Optimal. Leaf size=296 \[ -\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac {f x}{e}\right )}{a e^2 (1+n)} \]
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Rubi [A]
time = 0.31, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {974, 67, 844,
70} \begin {gather*} -\frac {c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {b (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {f x}{e}+1\right )}{a^2 e (n+1)}+\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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 67
Rule 70
Rule 844
Rule 974
Rubi steps
\begin {align*} \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx &=\int \left (\frac {(e+f x)^n}{a x^2}-\frac {b (e+f x)^n}{a^2 x}+\frac {\left (b^2-a c+b c x\right ) (e+f x)^n}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {\left (b^2-a c+b c x\right ) (e+f x)^n}{a+b x+c x^2} \, dx}{a^2}+\frac {\int \frac {(e+f x)^n}{x^2} \, dx}{a}-\frac {b \int \frac {(e+f x)^n}{x} \, dx}{a^2}\\ &=\frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (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)}+\frac {\int \left (\frac {\left (b c+\frac {c \left (b^2-2 a c\right )}{\sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (b c-\frac {c \left (b^2-2 a c\right )}{\sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{a^2}\\ &=\frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (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)}+\frac {\left (c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{a^2}+\frac {\left (c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{a^2}\\ &=-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (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*}
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Mathematica [A]
time = 1.39, size = 431, normalized size = 1.46 \begin {gather*} \frac {\left (1+\frac {e}{f x}\right )^{-n} (e+f x)^n \left (\frac {2 a \, _2F_1\left (1-n,-n;2-n;-\frac {e}{f x}\right )}{(-1+n) x}-\frac {2 b \, _2F_1\left (-n,-n;1-n;-\frac {e}{f x}\right )}{n}+\frac {2^{-n} \left (b^2 f-2 a c f+b \sqrt {\left (b^2-4 a c\right ) f^2}\right ) \left (1+\frac {e}{f x}\right )^n \left (\frac {c (e+f x)}{b f-\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {2 c e-b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{-b f+\sqrt {\left (b^2-4 a c\right ) f^2}-2 c f x}\right )}{\sqrt {\left (b^2-4 a c\right ) f^2} n}+\frac {2^{-n} \left (-b^2 f+2 a c f+b \sqrt {\left (b^2-4 a c\right ) f^2}\right ) \left (1+\frac {e}{f x}\right )^n \left (\frac {c (e+f x)}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {-2 c e+b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )}{\sqrt {\left (b^2-4 a c\right ) f^2} n}\right )}{2 a^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (c \,x^{2}+b x +a \right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (c\,x^2+b\,x+a\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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