/2} \left (\sqrt{c} d+\sqrt{-a} e\right ) (1+n)}+\frac{c \left (c d^2+a e^2 (1-n)-\sqrt{-a} \sqrt{c} d e 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 (-a)^{5/2} \left (\sqrt{c} d+\sqrt{-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}+\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a^2 d^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.760214, size = 437, normalized size = 0.85 \[ \frac{1}{4} (d+e x)^{n+1} \left (-\frac{2 c (a e+c d x)}{a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )}+\frac{4 e \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{a^2 d^2 (n+1)}+\frac{a c \left (\frac{\left (\sqrt{-a} \sqrt{c} d e n-a e^2 (n-1)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{c} d-\sqrt{-a} e}-\frac{\left (-\sqrt{-a} \sqrt{c} d e n-a e^2 (n-1)+c d^2\right ) \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} e+\sqrt{c} d}\right )}{(-a)^{7/2} (n+1) \left (a e^2+c d^2\right )}+\frac{2 c \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{(-a)^{5/2} (n+1) \left (\sqrt{-a} e-\sqrt{c} d\right )}+\frac{2 c \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{(-a)^{5/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^(1 + n)*((-2*c*(a*e + c*d*x))/(a^2*(c*d^2 + a*e^2)*(a + c*x^2)) + (2*c*Hypergeometric2F1[1, 1 + n,
2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/((-a)^(5/2)*(-(Sqrt[c]*d) + Sqrt[-a]*e)*(1 + n)) + (2*c*
Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/((-a)^(5/2)*(Sqrt[c]*d + Sqr
t[-a]*e)*(1 + n)) + (a*c*(((c*d^2 - a*e^2*(-1 + n) + Sqrt[-a]*Sqrt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n
, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d - Sqrt[-a]*e)])/(Sqrt[c]*d - Sqrt[-a]*e) - ((c*d^2 - a*e^2*(-1 + n) - Sqrt[-a
]*Sqrt[c]*d*e*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (Sqrt[c]*(d + e*x))/(Sqrt[c]*d + Sqrt[-a]*e)])/(Sqrt[c]*d
+ Sqrt[-a]*e)))/((-a)^(7/2)*(c*d^2 + a*e^2)*(1 + n)) + (4*e*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (e*x)/d])/(
a^2*d^2*(1 + n))))/4

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Maple [F]  time = 0.753, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{n}}{{x}^{2} \left ( c{x}^{2}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^n/((c*x^2 + a)^2*x^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}}{c^{2} x^{6} + 2 \, a c x^{4} + a^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e*x + d)^n/(c^2*x^6 + 2*a*c*x^4 + a^2*x^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((e*x+d)**n/x**2/(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}}{{\left (c x^{2} + a\right )}^{2} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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