3.193 \(\int \frac{(e+f x)^n}{x^2 (a+b x^3)} \, dx\)

Optimal. Leaf size=326 \[ -\frac{b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

[Out]

-(b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(
3*a^(4/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(1/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n,
2 + n, ((-1)^(2/3)*b^(1/3)*(e + f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/(3*a^(4/3)*((-1)^(2/3)*b^(1/3)*e -
a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^
(1/3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/(3*a^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n)) +
(f*(e + f*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))

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Rubi [A]  time = 0.611083, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6725, 65, 68} \[ -\frac{b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac{f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{a e^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^n/(x^2*(a + b*x^3)),x]

[Out]

-(b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)])/(
3*a^(4/3)*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + ((-1)^(1/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n,
2 + n, ((-1)^(2/3)*b^(1/3)*(e + f*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/(3*a^(4/3)*((-1)^(2/3)*b^(1/3)*e -
a^(1/3)*f)*(1 + n)) + ((-1)^(2/3)*b^(2/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^
(1/3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/(3*a^(4/3)*((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)*(1 + n)) +
(f*(e + f*x)^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && 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 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{(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx &=\int \left (\frac{(e+f x)^n}{a x^2}-\frac{b x (e+f x)^n}{a \left (a+b x^3\right )}\right ) \, dx\\ &=\frac{\int \frac{(e+f x)^n}{x^2} \, dx}{a}-\frac{b \int \frac{x (e+f x)^n}{a+b x^3} \, dx}{a}\\ &=\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}-\frac{b \int \left (-\frac{(e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac{(-1)^{2/3} (e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x\right )}+\frac{\sqrt [3]{-1} (e+f x)^n}{3 \sqrt [3]{a} \sqrt [3]{b} \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}\right ) \, dx}{a}\\ &=\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}+\frac{b^{2/3} \int \frac{(e+f x)^n}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{4/3}}-\frac{\left (\sqrt [3]{-1} b^{2/3}\right ) \int \frac{(e+f x)^n}{\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x} \, dx}{3 a^{4/3}}+\frac{\left ((-1)^{2/3} b^{2/3}\right ) \int \frac{(e+f x)^n}{\sqrt [3]{a}-\sqrt [3]{-1} \sqrt [3]{b} x} \, dx}{3 a^{4/3}}\\ &=-\frac{b^{2/3} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac{\sqrt [3]{-1} b^{2/3} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac{(-1)^{2/3} b^{2/3} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}+\frac{f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac{f x}{e}\right )}{a e^2 (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.264991, size = 273, normalized size = 0.84 \[ \frac{(e+f x)^{n+1} \left (-\frac{b^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac{\sqrt [3]{-1} b^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}+\frac{(-1)^{2/3} b^{2/3} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e}+\frac{3 \sqrt [3]{a} f \, _2F_1\left (2,n+1;n+2;\frac{f x}{e}+1\right )}{e^2}\right )}{3 a^{4/3} (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + f*x)^(1 + n)*(-((b^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - a^(1/3)*f)]
)/(b^(1/3)*e - a^(1/3)*f)) + ((-1)^(1/3)*b^(2/3)*Hypergeometric2F1[1, 1 + n, 2 + n, ((-1)^(2/3)*b^(1/3)*(e + f
*x))/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f)])/((-1)^(2/3)*b^(1/3)*e - a^(1/3)*f) + ((-1)^(2/3)*b^(2/3)*Hypergeomet
ric2F1[1, 1 + n, 2 + n, ((-1)^(1/3)*b^(1/3)*(e + f*x))/((-1)^(1/3)*b^(1/3)*e + a^(1/3)*f)])/((-1)^(1/3)*b^(1/3
)*e + a^(1/3)*f) + (3*a^(1/3)*f*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/e^2))/(3*a^(4/3)*(1 + n))

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{{x}^{2} \left ( b{x}^{3}+a \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^n/x^2/(b*x^3+a),x)

[Out]

int((f*x+e)^n/x^2/(b*x^3+a),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^n/((b*x^3 + a)*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n}}{b x^{5} + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x + e)^n/(b*x^5 + a*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((f*x+e)**n/x**2/(b*x**3+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^n/((b*x^3 + a)*x^2), x)