Optimal. Leaf size=526 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {1}{a d \sqrt {a+c x^2}} \]
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Rubi [A] time = 2.18, antiderivative size = 526, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6728, 266, 51, 63, 208, 1017, 1034, 725, 206} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \sqrt {a+c x^2} \left ((c d-a f)^2+a c e^2\right )}+\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (\sqrt {e^2-4 d f}+e\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d \sqrt {e^2-4 d f} \left ((c d-a f)^2+a c e^2\right ) \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {1}{a d \sqrt {a+c x^2}} \]
Antiderivative was successfully verified.
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Rule 51
Rule 63
Rule 206
Rule 208
Rule 266
Rule 725
Rule 1017
Rule 1034
Rule 6728
Rubi steps
\begin {align*} \int \frac {1}{x \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx &=\int \left (\frac {1}{d x \left (a+c x^2\right )^{3/2}}+\frac {-e-f x}{d \left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {1}{x \left (a+c x^2\right )^{3/2}} \, dx}{d}+\frac {\int \frac {-e-f x}{\left (a+c x^2\right )^{3/2} \left (d+e x+f x^2\right )} \, dx}{d}\\ &=-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x (a+c x)^{3/2}} \, dx,x,x^2\right )}{2 d}+\frac {\int \frac {-2 a c e \left (a f^2+c \left (e^2-2 d f\right )\right )-2 a c f \left (a f^2+c \left (e^2-d f\right )\right ) x}{\sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx}{2 a c d \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {1}{a d \sqrt {a+c x^2}}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^2\right )}{2 a d}-\frac {\left (f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right )\right ) \int \frac {1}{\left (e-\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{d \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}+\frac {\left (f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\right )\right ) \int \frac {1}{\left (e+\sqrt {e^2-4 d f}+2 f x\right ) \sqrt {a+c x^2}} \, dx}{d \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {1}{a d \sqrt {a+c x^2}}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^2}\right )}{a c d}+\frac {\left (f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\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 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}-\frac {\left (f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\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 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right )}\\ &=\frac {1}{a d \sqrt {a+c x^2}}-\frac {a \left (a f^2+c \left (e^2-d f\right )\right )+c^2 d e x}{a d \left (a c e^2+(c d-a f)^2\right ) \sqrt {a+c x^2}}+\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e-\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\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 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {f \left (2 e \left (a f^2+c \left (e^2-2 d f\right )\right )-\left (e+\sqrt {e^2-4 d f}\right ) \left (a f^2+c \left (e^2-d f\right )\right )\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 \sqrt {e^2-4 d f} \left (a c e^2+(c d-a f)^2\right ) \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2} d}\\ \end {align*}
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Mathematica [C] time = 3.50, size = 497, normalized size = 0.94 \[ \frac {-\frac {f \left (\frac {e}{\sqrt {e^2-4 d f}}+1\right ) \left (2 a f+c x \left (e-\sqrt {e^2-4 d f}\right )\right )}{a \sqrt {a+c x^2} \left (4 a f^2+c \left (e-\sqrt {e^2-4 d f}\right )^2\right )}-\frac {f \left (1-\frac {e}{\sqrt {e^2-4 d f}}\right ) \left (2 a f+c x \left (\sqrt {e^2-4 d f}+e\right )\right )}{a \sqrt {a+c x^2} \left (4 a f^2+c \left (\sqrt {e^2-4 d f}+e\right )^2\right )}+\frac {\sqrt {2} f^3 \left (\sqrt {e^2-4 d f}+e\right ) \tanh ^{-1}\left (\frac {2 a f+c x \left (\sqrt {e^2-4 d f}-e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2-2 c \left (e \sqrt {e^2-4 d f}+2 d f-e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac {\sqrt {2} f^3 \left (\sqrt {e^2-4 d f}-e\right ) \tanh ^{-1}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {a+c x^2} \sqrt {4 a f^2+2 c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {e^2-4 d f} \left (2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )\right )^{3/2}}+\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c x^2}{a}+1\right )}{a \sqrt {a+c x^2}}}{d} \]
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(-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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maple [B] time = 0.02, size = 1945, normalized size = 3.70 \[ \text {result too large to display} \]
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 \[ \int \frac {1}{{\left (c x^{2} + a\right )}^{\frac {3}{2}} {\left (f x^{2} + e x + d\right )} x}\,{d x} \]
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 \[ \int \frac {1}{x\,{\left (c\,x^2+a\right )}^{3/2}\,\left (f\,x^2+e\,x+d\right )} \,d x \]
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 \[ \int \frac {1}{x \left (a + c x^{2}\right )^{\frac {3}{2}} \left (d + e x + f x^{2}\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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