Optimal. Leaf size=227 \[ \frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}-\frac {a^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (b^2-12 a c\right ) x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]
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Rubi [A]
time = 0.16, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1935, 1959,
1947, 857, 635, 212, 738} \begin {gather*} -\frac {a^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b x \left (b^2-12 a c\right ) \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {\left (8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 635
Rule 738
Rule 857
Rule 1935
Rule 1947
Rule 1959
Rubi steps
\begin {align*} \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^4} \, dx &=\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}+\frac {1}{2} \int \frac {(2 a+b x) \sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx\\ &=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}+\frac {\int \frac {8 a^2 c-\frac {1}{2} b \left (b^2-12 a c\right ) x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{8 c}\\ &=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}+\frac {\left (x \sqrt {a+b x+c x^2}\right ) \int \frac {8 a^2 c-\frac {1}{2} b \left (b^2-12 a c\right ) x}{x \sqrt {a+b x+c x^2}} \, dx}{8 c \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}+\frac {\left (a^2 x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{\sqrt {a x^2+b x^3+c x^4}}-\frac {\left (b \left (b^2-12 a c\right ) x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}-\frac {\left (2 a^2 x \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x}{\sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {\left (b \left (b^2-12 a c\right ) x \sqrt {a+b x+c x^2}\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 )}{8 c \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}-\frac {a^{3/2} x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (b^2-12 a c\right ) x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}\\ \end {align*}
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Mathematica [A]
time = 0.53, size = 166, normalized size = 0.73 \begin {gather*} \frac {x \sqrt {a+x (b+c x)} \left (2 \sqrt {c} \sqrt {a+x (b+c x)} \left (3 b^2+14 b c x+8 c \left (4 a+c x^2\right )\right )+96 a^{3/2} c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+3 \left (b^3-12 a b c\right ) \log \left (c \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{48 c^{3/2} \sqrt {x^2 (a+x (b+c x))}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.03, size = 222, normalized size = 0.98
method | result | size |
default | \(-\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 c^{\frac {5}{2}} a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )-16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}-12 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b x -48 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a -6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{2}-36 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a b \,c^{2}+3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{3} c \right )}{48 x^{3} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}}\) | \(222\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 791, normalized size = 3.48 \begin {gather*} \left [\frac {48 \, a^{\frac {3}{2}} c^{2} x \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{2} x}, \frac {24 \, a^{\frac {3}{2}} c^{2} x \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{2} x}, \frac {96 \, \sqrt {-a} a c^{2} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{2} x}, \frac {48 \, \sqrt {-a} a c^{2} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 3 \, {\left (b^{3} - 12 \, a b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, {\left (8 \, c^{3} x^{2} + 14 \, b c^{2} x + 3 \, b^{2} c + 32 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{2} x}\right ] \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 (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{4}}\, dx \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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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