Optimal. Leaf size=642 \[ \frac{g^2 \sqrt{a+b x+c x^2} \left (-4 c g (2 a g+b f)+3 b^2 g^2+4 c^2 f^2\right )}{\left (b^2-4 a c\right ) (f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )^2}-\frac{2 e^2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g)^2 \left (a e^2-b d e+c d^2\right )}+\frac{2 e g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac{2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) (f+g x) \sqrt{a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e^4 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g)^2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{e g^3 \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g)^2 \left (a g^2-b f g+c f^2\right )^{3/2}}-\frac{3 g^3 (2 c f-b g) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 (e f-d g) \left (a g^2-b f g+c f^2\right )^{5/2}} \]
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Rubi [A] time = 0.910685, antiderivative size = 642, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {960, 740, 12, 724, 206, 806} \[ \frac{g^2 \sqrt{a+b x+c x^2} \left (-4 c g (2 a g+b f)+3 b^2 g^2+4 c^2 f^2\right )}{\left (b^2-4 a c\right ) (f+g x) (e f-d g) \left (a g^2-b f g+c f^2\right )^2}-\frac{2 e^2 \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g)^2 \left (a e^2-b d e+c d^2\right )}+\frac{2 e g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} (e f-d g)^2 \left (a g^2-b f g+c f^2\right )}+\frac{2 g \left (2 a c g+b^2 (-g)+c x (2 c f-b g)+b c f\right )}{\left (b^2-4 a c\right ) (f+g x) \sqrt{a+b x+c x^2} (e f-d g) \left (a g^2-b f g+c f^2\right )}+\frac{e^4 \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{(e f-d g)^2 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac{e g^3 \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{(e f-d g)^2 \left (a g^2-b f g+c f^2\right )^{3/2}}-\frac{3 g^3 (2 c f-b g) \tanh ^{-1}\left (\frac{-2 a g+x (2 c f-b g)+b f}{2 \sqrt{a+b x+c x^2} \sqrt{a g^2-b f g+c f^2}}\right )}{2 (e f-d g) \left (a g^2-b f g+c f^2\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 960
Rule 740
Rule 12
Rule 724
Rule 206
Rule 806
Rubi steps
\begin{align*} \int \frac{1}{(d+e x) (f+g x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx &=\int \left (\frac{e^2}{(e f-d g)^2 (d+e x) \left (a+b x+c x^2\right )^{3/2}}-\frac{g}{(e f-d g) (f+g x)^2 \left (a+b x+c x^2\right )^{3/2}}-\frac{e g}{(e f-d g)^2 (f+g x) \left (a+b x+c x^2\right )^{3/2}}\right ) \, dx\\ &=\frac{e^2 \int \frac{1}{(d+e x) \left (a+b x+c x^2\right )^{3/2}} \, dx}{(e f-d g)^2}-\frac{(e g) \int \frac{1}{(f+g x) \left (a+b x+c x^2\right )^{3/2}} \, dx}{(e f-d g)^2}-\frac{g \int \frac{1}{(f+g x)^2 \left (a+b x+c x^2\right )^{3/2}} \, dx}{e f-d g}\\ &=-\frac{2 e^2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g)^2 \sqrt{a+b x+c x^2}}+\frac{2 e g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt{a+b x+c x^2}}+\frac{2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x) \sqrt{a+b x+c x^2}}-\frac{\left (2 e^2\right ) \int -\frac{\left (b^2-4 a c\right ) e^2}{2 (d+e x) \sqrt{a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g)^2}+\frac{(2 e g) \int -\frac{\left (b^2-4 a c\right ) g^2}{2 (f+g x) \sqrt{a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-b f g+a g^2\right )}+\frac{(2 g) \int \frac{\frac{1}{2} g \left (2 b c f-3 b^2 g+8 a c g\right )+c g (2 c f-b g) x}{(f+g x)^2 \sqrt{a+b x+c x^2}} \, dx}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right )}\\ &=-\frac{2 e^2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g)^2 \sqrt{a+b x+c x^2}}+\frac{2 e g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt{a+b x+c x^2}}+\frac{2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x) \sqrt{a+b x+c x^2}}+\frac{g^2 \left (4 c^2 f^2+3 b^2 g^2-4 c g (b f+2 a g)\right ) \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac{e^4 \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{\left (c d^2-b d e+a e^2\right ) (e f-d g)^2}-\frac{\left (3 g^3 (2 c f-b g)\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{2 (e f-d g) \left (c f^2-b f g+a g^2\right )^2}-\frac{\left (e g^3\right ) \int \frac{1}{(f+g x) \sqrt{a+b x+c x^2}} \, dx}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )}\\ &=-\frac{2 e^2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g)^2 \sqrt{a+b x+c x^2}}+\frac{2 e g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt{a+b x+c x^2}}+\frac{2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x) \sqrt{a+b x+c x^2}}+\frac{g^2 \left (4 c^2 f^2+3 b^2 g^2-4 c g (b f+2 a g)\right ) \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}-\frac{\left (2 e^4\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 )}{\left (c d^2-b d e+a e^2\right ) (e f-d g)^2}+\frac{\left (3 g^3 (2 c f-b 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 )}{(e f-d g) \left (c f^2-b f g+a g^2\right )^2}+\frac{\left (2 e g^3\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 )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )}\\ &=-\frac{2 e^2 \left (b c d-b^2 e+2 a c e+c (2 c d-b e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) (e f-d g)^2 \sqrt{a+b x+c x^2}}+\frac{2 e g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-b f g+a g^2\right ) \sqrt{a+b x+c x^2}}+\frac{2 g \left (b c f-b^2 g+2 a c g+c (2 c f-b g) x\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right ) (f+g x) \sqrt{a+b x+c x^2}}+\frac{g^2 \left (4 c^2 f^2+3 b^2 g^2-4 c g (b f+2 a g)\right ) \sqrt{a+b x+c x^2}}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-b f g+a g^2\right )^2 (f+g x)}+\frac{e^4 \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 )}{\left (c d^2-b d e+a e^2\right )^{3/2} (e f-d g)^2}-\frac{3 g^3 (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 (e f-d g) \left (c f^2-b f g+a g^2\right )^{5/2}}-\frac{e g^3 \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 )}{(e f-d g)^2 \left (c f^2-b f g+a g^2\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 6.21851, size = 868, normalized size = 1.35 \[ \frac{\left (c x^2+b x+a\right )^{3/2} \tanh ^{-1}\left (\frac{b d-2 a e+(2 c d-b e) x}{2 \sqrt{c d^2-e (b d-a e)} \sqrt{c x^2+b x+a}}\right ) e^4}{\left (c d^2-e (b d-a e)\right )^{3/2} (e f-d g)^2 (a+x (b+c x))^{3/2}}-\frac{2 \left (-e b^2+c d b+2 a c e+c (2 c d-b e) x\right ) \left (c x^2+b x+a\right ) e^2}{\left (b^2-4 a c\right ) \left (c d^2-e (b d-a e)\right ) (e f-d g)^2 (a+x (b+c x))^{3/2}}-\frac{g^3 \left (c x^2+b x+a\right )^{3/2} \tanh ^{-1}\left (\frac{b f-2 a g+(2 c f-b g) x}{2 \sqrt{c f^2-g (b f-a g)} \sqrt{c x^2+b x+a}}\right ) e}{(e f-d g)^2 \left (c f^2-g (b f-a g)\right )^{3/2} (a+x (b+c x))^{3/2}}+\frac{2 g \left (-g b^2+c f b+2 a c g+c (2 c f-b g) x\right ) \left (c x^2+b x+a\right ) e}{\left (b^2-4 a c\right ) (e f-d g)^2 \left (c f^2-g (b f-a g)\right ) (a+x (b+c x))^{3/2}}-\frac{2 g \left (c x^2+b x+a\right )^{3/2} \left (\frac{2 \left (b \left (\frac{1}{2} \left (-3 g b^2+2 c f b+8 a c g\right ) g^2+c f (2 c f-b g) g\right )-2 \left (a c (2 c f-b g) g^2+\frac{1}{2} c f \left (-3 g b^2+2 c f b+8 a c g\right ) g\right )\right ) \tanh ^{-1}\left (\frac{-b f+2 a g-(2 c f-b g) x}{2 \sqrt{c f^2-b g f+a g^2} \sqrt{c x^2+b x+a}}\right )}{\sqrt{c f^2-b g f+a g^2} \left (4 c f^2-4 b g f+4 a g^2\right )}-\frac{\left (\frac{1}{2} g^2 \left (-3 g b^2+2 c f b+8 a c g\right )-c f g (2 c f-b g)\right ) \sqrt{c x^2+b x+a}}{\left (c f^2-b g f+a g^2\right ) (f+g x)}\right )}{\left (b^2-4 a c\right ) (d g-e f) \left (c f^2-b g f+a g^2\right ) (a+x (b+c x))^{3/2}}+\frac{2 g \left (-g b^2+c f b+2 a c g+c (2 c f-b g) x\right ) \left (c x^2+b x+a\right )}{\left (b^2-4 a c\right ) (e f-d g) \left (c f^2-g (b f-a g)\right ) (f+g x) (a+x (b+c x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.416, size = 2807, normalized size = 4.4 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (e x + d\right )}{\left (g x + f\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}{\left (e x + d\right )}{\left (g x + f\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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