Integrand size = 16, antiderivative size = 165 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=-\frac {\sqrt {3} \sqrt [3]{e} q \arctan \left (\frac {\sqrt [3]{e}-2 \sqrt [3]{d} (f+g x)}{\sqrt {3} \sqrt [3]{e}}\right )}{\sqrt [3]{d} g}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{e}+\sqrt [3]{d} (f+g x)\right )}{\sqrt [3]{d} g}-\frac {\sqrt [3]{e} q \log \left (e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+d^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{d} g} \]
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Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {2533, 2498, 269, 206, 31, 648, 631, 210, 642} \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=-\frac {\sqrt {3} \sqrt [3]{e} q \arctan \left (\frac {\sqrt [3]{e}-2 \sqrt [3]{d} (f+g x)}{\sqrt {3} \sqrt [3]{e}}\right )}{\sqrt [3]{d} g}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}-\frac {\sqrt [3]{e} q \log \left (d^{2/3} (f+g x)^2-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+e^{2/3}\right )}{2 \sqrt [3]{d} g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{d} (f+g x)+\sqrt [3]{e}\right )}{\sqrt [3]{d} g} \]
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Rule 31
Rule 206
Rule 210
Rule 269
Rule 631
Rule 642
Rule 648
Rule 2498
Rule 2533
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \log \left (c \left (d+\frac {e}{x^3}\right )^q\right ) \, dx,x,f+g x\right )}{g} \\ & = \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {(3 e q) \text {Subst}\left (\int \frac {1}{\left (d+\frac {e}{x^3}\right ) x^3} \, dx,x,f+g x\right )}{g} \\ & = \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {(3 e q) \text {Subst}\left (\int \frac {1}{e+d x^3} \, dx,x,f+g x\right )}{g} \\ & = \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {\left (\sqrt [3]{e} q\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{e}+\sqrt [3]{d} x} \, dx,x,f+g x\right )}{g}+\frac {\left (\sqrt [3]{e} q\right ) \text {Subst}\left (\int \frac {2 \sqrt [3]{e}-\sqrt [3]{d} x}{e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3} x^2} \, dx,x,f+g x\right )}{g} \\ & = \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{e}+\sqrt [3]{d} (f+g x)\right )}{\sqrt [3]{d} g}-\frac {\left (\sqrt [3]{e} q\right ) \text {Subst}\left (\int \frac {-\sqrt [3]{d} \sqrt [3]{e}+2 d^{2/3} x}{e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3} x^2} \, dx,x,f+g x\right )}{2 \sqrt [3]{d} g}+\frac {\left (3 e^{2/3} q\right ) \text {Subst}\left (\int \frac {1}{e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+d^{2/3} x^2} \, dx,x,f+g x\right )}{2 g} \\ & = \frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{e}+\sqrt [3]{d} (f+g x)\right )}{\sqrt [3]{d} g}-\frac {\sqrt [3]{e} q \log \left (e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+d^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{d} g}+\frac {\left (3 \sqrt [3]{e} q\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{d} (f+g x)}{\sqrt [3]{e}}\right )}{\sqrt [3]{d} g} \\ & = -\frac {\sqrt {3} \sqrt [3]{e} q \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{d} (f+g x)}{\sqrt [3]{e}}}{\sqrt {3}}\right )}{\sqrt [3]{d} g}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g}+\frac {\sqrt [3]{e} q \log \left (\sqrt [3]{e}+\sqrt [3]{d} (f+g x)\right )}{\sqrt [3]{d} g}-\frac {\sqrt [3]{e} q \log \left (e^{2/3}-\sqrt [3]{d} \sqrt [3]{e} (f+g x)+d^{2/3} (f+g x)^2\right )}{2 \sqrt [3]{d} g} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.40 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=-\frac {3 e q \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {e}{d (f+g x)^3}\right )}{2 d g (f+g x)^2}+\frac {(f+g x) \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right )}{g} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.96 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.84
method | result | size |
parts | \(\ln \left (c \left (d +\frac {e}{\left (g x +f \right )^{3}}\right )^{q}\right ) x +3 q e g \left (\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,g^{3} \textit {\_Z}^{3}+3 d \,g^{2} f \,\textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )}{\sum }\frac {\left (\textit {\_R}^{2} d f \,g^{2}+2 \textit {\_R} d \,f^{2} g +d \,f^{3}+e \right ) \ln \left (x -\textit {\_R} \right )}{g^{2} \textit {\_R}^{2}+2 f g \textit {\_R} +f^{2}}}{3 d \,g^{2} e}-\frac {f \ln \left (g x +f \right )}{g^{2} e}\right )\) | \(138\) |
default | \(\ln \left (c \left (\frac {d \,g^{3} x^{3}+3 d f \,g^{2} x^{2}+3 d \,f^{2} g x +d \,f^{3}+e}{\left (g x +f \right )^{3}}\right )^{q}\right ) x +3 q e g \left (\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (d \,g^{3} \textit {\_Z}^{3}+3 d \,g^{2} f \,\textit {\_Z}^{2}+3 d \,f^{2} g \textit {\_Z} +d \,f^{3}+e \right )}{\sum }\frac {\left (\textit {\_R}^{2} d f \,g^{2}+2 \textit {\_R} d \,f^{2} g +d \,f^{3}+e \right ) \ln \left (x -\textit {\_R} \right )}{g^{2} \textit {\_R}^{2}+2 f g \textit {\_R} +f^{2}}}{3 d \,g^{2} e}-\frac {f \ln \left (g x +f \right )}{g^{2} e}\right )\) | \(168\) |
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Result contains complex when optimal does not.
Time = 1.06 (sec) , antiderivative size = 1169, normalized size of antiderivative = 7.08 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=\text {Too large to display} \]
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Timed out. \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=\text {Timed out} \]
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\[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=\int { \log \left (c {\left (d + \frac {e}{{\left (g x + f\right )}^{3}}\right )}^{q}\right ) \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (132) = 264\).
Time = 0.74 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.93 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=\frac {1}{2} \, d e g^{5} q {\left (\frac {2 \, f \log \left ({\left | d g^{3} x^{3} + 3 \, d f g^{2} x^{2} + 3 \, d f^{2} g x + d f^{3} + e \right |}\right )}{d e g^{6}} - \frac {6 \, f \log \left ({\left | g x + f \right |}\right )}{d e g^{6}} + \frac {2 \, \sqrt {3} \left (d^{5} e^{4} g^{21}\right )^{\frac {1}{3}} \arctan \left (-\frac {d g x + d f + \left (d^{2} e\right )^{\frac {1}{3}}}{\sqrt {3} d g x + \sqrt {3} d f - \sqrt {3} \left (d^{2} e\right )^{\frac {1}{3}}}\right ) - \left (d^{5} e^{4} g^{21}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} d g x + \sqrt {3} d f - \sqrt {3} \left (d^{2} e\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (d g x + d f + \left (d^{2} e\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (d^{5} e^{4} g^{21}\right )^{\frac {1}{3}} \log \left ({\left | d g x + d f + \left (d^{2} e\right )^{\frac {1}{3}} \right |}\right )}{d^{3} e^{2} g^{13}}\right )} + q x \log \left (d g^{3} x^{3} + 3 \, d f g^{2} x^{2} + 3 \, d f^{2} g x + d f^{3} + e\right ) - q x \log \left (g^{3} x^{3} + 3 \, f g^{2} x^{2} + 3 \, f^{2} g x + f^{3}\right ) + x \log \left (c\right ) \]
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Time = 1.69 (sec) , antiderivative size = 499, normalized size of antiderivative = 3.02 \[ \int \log \left (c \left (d+\frac {e}{(f+g x)^3}\right )^q\right ) \, dx=x\,\ln \left (c\,{\left (d+\frac {e}{{\left (f+g\,x\right )}^3}\right )}^q\right )-\left (\sum _{k=1}^3\ln \left (-d^2\,e^2\,g^{11}\,\left (3\,e\,q^3\,x+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,e\,q^2+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^3\,d\,f\,g^2\,4+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,d\,f^3\,q^2\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^3\,d\,g^3\,x\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^2\,d\,f^2\,g\,q\,8+\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\,d\,f^2\,g\,q^2\,x\,4+{\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )}^2\,d\,f\,g^2\,q\,x\,8\right )\,9\right )\,\mathrm {root}\left (d\,g^3\,z^3+3\,d\,f\,g^2\,q\,z^2+3\,d\,f^2\,g\,q^2\,z+d\,f^3\,q^3+e\,q^3,z,k\right )\right )-\frac {3\,f\,q\,\ln \left (f+g\,x\right )}{g} \]
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