Optimal. Leaf size=617 \[ \frac {\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt {e^2-4 d f}\right ) \left (B \left (f (b e-a f)-c \left (e^2-d f\right )\right )+A f (c e-b f)\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (e-\sqrt {e^2-4 d f}\right )\right )-b \left (e-\sqrt {e^2-4 d f}\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2-\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (\sqrt {e^2-4 d f}+e\right ) \left (B \left (f (b e-a f)-c \left (e^2-d f\right )\right )+A f (c e-b f)\right )\right ) \tanh ^{-1}\left (\frac {4 a f+2 x \left (b f-c \left (\sqrt {e^2-4 d f}+e\right )\right )-b \left (\sqrt {e^2-4 d f}+e\right )}{2 \sqrt {2} \sqrt {a+b x+c x^2} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}\right )}{\sqrt {2} f^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+\sqrt {e^2-4 d f} (c e-b f)-b e f-2 c d f+c e^2}}-\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (-2 A c f-b B f+2 B c e)}{2 \sqrt {c} f^2}+\frac {B \sqrt {a+b x+c x^2}}{f} \]
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Rubi [A] time = 9.00, antiderivative size = 615, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1019, 1076, 621, 206, 1032, 724} \begin {gather*} \frac {\left (2 f (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt {e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )\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 (2 f (A f (c d-a f)-B d (c e-b f))-\left (\sqrt {e^2-4 d f}+e\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )\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 {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) (-2 A c f-b B f+2 B c e)}{2 \sqrt {c} f^2}+\frac {B \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 {(A+B x) \sqrt {a+b x+c x^2}}{d+e x+f x^2} \, dx &=\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {\int \frac {\frac {1}{2} (b B d-2 a A f)-\frac {1}{2} (2 A b f-B (2 c d+b e-2 a f)) x+\frac {1}{2} (2 B c e-b B f-2 A c f) x^2}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f}\\ &=\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {\int \frac {\frac {1}{2} f (b B d-2 a A f)-\frac {1}{2} d (2 B c e-b B f-2 A c f)+\left (-\frac {1}{2} e (2 B c e-b B f-2 A c f)+\frac {1}{2} f (-2 A b f+B (2 c d+b e-2 a f))\right ) x}{\sqrt {a+b x+c x^2} \left (d+e x+f x^2\right )} \, dx}{f^2}-\frac {(2 B c e-b B f-2 A c f) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 f^2}\\ &=\frac {B \sqrt {a+b x+c x^2}}{f}-\frac {(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt {e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-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}{f^2 \sqrt {e^2-4 d f}}-\frac {\left (2 f \left (\frac {1}{2} f (b B d-2 a A f)-\frac {1}{2} d (2 B c e-b B f-2 A c f)\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (-\frac {1}{2} e (2 B c e-b B f-2 A c f)+\frac {1}{2} f (-2 A b f+B (2 c d+b e-2 a 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 {B \sqrt {a+b x+c x^2}}{f}-\frac {(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt {e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-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 )}{f^2 \sqrt {e^2-4 d f}}+\frac {\left (2 \left (2 f \left (\frac {1}{2} f (b B d-2 a A f)-\frac {1}{2} d (2 B c e-b B f-2 A c f)\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (-\frac {1}{2} e (2 B c e-b B f-2 A c f)+\frac {1}{2} f (-2 A b f+B (2 c d+b e-2 a 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 {B \sqrt {a+b x+c x^2}}{f}-\frac {(2 B c e-b B f-2 A c 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 (A f (c d-a f)-B d (c e-b f))-\left (e-\sqrt {e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B 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 (A f (c d-a f)-B d (c e-b f))-\left (e+\sqrt {e^2-4 d f}\right ) \left (B f (b e-a f)+A f (c e-b f)-B 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 = 2.09, size = 517, normalized size = 0.84 \begin {gather*} \frac {-\sqrt {2} \left (B \left (\sqrt {e^2-4 d f}+e\right )-2 A f\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 (2 A f+B \left (\sqrt {e^2-4 d f}-e\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 )} \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 )+4 B f \sqrt {e^2-4 d f} \sqrt {a+x (b+c x)}}{4 f^2 \sqrt {e^2-4 d f}}+\frac {\tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) (2 A c f+b B f-2 B c e)}{2 \sqrt {c} f^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [C] time = 1.21, size = 893, normalized size = 1.45 \begin {gather*} \frac {\sqrt {c x^2+b x+a} B}{f}+\frac {(2 B c e-b B f-2 A c f) \log \left (b+2 c x-2 \sqrt {c} \sqrt {c x^2+b x+a}\right )}{2 \sqrt {c} f^2}+\frac {\text {RootSum}\left [f \text {$\#$1}^4-2 \sqrt {c} e \text {$\#$1}^3+4 c d \text {$\#$1}^2+b e \text {$\#$1}^2-2 a f \text {$\#$1}^2-4 b \sqrt {c} d \text {$\#$1}+2 a \sqrt {c} e \text {$\#$1}+b^2 d-a b e+a^2 f\&,\frac {B d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b^2-A f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2 b+B e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2 b-B c d e \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b+A c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b-a B e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) b-2 B \sqrt {c} d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1} b-B c e^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2-a B f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+B c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+A c e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}^2+a B c e^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+a^2 B f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )-a B c d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )-a A c e f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right )+2 a A \sqrt {c} f^2 \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}+2 B c^{3/2} d e \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}-2 A c^{3/2} d f \log \left (-\sqrt {c} x-\text {$\#$1}+\sqrt {c x^2+b x+a}\right ) \text {$\#$1}}{-2 f \text {$\#$1}^3+3 \sqrt {c} e \text {$\#$1}^2-4 c d \text {$\#$1}-b e \text {$\#$1}+2 a f \text {$\#$1}+2 b \sqrt {c} d-a \sqrt {c} e}\&\right ]}{f^2} \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 = 0.05, size = 16209, normalized size = 26.27 \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 {\left (A+B\,x\right )\,\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 {\left (A + B x\right ) \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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