Optimal. Leaf size=549 \[ -\frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\left (\left (\sqrt {e^2-4 d f}+e\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {(2 c e-b f) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}+\frac {\sqrt {a+b x+c x^2}}{f} \]
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Rubi [A] time = 7.03, antiderivative size = 549, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1019, 1076, 621, 206, 1032, 724} \begin {gather*} -\frac {\left (\left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b f)\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} f^2 \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 {\left (\left (\sqrt {e^2-4 d f}+e\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )+2 d f (c e-b 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} f^2 \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 {(2 c e-b f) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}+\frac {\sqrt {a+b x+c x^2}}{f} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 621
Rule 724
Rule 1019
Rule 1032
Rule 1076
Rubi steps
\begin {align*} \int \frac {x \sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx &=\frac {\sqrt {a+b x+c x^2}}{f}-\frac {\int \frac {\frac {b d}{2}+\frac {1}{2} (2 c d+b e-2 a f) x+\frac {1}{2} (2 c e-b f) x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}\\ &=\frac {\sqrt {a+b x+c x^2}}{f}-\frac {\int \frac {\frac {b d f}{2}-\frac {1}{2} d (2 c e-b f)+\left (\frac {1}{2} f (2 c d+b e-2 a f)-\frac {1}{2} e (2 c e-b f)\right ) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac {(2 c e-b f) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 f^2}\\ &=\frac {\sqrt {a+b x+c x^2}}{f}-\frac {(2 c e-b f) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{f^2}-\frac {\left (2 f \left (\frac {b d f}{2}-\frac {1}{2} d (2 c e-b f)\right )-\left (\frac {1}{2} f (2 c d+b e-2 a f)-\frac {1}{2} e (2 c e-b f)\right ) \left (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}{f^2 \sqrt {e^2-4 d f}}+\frac {\left (2 f \left (\frac {b d f}{2}-\frac {1}{2} d (2 c e-b f)\right )-\left (\frac {1}{2} f (2 c d+b e-2 a f)-\frac {1}{2} e (2 c e-b f)\right ) \left (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}{f^2 \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {a+b x+c x^2}}{f}-\frac {(2 c e-b f) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}+\frac {\left (2 \left (2 f \left (\frac {b d f}{2}-\frac {1}{2} d (2 c e-b f)\right )-\left (\frac {1}{2} f (2 c d+b e-2 a f)-\frac {1}{2} e (2 c e-b f)\right ) \left (e-\sqrt {e^2-4 d f}\right )\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 )}{f^2 \sqrt {e^2-4 d f}}-\frac {\left (2 \left (2 f \left (\frac {b d f}{2}-\frac {1}{2} d (2 c e-b f)\right )-\left (\frac {1}{2} f (2 c d+b e-2 a f)-\frac {1}{2} e (2 c e-b f)\right ) \left (e+\sqrt {e^2-4 d f}\right )\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 )}{f^2 \sqrt {e^2-4 d f}}\\ &=\frac {\sqrt {a+b x+c x^2}}{f}-\frac {(2 c e-b f) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} f^2}-\frac {\left (2 f (c d e-b d f)+\left (e-\sqrt {e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )\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} f^2 \sqrt {e^2-4 d f} \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 {\left (2 f (c d e-b d f)+\left (e+\sqrt {e^2-4 d f}\right ) \left (f (b e-a f)-c \left (e^2-d f\right )\right )\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} f^2 \sqrt {e^2-4 d f} \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 = 1.78, size = 496, normalized size = 0.90 \begin {gather*} \frac {4 f \sqrt {e^2-4 d f} \sqrt {a+x (b+c x)}-\sqrt {2} \left (\sqrt {e^2-4 d f}+e\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 )} \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 {2} \left (\sqrt {e^2-4 d f}-e\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 )} \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 )}{4 f^2 \sqrt {e^2-4 d f}}-\frac {(2 c e-b f) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 \sqrt {c} f^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [C] time = 0.86, size = 646, normalized size = 1.18 \begin {gather*} \frac {\text {RootSum}\left [\text {$\#$1}^4 f-2 \text {$\#$1}^3 \sqrt {c} e-2 \text {$\#$1}^2 a f+\text {$\#$1}^2 b e+4 \text {$\#$1}^2 c d+2 \text {$\#$1} a \sqrt {c} e-4 \text {$\#$1} b \sqrt {c} d+a^2 f-a b e+b^2 d\&,\frac {\text {$\#$1}^2 c d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 c e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+\text {$\#$1}^2 b e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-\text {$\#$1}^2 a f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a^2 f^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+b^2 d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+2 \text {$\#$1} c^{3/2} d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-b c d e \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-2 \text {$\#$1} b \sqrt {c} d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a c d f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )+a c e^2 \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )-a b e f \log \left (-\text {$\#$1}+\sqrt {a+b x+c x^2}-\sqrt {c} x\right )}{-2 \text {$\#$1}^3 f+3 \text {$\#$1}^2 \sqrt {c} e+2 \text {$\#$1} a f-\text {$\#$1} b e-4 \text {$\#$1} c d-a \sqrt {c} e+2 b \sqrt {c} d}\&\right ]}{f^2}+\frac {(2 c e-b f) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{2 \sqrt {c} f^2}+\frac {\sqrt {a+b x+c x^2}}{f} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] 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(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 10138, normalized size = 18.47 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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 {x\,\sqrt {c\,x^2+b\,x+a}}{f\,x^2+e\,x+d} \,d x \end {gather*}
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 \begin {gather*} \int \frac {x \sqrt {a + b x + c x^{2}}}{d + e x + f x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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