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

Optimal. Leaf size=300 \[ \frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}-\frac{(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a e (n+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.559143, antiderivative size = 300, 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} \[ \frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e\right )}+\frac{\sqrt [3]{b} (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 a (n+1) \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right )}-\frac{(e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{a e (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^n/(x*(a + b*x^3)),x]

[Out]

(b^(1/3)*(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*(b^(1/3)*e - a^(1/3)*f)*(1 + n)) + (b^(1/3)*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(
e + f*x))/(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)])/(3*a*(b^(1/3)*e + (-1)^(1/3)*a^(1/3)*f)*(1 + n)) + (b^(1/3)*(e
+ f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b^(1/3)*(e + f*x))/(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)])/(3*
a*(b^(1/3)*e - (-1)^(2/3)*a^(1/3)*f)*(1 + n)) - ((e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x
)/e])/(a*e*(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 \left (a+b x^3\right )} \, dx &=\int \left (\frac{(e+f x)^n}{a x}-\frac{b x^2 (e+f x)^n}{a \left (a+b x^3\right )}\right ) \, dx\\ &=\frac{\int \frac{(e+f x)^n}{x} \, dx}{a}-\frac{b \int \frac{x^2 (e+f x)^n}{a+b x^3} \, dx}{a}\\ &=-\frac{(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a e (1+n)}-\frac{b \int \left (\frac{(e+f x)^n}{3 b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac{(e+f x)^n}{3 b^{2/3} \left (-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x\right )}+\frac{(e+f x)^n}{3 b^{2/3} \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}\right ) \, dx}{a}\\ &=-\frac{(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a e (1+n)}-\frac{\sqrt [3]{b} \int \frac{(e+f x)^n}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a}-\frac{\sqrt [3]{b} \int \frac{(e+f x)^n}{-\sqrt [3]{-1} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a}-\frac{\sqrt [3]{b} \int \frac{(e+f x)^n}{(-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a}\\ &=\frac{\sqrt [3]{b} (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 \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac{\sqrt [3]{b} (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]{-1} \sqrt [3]{a} f}\right )}{3 a \left (\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f\right ) (1+n)}+\frac{\sqrt [3]{b} (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{3 a \left (\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f\right ) (1+n)}-\frac{(e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac{f x}{e}\right )}{a e (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.246914, size = 244, normalized size = 0.81 \[ \frac{(e+f x)^{n+1} \left (\frac{\sqrt [3]{b} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac{\sqrt [3]{b} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e+\sqrt [3]{-1} \sqrt [3]{a} f}\right )}{\sqrt [3]{-1} \sqrt [3]{a} f+\sqrt [3]{b} e}+\frac{\sqrt [3]{b} \, _2F_1\left (1,n+1;n+2;\frac{\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-(-1)^{2/3} \sqrt [3]{a} f}-\frac{3 \, _2F_1\left (1,n+1;n+2;\frac{f x}{e}+1\right )}{e}\right )}{3 a (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}}{x \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/(b*x^3+a),x)

[Out]

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

________________________________________________________________________________________

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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n}}{b x^{4} + a x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x + e)^n/(b*x^4 + a*x), x)

________________________________________________________________________________________

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/(b*x**3+a),x)

[Out]

Timed out

________________________________________________________________________________________

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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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