Optimal. Leaf size=367 \[ -\frac {f \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2}-\frac {\sqrt {a+c x^2}}{a d x} \]
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Rubi [A] time = 1.20, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {6728, 264, 266, 63, 208, 1034, 725, 206} \[ -\frac {f \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {f \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2}-\frac {\sqrt {a+c x^2}}{a d x} \]
Antiderivative was successfully verified.
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Rule 63
Rule 206
Rule 208
Rule 264
Rule 266
Rule 725
Rule 1034
Rule 6728
Rubi steps
\begin {align*} \int \frac {1}{x^2 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac {1}{d x^2 \sqrt {a+c x^2}}-\frac {e}{d^2 x \sqrt {a+c x^2}}+\frac {e^2-d f+e f x}{d^2 \sqrt {a+c x^2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {e^2-d f+e f x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{d^2}+\frac {\int \frac {1}{x^2 \sqrt {a+c x^2}} \, dx}{d}-\frac {e \int \frac {1}{x \sqrt {a+c x^2}} \, dx}{d^2}\\ &=-\frac {\sqrt {a+c x^2}}{a d x}-\frac {e \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (f \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{d^2 \sqrt {e^2-4 d f}}+\frac {\left (f \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{d^2 \sqrt {e^2-4 d f}}\\ &=-\frac {\sqrt {a+c x^2}}{a d x}-\frac {e \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{c d^2}+\frac {\left (f \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e+\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{d^2 \sqrt {e^2-4 d f}}-\frac {\left (f \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2-x^2} \, dx,x,\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {a+c x^2}}\right )}{d^2 \sqrt {e^2-4 d f}}\\ &=-\frac {\sqrt {a+c x^2}}{a d x}-\frac {f \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}+\frac {f \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right ) \tanh ^{-1}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^2 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^2}\\ \end {align*}
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Mathematica [A] time = 0.84, size = 356, normalized size = 0.97 \[ -\frac {\frac {\sqrt {2} f \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right ) \tanh ^{-1}\left (\frac {2 a f+c x \left (\sqrt {e^2-4 d f}-e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {\sqrt {2} f \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {2 d \sqrt {a+c x^2}}{a x}-\frac {2 e \tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}}{2 d^2} \]
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 [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 736, normalized size = 2.01 \[ \frac {16 e \,f^{2} \ln \left (\frac {2 a +2 \sqrt {c \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{2} \left (e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {a}}-\frac {4 \sqrt {2}\, f^{2} \ln \left (\frac {-\frac {\left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right ) c}{f}+\frac {2 a \,f^{2}-2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}+\frac {\sqrt {2}\, \sqrt {\frac {2 a \,f^{2}-2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}\, \sqrt {4 \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right ) c}{f}+\frac {4 a \,f^{2}-4 c d f +2 c \,e^{2}-2 \sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {2 a \,f^{2}-2 c d f +c \,e^{2}-\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}+\frac {4 \sqrt {2}\, f^{2} \ln \left (\frac {-\frac {\left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right ) c}{f}+\frac {2 a \,f^{2}-2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}+\frac {\sqrt {2}\, \sqrt {\frac {2 a \,f^{2}-2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}\, \sqrt {4 \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )^{2} c -\frac {4 \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right ) c}{f}+\frac {4 a \,f^{2}-4 c d f +2 c \,e^{2}+2 \sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right )^{2} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {2 a \,f^{2}-2 c d f +c \,e^{2}+\sqrt {-4 d f +e^{2}}\, c e}{f^{2}}}}+\frac {4 \sqrt {c \,x^{2}+a}\, f}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) a x} \]
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 \[ \int \frac {1}{\sqrt {c x^{2} + a} {\left (f x^{2} + e x + d\right )} x^{2}}\,{d x} \]
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 \[ \int \frac {1}{x^2\,\sqrt {c\,x^2+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \]
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 \[ \int \frac {1}{x^{2} \sqrt {a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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