Integrand size = 29, antiderivative size = 270 \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} e^3}+\frac {(e f-d g)^3 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{e^3 \sqrt {c d^2-b d e+a e^2}} \]
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Time = 0.42 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {1667, 857, 635, 212, 738} \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\frac {g \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e g (a e g-b d g+3 b e f)+3 b^2 e^2 g^2+8 c^2 \left (d^2 g^2-3 d e f g+3 e^2 f^2\right )\right )}{8 c^{5/2} e^3}+\frac {(e f-d g)^3 \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{e^3 \sqrt {a e^2-b d e+c d^2}}+\frac {3 g^2 \sqrt {a+b x+c x^2} (-b e g-2 c d g+4 c e f)}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2} \]
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Rule 212
Rule 635
Rule 738
Rule 857
Rule 1667
Rubi steps \begin{align*} \text {integral}& = \frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {\int \frac {\frac {1}{2} e \left (4 c e^2 f^3-d (b d+2 a e) g^3\right )-e g \left (e (2 b d+a e) g^2-c \left (6 e^2 f^2-d^2 g^2\right )\right ) x+\frac {3}{2} e^2 g^2 (4 c e f-2 c d g-b e g) x^2}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 c e^3} \\ & = \frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {\int \frac {\frac {1}{4} e^3 \left (8 c^2 e^2 f^3+3 b^2 d e g^3-4 c d g^2 (3 b e f-b d g+a e g)\right )+\frac {1}{4} e^3 g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{2 c^2 e^5} \\ & = \frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {(e f-d g)^3 \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{e^3}+\frac {\left (g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c^2 e^3} \\ & = \frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}-\frac {\left (2 (e f-d g)^3\right ) \text {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^3}+\frac {\left (g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right )\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{4 c^2 e^3} \\ & = \frac {3 g^2 (4 c e f-2 c d g-b e g) \sqrt {a+b x+c x^2}}{4 c^2 e^2}+\frac {g^3 (d+e x) \sqrt {a+b x+c x^2}}{2 c e^2}+\frac {g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} e^3}+\frac {(e f-d g)^3 \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^3 \sqrt {c d^2-b d e+a e^2}} \\ \end{align*}
Time = 1.15 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.91 \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=-\frac {-\frac {2 e g^2 \sqrt {a+x (b+c x)} (-3 b e g+2 c (6 e f-2 d g+e g x))}{c^2}+\frac {16 \sqrt {-c d^2+b d e-a e^2} (-e f+d g)^3 \arctan \left (\frac {\sqrt {c} (d+e x)-e \sqrt {a+x (b+c x)}}{\sqrt {-c d^2+e (b d-a e)}}\right )}{c d^2+e (-b d+a e)}+\frac {g \left (3 b^2 e^2 g^2-4 c e g (3 b e f-b d g+a e g)+8 c^2 \left (3 e^2 f^2-3 d e f g+d^2 g^2\right )\right ) \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{c^{5/2}}}{8 e^3} \]
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Time = 0.89 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.31
method | result | size |
risch | \(-\frac {g^{2} \left (-2 c e g x +3 b e g +4 c d g -12 c e f \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{2} e^{2}}-\frac {-\frac {8 \left (d^{3} g^{3}-3 d^{2} e f \,g^{2}+3 d \,e^{2} f^{2} g -e^{3} f^{3}\right ) c^{2} \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}+\frac {g \left (4 a c \,e^{2} g^{2}-3 b^{2} e^{2} g^{2}-4 b c d e \,g^{2}+12 b c \,e^{2} f g -8 c^{2} d^{2} g^{2}+24 c^{2} d e f g -24 c^{2} e^{2} f^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{e \sqrt {c}}}{8 c^{2} e^{2}}\) | \(355\) |
default | \(-\frac {\left (-d^{3} g^{3}+3 d^{2} e f \,g^{2}-3 d \,e^{2} f^{2} g +e^{3} f^{3}\right ) \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {\left (x +\frac {d}{e}\right )^{2} c +\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{4} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}+\frac {g \left (\frac {d^{2} g^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+e^{2} g^{2} \left (\frac {x \sqrt {c \,x^{2}+b x +a}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}-\frac {a \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )+\frac {3 e^{2} f^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}-\frac {3 d e f g \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+\left (-d e \,g^{2}+3 e^{2} f g \right ) \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\right )}{e^{3}}\) | \(480\) |
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Timed out. \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\text {Timed out} \]
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\[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\int \frac {\left (f + g x\right )^{3}}{\left (d + e x\right ) \sqrt {a + b x + c x^{2}}}\, dx \]
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Exception generated. \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \]
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Exception generated. \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {(f+g x)^3}{(d+e x) \sqrt {a+b x+c x^2}} \, dx=\int \frac {{\left (f+g\,x\right )}^3}{\left (d+e\,x\right )\,\sqrt {c\,x^2+b\,x+a}} \,d x \]
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