Optimal. Leaf size=113 \[ \frac {\left (b^2-4 a c\right ) (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}}-\frac {(b+2 c x) \sqrt {a+b x+c x^2} (b B-2 A c)}{8 c^2}+\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c} \]
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Rubi [A] time = 0.05, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {640, 612, 621, 206} \begin {gather*} \frac {\left (b^2-4 a c\right ) (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}}-\frac {(b+2 c x) \sqrt {a+b x+c x^2} (b B-2 A c)}{8 c^2}+\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 612
Rule 621
Rule 640
Rubi steps
\begin {align*} \int (A+B x) \sqrt {a+b x+c x^2} \, dx &=\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c}+\frac {(-b B+2 A c) \int \sqrt {a+b x+c x^2} \, dx}{2 c}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c}+\frac {\left (\left (b^2-4 a c\right ) (b B-2 A c)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c}+\frac {\left (\left (b^2-4 a c\right ) (b B-2 A c)\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 )}{8 c^2}\\ &=-\frac {(b B-2 A c) (b+2 c x) \sqrt {a+b x+c x^2}}{8 c^2}+\frac {B \left (a+b x+c x^2\right )^{3/2}}{3 c}+\frac {\left (b^2-4 a c\right ) (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.11, size = 114, normalized size = 1.01 \begin {gather*} \frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (4 c (2 a B+c x (3 A+2 B x))+2 b c (3 A+B x)-3 b^2 B\right )+3 \left (b^2-4 a c\right ) (b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{48 c^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.52, size = 125, normalized size = 1.11 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (8 a B c+6 A b c+12 A c^2 x-3 b^2 B+2 b B c x+8 B c^2 x^2\right )}{24 c^2}+\frac {\left (-8 a A c^2+4 a b B c+2 A b^2 c+b^3 (-B)\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{16 c^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 291, normalized size = 2.58 \begin {gather*} \left [\frac {3 \, {\left (B b^{3} + 8 \, A a c^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 2 \, {\left (4 \, B a + 3 \, A b\right )} c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, c^{3}}, -\frac {3 \, {\left (B b^{3} + 8 \, A a c^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} - 3 \, B b^{2} c + 2 \, {\left (4 \, B a + 3 \, A b\right )} c^{2} + 2 \, {\left (B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, c^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.25, size = 123, normalized size = 1.09 \begin {gather*} \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, B x + \frac {B b c + 6 \, A c^{2}}{c^{2}}\right )} x - \frac {3 \, B b^{2} - 8 \, B a c - 6 \, A b c}{c^{2}}\right )} - \frac {{\left (B b^{3} - 4 \, B a b c - 2 \, A b^{2} c + 8 \, A a c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 229, normalized size = 2.03 \begin {gather*} \frac {A a \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\frac {A \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}-\frac {B a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\frac {B \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, A x}{2}-\frac {\sqrt {c \,x^{2}+b x +a}\, B b x}{4 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, A b}{4 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, B \,b^{2}}{8 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{3 c} \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 [B] time = 1.43, size = 145, normalized size = 1.28 \begin {gather*} A\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {A\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {B\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {B\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2} \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 \left (A + B x\right ) \sqrt {a + b x + c x^{2}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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