Optimal. Leaf size=451 \[ \frac {f \left (\sqrt {e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {f \left (e-\sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d} \]
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Rubi [A] time = 2.63, antiderivative size = 451, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6728, 724, 206, 1032} \[ \frac {f \left (\sqrt {e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {f \left (e-\sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 724
Rule 1032
Rule 6728
Rubi steps
\begin {align*} \int \frac {1}{x \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac {1}{d x \sqrt {a+b x+c x^2}}+\frac {-e-f x}{d \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{d}+\frac {\int \frac {-e-f x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{d}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )}{d}-\frac {\left (f \left (1-\frac {e}{\sqrt {e^2-4 d f}}\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+b x+c x^2}} \, dx}{d}-\frac {\left (f \left (1+\frac {e}{\sqrt {e^2-4 d f}}\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+b x+c x^2}} \, dx}{d}\\ &=-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d}+\frac {\left (2 f \left (1-\frac {e}{\sqrt {e^2-4 d f}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{16 a f^2-8 b f \left (e+\sqrt {e^2-4 d f}\right )+4 c \left (e+\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )-\left (-2 b f+2 c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d}+\frac {\left (2 f \left (1+\frac {e}{\sqrt {e^2-4 d f}}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{16 a f^2-8 b f \left (e-\sqrt {e^2-4 d f}\right )+4 c \left (e-\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )-\left (-2 b f+2 c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{\sqrt {a+b x+c x^2}}\right )}{d}\\ &=-\frac {\tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a} d}+\frac {f \left (1+\frac {e}{\sqrt {e^2-4 d f}}\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e-\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {c e^2-2 c d f-b e f+2 a f^2-(c e-b f) \sqrt {e^2-4 d f}}}+\frac {f \left (1-\frac {e}{\sqrt {e^2-4 d f}}\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (e+\sqrt {e^2-4 d f}\right )+2 \left (b f-c \left (e+\sqrt {e^2-4 d f}\right )\right ) x}{2 \sqrt {2} \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}} \sqrt {a+b x+c x^2}}\right )}{\sqrt {2} d \sqrt {c e^2-2 c d f-b e f+2 a f^2+(c e-b f) \sqrt {e^2-4 d f}}}\\ \end {align*}
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Mathematica [A] time = 2.41, size = 450, normalized size = 1.00 \[ \frac {\frac {\sqrt {2} f \left (\frac {\left (\sqrt {e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac {4 a f-b \left (\sqrt {e^2-4 d f}+e-2 f x\right )-2 c x \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+x (b+c x)} \sqrt {f \left (2 a f-b \left (\sqrt {e^2-4 d f}+e\right )\right )+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {f \left (2 a f-b \left (\sqrt {e^2-4 d f}+e\right )\right )+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\left (\sqrt {e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac {4 a f+b \left (\sqrt {e^2-4 d f}-e+2 f x\right )+2 c x \left (\sqrt {e^2-4 d f}-e\right )}{2 \sqrt {2} \sqrt {a+x (b+c x)} \sqrt {f \left (2 a f+b \sqrt {e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {f \left (2 a f+b \sqrt {e^2-4 d f}+b (-e)\right )+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f}}-\frac {2 \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{\sqrt {a}}}{2 d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 859, normalized size = 1.90 \[ -\frac {2 \sqrt {2}\, f \ln \left (\frac {\frac {\left (b f -c e -\sqrt {-4 d f +e^{2}}\, c \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, b f +\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}+\frac {\sqrt {2}\, \sqrt {\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, b f +\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c +\frac {4 \left (b f -c e -\sqrt {-4 d f +e^{2}}\, c \right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}-2 \sqrt {-4 d f +e^{2}}\, b f +2 \sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, b f +\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}-\frac {2 \sqrt {2}\, f \ln \left (\frac {\frac {\left (b f -c e +\sqrt {-4 d f +e^{2}}\, c \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, b f -\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}+\frac {\sqrt {2}\, \sqrt {\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, b f -\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c +\frac {4 \left (b f -c e +\sqrt {-4 d f +e^{2}}\, c \right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {4 a \,f^{2}-2 b e f -4 c d f +2 c \,e^{2}+2 \sqrt {-4 d f +e^{2}}\, b f -2 \sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {2 a \,f^{2}-b e f -2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, b f -\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}+\frac {4 f \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {a}} \]
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 \[ \int \frac {1}{\sqrt {c x^{2} + b x + a} {\left (f x^{2} + e x + d\right )} x}\,{d x} \]
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 \[ \int \frac {1}{x\,\sqrt {c\,x^2+b\,x+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sqrt {a + b x + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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