Optimal. Leaf size=488 \[ -\frac {\left (f (b e-4 a f)-\left (e-\sqrt {e^2-4 d f}\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \left (e^2-4 d f\right )^{3/2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}+\frac {\left (f (b e-4 a f)-\left (\sqrt {e^2-4 d f}+e\right ) (c e-b f)\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} \left (e^2-4 d f\right )^{3/2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )} \]
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Rubi [A] time = 2.93, antiderivative size = 488, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {971, 1032, 724, 206} \begin {gather*} -\frac {\left (f (b e-4 a f)-\left (e-\sqrt {e^2-4 d f}\right ) (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} \left (e^2-4 d f\right )^{3/2} \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}}+\frac {\left (f (b e-4 a f)-\left (\sqrt {e^2-4 d f}+e\right ) (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} \left (e^2-4 d f\right )^{3/2} \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}}-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 724
Rule 971
Rule 1032
Rubi steps
\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{\left (d+e x+f x^2\right )^2} \, dx &=-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}-\frac {\int \frac {\frac {1}{2} (b e-4 a f)+(c e-b f) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{-e^2+4 d f}\\ &=-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}-\frac {\left (c e \left (e-\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e-\sqrt {e^2-4 d f}\right )\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}{\left (e^2-4 d f\right )^{3/2}}+\frac {\left (c e \left (e+\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e+\sqrt {e^2-4 d f}\right )\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}{\left (e^2-4 d f\right )^{3/2}}\\ &=-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {\left (2 \left (c e \left (e-\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e-\sqrt {e^2-4 d f}\right )\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 )}{\left (e^2-4 d f\right )^{3/2}}-\frac {\left (2 \left (c e \left (e+\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e+\sqrt {e^2-4 d f}\right )\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 )}{\left (e^2-4 d f\right )^{3/2}}\\ &=-\frac {(e+2 f x) \sqrt {a+b x+c x^2}}{\left (e^2-4 d f\right ) \left (d+e x+f x^2\right )}+\frac {\left (c e \left (e-\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e-\sqrt {e^2-4 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} \left (e^2-4 d f\right )^{3/2} \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}}}-\frac {\left (c e \left (e+\sqrt {e^2-4 d f}\right )+f \left (4 a f-b \left (2 e+\sqrt {e^2-4 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} \left (e^2-4 d f\right )^{3/2} \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}}}\\ \end {align*}
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Mathematica [A] time = 5.09, size = 555, normalized size = 1.14 \begin {gather*} \frac {4 f (e+2 f x) \sqrt {a+x (b+c x)}}{\left (e^2-4 d f\right ) \left (\sqrt {e^2-4 d f}-e-2 f x\right ) \left (\sqrt {e^2-4 d f}+e+2 f x\right )}+\frac {\left (c e \left (\sqrt {e^2-4 d f}-e\right )-f \left (4 a f+b \left (\sqrt {e^2-4 d f}-2 e\right )\right )\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 (e-\sqrt {e^2-4 d f}\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 (e^2-4 d f\right )^{3/2} \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 (f \left (4 a f-b \left (\sqrt {e^2-4 d f}+2 e\right )\right )+c e \left (\sqrt {e^2-4 d f}+e\right )\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 (e^2-4 d f\right )^{3/2} \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 )}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [C] time = 2.55, size = 946, normalized size = 1.94 \begin {gather*} \frac {\sqrt {c x^2+b x+a} (-e-2 f x)}{\left (e^2-4 d f\right ) \left (f x^2+e x+d\right )}-\frac {c \text {RootSum}\left [f \text {$\#$1}^4-2 \sqrt {c} e \text {$\#$1}^3+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+b^2 d-a b e+a^2 f\&,\frac {4 c e \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )-5 b f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+2 \sqrt {c} f \text {$\#$1} \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )}{2 f \text {$\#$1}^3-3 \sqrt {c} e \text {$\#$1}^2+4 c d \text {$\#$1}+b e \text {$\#$1}-2 a f \text {$\#$1}-2 b \sqrt {c} d+a \sqrt {c} e}\&\right ]}{f^2}-\frac {\text {RootSum}\left [f \text {$\#$1}^4-2 \sqrt {c} e \text {$\#$1}^3+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+b^2 d-a b e+a^2 f\&,\frac {8 c^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) e^3-10 b c f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) e^2+4 c^{3/2} f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1} e^2+2 c f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2 e+b^2 f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) e-2 a c f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) e-32 c^2 d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) e-2 b \sqrt {c} f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1} e-2 b f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2-2 a b f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+40 b c d f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+8 a \sqrt {c} f^3 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}-16 c^{3/2} d f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}}{2 f \text {$\#$1}^3-3 \sqrt {c} e \text {$\#$1}^2+4 c d \text {$\#$1}+b e \text {$\#$1}-2 a f \text {$\#$1}-2 b \sqrt {c} d+a \sqrt {c} e}\&\right ]}{2 f^2 \left (4 d f-e^2\right )} \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(-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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maple [B] time = 0.04, size = 22287, normalized size = 45.67 \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] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {c x^{2} + b x + a}}{{\left (f x^{2} + e x + d\right )}^{2}}\,{d x} \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 {\sqrt {c\,x^2+b\,x+a}}{{\left (f\,x^2+e\,x+d\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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