Optimal. Leaf size=306 \[ -\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e p \text {Li}_2\left (\frac {b x^2}{a}+1\right )}{2 d^2}-\frac {e p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^2}-\frac {e p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{d^2}-\frac {e p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{d^2}-\frac {e p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^2}+\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d} \]
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Rubi [A] time = 0.35, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {2466, 2455, 205, 2454, 2394, 2315, 2462, 260, 2416, 2393, 2391} \[ -\frac {e p \text {PolyLog}\left (2,\frac {b x^2}{a}+1\right )}{2 d^2}-\frac {e p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^2}-\frac {e p \text {PolyLog}\left (2,\frac {\sqrt {b} (d+e x)}{\sqrt {-a} e+\sqrt {b} d}\right )}{d^2}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e p \log (d+e x) \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )}{d^2}-\frac {e p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^2}+\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
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Rule 205
Rule 260
Rule 2315
Rule 2391
Rule 2393
Rule 2394
Rule 2416
Rule 2454
Rule 2455
Rule 2462
Rule 2466
Rubi steps
\begin {align*} \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2 (d+e x)} \, dx &=\int \left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x^2}-\frac {e \log \left (c \left (a+b x^2\right )^p\right )}{d^2 x}+\frac {e^2 \log \left (c \left (a+b x^2\right )^p\right )}{d^2 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x^2} \, dx}{d}-\frac {e \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx}{d^2}+\frac {e^2 \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{d+e x} \, dx}{d^2}\\ &=-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {e \operatorname {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,x^2\right )}{2 d^2}+\frac {(2 b p) \int \frac {1}{a+b x^2} \, dx}{d}-\frac {(2 b e p) \int \frac {x \log (d+e x)}{a+b x^2} \, dx}{d^2}\\ &=\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}+\frac {(b e p) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,x^2\right )}{2 d^2}-\frac {(2 b e p) \int \left (-\frac {\log (d+e x)}{2 \sqrt {b} \left (\sqrt {-a}-\sqrt {b} x\right )}+\frac {\log (d+e x)}{2 \sqrt {b} \left (\sqrt {-a}+\sqrt {b} x\right )}\right ) \, dx}{d^2}\\ &=\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {e p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^2}+\frac {\left (\sqrt {b} e p\right ) \int \frac {\log (d+e x)}{\sqrt {-a}-\sqrt {b} x} \, dx}{d^2}-\frac {\left (\sqrt {b} e p\right ) \int \frac {\log (d+e x)}{\sqrt {-a}+\sqrt {b} x} \, dx}{d^2}\\ &=\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {e p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {e p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^2}+\frac {\left (e^2 p\right ) \int \frac {\log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right )}{d+e x} \, dx}{d^2}+\frac {\left (e^2 p\right ) \int \frac {\log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{-\sqrt {b} d+\sqrt {-a} e}\right )}{d+e x} \, dx}{d^2}\\ &=\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {e p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {e p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^2}+\frac {(e p) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {b} x}{-\sqrt {b} d+\sqrt {-a} e}\right )}{x} \, dx,x,d+e x\right )}{d^2}+\frac {(e p) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {b} x}{\sqrt {b} d+\sqrt {-a} e}\right )}{x} \, dx,x,d+e x\right )}{d^2}\\ &=\frac {2 \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {e p \log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {b} d+\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {e p \log \left (-\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {b} d-\sqrt {-a} e}\right ) \log (d+e x)}{d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{d x}-\frac {e \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )}{2 d^2}+\frac {e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )}{d^2}-\frac {e p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )}{d^2}-\frac {e p \text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )}{d^2}-\frac {e p \text {Li}_2\left (1+\frac {b x^2}{a}\right )}{2 d^2}\\ \end {align*}
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Mathematica [A] time = 0.21, size = 268, normalized size = 0.88 \[ -\frac {-2 e \log (d+e x) \log \left (c \left (a+b x^2\right )^p\right )+\frac {2 d \log \left (c \left (a+b x^2\right )^p\right )}{x}+e \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \text {Li}_2\left (\frac {b x^2}{a}+1\right )\right )+2 e p \left (\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d-\sqrt {-a} e}\right )+\text {Li}_2\left (\frac {\sqrt {b} (d+e x)}{\sqrt {b} d+\sqrt {-a} e}\right )+\log (d+e x) \left (\log \left (\frac {e \left (\sqrt {-a}-\sqrt {b} x\right )}{\sqrt {-a} e+\sqrt {b} d}\right )+\log \left (\frac {e \left (\sqrt {-a}+\sqrt {b} x\right )}{\sqrt {-a} e-\sqrt {b} d}\right )\right )\right )-\frac {4 \sqrt {b} d p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a}}}{2 d^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{e x^{3} + d x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.28, size = 831, normalized size = 2.72 \[ \text {result too large to display} \]
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 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} x^{2}}\,{d x} \]
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 \[ \int \frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{x^2\,\left (d+e\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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