Optimal. Leaf size=673 \[ \frac {g \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac {e \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^3}-\frac {e^2 \sqrt {a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{g (e f-d g)^3}+\frac {e^2 (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^3}-\frac {g \sqrt {a+b x+c x^2} (-2 a g+x (2 c f-b g)+b f)}{4 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e (2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 g (e f-d g)^2 \sqrt {a g^2-b f g+c f^2}}+\frac {e \sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)^2}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac {e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^3} \]
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Rubi [A] time = 0.86, antiderivative size = 673, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {960, 734, 843, 621, 206, 724, 720, 732} \[ \frac {g \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{8 (e f-d g) \left (a g^2-b f g+c f^2\right )^{3/2}}+\frac {e \sqrt {a e^2-b d e+c d^2} \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{(e f-d g)^3}-\frac {e^2 \sqrt {a g^2-b f g+c f^2} \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{g (e f-d g)^3}+\frac {e^2 (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^3}-\frac {g \sqrt {a+b x+c x^2} (-2 a g+x (2 c f-b g)+b f)}{4 (f+g x)^2 (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac {e (2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+x (2 c f-b g)+b f}{2 \sqrt {a+b x+c x^2} \sqrt {a g^2-b f g+c f^2}}\right )}{2 g (e f-d g)^2 \sqrt {a g^2-b f g+c f^2}}+\frac {e \sqrt {a+b x+c x^2}}{(f+g x) (e f-d g)^2}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}-\frac {e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^3} \]
Antiderivative was successfully verified.
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Rule 206
Rule 621
Rule 720
Rule 724
Rule 732
Rule 734
Rule 843
Rule 960
Rubi steps
\begin {align*} \int \frac {\sqrt {a+b x+c x^2}}{(d+e x) (f+g x)^3} \, dx &=\int \left (\frac {e^3 \sqrt {a+b x+c x^2}}{(e f-d g)^3 (d+e x)}-\frac {g \sqrt {a+b x+c x^2}}{(e f-d g) (f+g x)^3}-\frac {e g \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)^2}-\frac {e^2 g \sqrt {a+b x+c x^2}}{(e f-d g)^3 (f+g x)}\right ) \, dx\\ &=\frac {e^3 \int \frac {\sqrt {a+b x+c x^2}}{d+e x} \, dx}{(e f-d g)^3}-\frac {\left (e^2 g\right ) \int \frac {\sqrt {a+b x+c x^2}}{f+g x} \, dx}{(e f-d g)^3}-\frac {(e g) \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^2} \, dx}{(e f-d g)^2}-\frac {g \int \frac {\sqrt {a+b x+c x^2}}{(f+g x)^3} \, dx}{e f-d g}\\ &=\frac {e \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac {e^2 \int \frac {b d-2 a e+(2 c d-b e) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^3}+\frac {e^2 \int \frac {b f-2 a g+(2 c f-b g) x}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^3}-\frac {e \int \frac {b+2 c x}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^2}+\frac {\left (\left (b^2-4 a c\right ) g\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )}\\ &=\frac {e \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac {(e (2 c d-b e)) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 (e f-d g)^3}+\frac {\left (e \left (c d^2-b d e+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{(e f-d g)^3}+\frac {\left (e^2 (2 c f-b g)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 g (e f-d g)^3}-\frac {(c e) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{g (e f-d g)^2}+\frac {(e (2 c f-b g)) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{2 g (e f-d g)^2}-\frac {\left (\left (b^2-4 a c\right ) g\right ) \operatorname {Subst}\left (\int \frac {1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac {-b f+2 a g-(2 c f-b g) x}{\sqrt {a+b x+c x^2}}\right )}{4 (e f-d g) \left (c f^2-b f g+a g^2\right )}-\frac {\left (e^2 \left (c f^2-b f g+a g^2\right )\right ) \int \frac {1}{(f+g x) \sqrt {a+b x+c x^2}} \, dx}{g (e f-d g)^3}\\ &=\frac {e \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}+\frac {\left (b^2-4 a c\right ) g \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}-\frac {(e (2 c d-b e)) \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 )}{(e f-d g)^3}-\frac {\left (2 e \left (c d^2-b d e+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{(e f-d g)^3}+\frac {\left (e^2 (2 c f-b g)\right ) \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 )}{g (e f-d g)^3}-\frac {(2 c e) \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 )}{g (e f-d g)^2}-\frac {(e (2 c f-b g)) \operatorname {Subst}\left (\int \frac {1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac {-b f+2 a g-(2 c f-b g) x}{\sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}+\frac {\left (2 e^2 \left (c f^2-b f g+a g^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c f^2-4 b f g+4 a g^2-x^2} \, dx,x,\frac {-b f+2 a g-(2 c f-b g) x}{\sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^3}\\ &=\frac {e \sqrt {a+b x+c x^2}}{(e f-d g)^2 (f+g x)}-\frac {g (b f-2 a g+(2 c f-b g) x) \sqrt {a+b x+c x^2}}{4 (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x)^2}-\frac {e (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} (e f-d g)^3}+\frac {e^2 (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} g (e f-d g)^3}-\frac {\sqrt {c} e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^2}+\frac {e \sqrt {c d^2-b d e+a e^2} \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{(e f-d g)^3}+\frac {\left (b^2-4 a c\right ) g \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{8 (e f-d g) \left (c f^2-b f g+a g^2\right )^{3/2}}+\frac {e (2 c f-b g) \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{2 g (e f-d g)^2 \sqrt {c f^2-b f g+a g^2}}-\frac {e^2 \sqrt {c f^2-b f g+a g^2} \tanh ^{-1}\left (\frac {b f-2 a g+(2 c f-b g) x}{2 \sqrt {c f^2-b f g+a g^2} \sqrt {a+b x+c x^2}}\right )}{g (e f-d g)^3}\\ \end {align*}
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Mathematica [A] time = 1.37, size = 609, normalized size = 0.90 \[ \frac {\frac {g \left (b^2-4 a c\right ) (e f-d g)^2 \tanh ^{-1}\left (\frac {-2 a g+b (f-g x)+2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )}{\left (g (a g-b f)+c f^2\right )^{3/2}}+8 e \sqrt {e (a e-b d)+c d^2} \tanh ^{-1}\left (\frac {-2 a e+b (d-e x)+2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )+\frac {2 g \sqrt {a+x (b+c x)} (e f-d g)^2 (2 a g-b f+b g x-2 c f x)}{(f+g x)^2 \left (g (a g-b f)+c f^2\right )}-\frac {4 e (e f-d g) \left (2 \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-\frac {(2 c f-b g) \tanh ^{-1}\left (\frac {-2 a g+b (f-g x)+2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )}{\sqrt {g (a g-b f)+c f^2}}\right )}{g}+\frac {8 e \sqrt {a+x (b+c x)} (e f-d g)}{f+g x}+\frac {4 e (b e-2 c d) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{\sqrt {c}}+\frac {4 e^2 \left ((2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} \sqrt {g (a g-b f)+c f^2} \tanh ^{-1}\left (\frac {-2 a g+b (f-g x)+2 c f x}{2 \sqrt {a+x (b+c x)} \sqrt {g (a g-b f)+c f^2}}\right )\right )}{\sqrt {c} g}}{8 (e f-d g)^3} \]
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 [B] time = 4.64, size = 1844, normalized size = 2.74 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 6714, normalized size = 9.98 \[ \text {output too large to display} \]
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 \[ \int \frac {\sqrt {c x^{2} + b x + a}}{{\left (e x + d\right )} {\left (g x + f\right )}^{3}}\,{d x} \]
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 \[ \int \frac {\sqrt {c\,x^2+b\,x+a}}{{\left (f+g\,x\right )}^3\,\left (d+e\,x\right )} \,d x \]
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 \[ \int \frac {\sqrt {a + b x + c x^{2}}}{\left (d + e x\right ) \left (f + g x\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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