Integrand size = 20, antiderivative size = 326 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=-\frac {b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right ) (1+n)}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} \left (\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f\right ) (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a e^2 (1+n)} \]
[Out]
Time = 0.47 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6857, 67, 70} \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=-\frac {b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {\sqrt [3]{-1} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left ((-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f\right )}+\frac {(-1)^{2/3} b^{2/3} (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{3 a^{4/3} (n+1) \left (\sqrt [3]{a} f+\sqrt [3]{-1} \sqrt [3]{b} e\right )}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a e^2 (n+1)} \]
[In]
[Out]
Rule 67
Rule 70
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \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*}
Time = 0.34 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.84 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{b} (e+f x)}{\sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{\sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {\sqrt [3]{-1} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {(-1)^{2/3} \sqrt [3]{b} (e+f x)}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}\right )}{(-1)^{2/3} \sqrt [3]{b} e-\sqrt [3]{a} f}+\frac {(-1)^{2/3} b^{2/3} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt [3]{-1} \sqrt [3]{b} (e+f x)}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}\right )}{\sqrt [3]{-1} \sqrt [3]{b} e+\sqrt [3]{a} f}+\frac {3 \sqrt [3]{a} f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{e^2}\right )}{3 a^{4/3} (1+n)} \]
[In]
[Out]
\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (b \,x^{3}+a \right )}d x\]
[In]
[Out]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x^{3} + a\right )} x^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x^3\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (b\,x^3+a\right )} \,d x \]
[In]
[Out]