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

Optimal. Leaf size=207 \[ \frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 (-a)^{3/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 (-a)^{3/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{e (d+e x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{a d^2 (n+1)} \]

[Out]

(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)
^(3/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(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)^(3/2)*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n)) + (e*(d + e*x)^(1 + n)*Hyp
ergeometric2F1[2, 1 + n, 2 + n, 1 + (e*x)/d])/(a*d^2*(1 + n))

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Rubi [A]  time = 0.220608, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {961, 65, 712, 68} \[ \frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{2 (-a)^{3/2} (n+1) \left (\sqrt{c} d-\sqrt{-a} e\right )}-\frac{c (d+e x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{2 (-a)^{3/2} (n+1) \left (\sqrt{-a} e+\sqrt{c} d\right )}+\frac{e (d+e x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{a d^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^n/(x^2*(a + c*x^2)),x]

[Out]

(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)
^(3/2)*(Sqrt[c]*d - Sqrt[-a]*e)*(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)^(3/2)*(Sqrt[c]*d + Sqrt[-a]*e)*(1 + n)) + (e*(d + e*x)^(1 + n)*Hyp
ergeometric2F1[2, 1 + n, 2 + n, 1 + (e*x)/d])/(a*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 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{(d+e x)^n}{x^2 \left (a+c x^2\right )} \, dx &=\int \left (\frac{(d+e x)^n}{a x^2}-\frac{c (d+e x)^n}{a \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{\int \frac{(d+e x)^n}{x^2} \, dx}{a}-\frac{c \int \frac{(d+e x)^n}{a+c x^2} \, dx}{a}\\ &=\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a 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}\\ &=\frac{e (d+e x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{e x}{d}\right )}{a d^2 (1+n)}-\frac{c \int \frac{(d+e x)^n}{\sqrt{-a}-\sqrt{c} x} \, dx}{2 (-a)^{3/2}}-\frac{c \int \frac{(d+e x)^n}{\sqrt{-a}+\sqrt{c} x} \, dx}{2 (-a)^{3/2}}\\ &=\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)^{3/2} \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)^{3/2} \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 d^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.290434, size = 167, normalized size = 0.81 \[ \frac{(d+e x)^{n+1} \left (-\frac{c \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d-\sqrt{-a} e}\right )}{\sqrt{-a} \sqrt{c} d+a e}+\frac{c \, _2F_1\left (1,n+1;n+2;\frac{\sqrt{c} (d+e x)}{\sqrt{c} d+\sqrt{-a} e}\right )}{\sqrt{-a} \sqrt{c} d-a e}+\frac{2 e \, _2F_1\left (2,n+1;n+2;\frac{e x}{d}+1\right )}{d^2}\right )}{2 a (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.752, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{n}}{{x}^{2} \left ( c{x}^{2}+a \right ) }}\, 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),x)

[Out]

int((e*x+d)^n/x^2/(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}}{{\left (c x^{2} + a\right )} 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),x, algorithm="maxima")

[Out]

integrate((e*x + d)^n/((c*x^2 + a)*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 x^{4} + a 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),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**n/(x**2*(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}}{{\left (c x^{2} + a\right )} 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),x, algorithm="giac")

[Out]

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