Optimal. Leaf size=496 \[ -\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {2 i b^2 m \text {Li}_4\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 i b^2 m \text {Li}_4\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c} \]
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Rubi [A] time = 0.79, antiderivative size = 496, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4642, 4780, 4742, 4520, 2190, 2531, 6609, 2282, 6589} \[ -\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,-\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {PolyLog}\left (2,-\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {PolyLog}\left (3,-\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {PolyLog}\left (3,-\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}+\frac {2 i b^2 m \text {PolyLog}\left (4,-\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 i b^2 m \text {PolyLog}\left (4,-\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {g e^{i \cos ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2282
Rule 2531
Rule 4520
Rule 4642
Rule 4742
Rule 4780
Rule 6589
Rule 6609
Rubi steps
\begin {align*} \int \frac {\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {(g m) \int \frac {\left (a+b \cos ^{-1}(c x)\right )^3}{f+g x} \, dx}{3 b c}\\ &=-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {(g m) \operatorname {Subst}\left (\int \frac {(a+b x)^3 \sin (x)}{c f+g \cos (x)} \, dx,x,\cos ^{-1}(c x)\right )}{3 b c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}+\frac {(i g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^3}{c f+e^{i x} g-\sqrt {c^2 f^2-g^2}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b c}+\frac {(i g m) \operatorname {Subst}\left (\int \frac {e^{i x} (a+b x)^3}{c f+e^{i x} g+\sqrt {c^2 f^2-g^2}} \, dx,x,\cos ^{-1}(c x)\right )}{3 b c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+\frac {e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}-\frac {m \operatorname {Subst}\left (\int (a+b x)^2 \log \left (1+\frac {e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {(2 i b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-\frac {e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}+\frac {(2 i b m) \operatorname {Subst}\left (\int (a+b x) \text {Li}_2\left (-\frac {e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {e^{i x} g}{c f-\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}-\frac {\left (2 b^2 m\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {e^{i x} g}{c f+\sqrt {c^2 f^2-g^2}}\right ) \, dx,x,\cos ^{-1}(c x)\right )}{c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {\left (2 i b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {g x}{-c f+\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c}+\frac {\left (2 i b^2 m\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {g x}{c f+\sqrt {c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \cos ^{-1}(c x)}\right )}{c}\\ &=-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^4}{12 b^2 c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{3 b c}+\frac {m \left (a+b \cos ^{-1}(c x)\right )^3 \log \left (1+\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{3 b c}-\frac {\left (a+b \cos ^{-1}(c x)\right )^3 \log \left (h (f+g x)^m\right )}{3 b c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}-\frac {i m \left (a+b \cos ^{-1}(c x)\right )^2 \text {Li}_2\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 b m \left (a+b \cos ^{-1}(c x)\right ) \text {Li}_3\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 i b^2 m \text {Li}_4\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{c}+\frac {2 i b^2 m \text {Li}_4\left (-\frac {e^{i \cos ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{c}\\ \end {align*}
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Mathematica [F] time = 27.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \cos ^{-1}(c x)\right )^2 \log \left (h (f+g x)^m\right )}{\sqrt {1-c^2 x^2}} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} x^{2} + 1} {\left (b^{2} \arccos \left (c x\right )^{2} + 2 \, a b \arccos \left (c x\right ) + a^{2}\right )} \log \left ({\left (g x + f\right )}^{m} h\right )}{c^{2} x^{2} - 1}, x\right ) \]
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 \[ \int \frac {{\left (b \arccos \left (c x\right ) + a\right )}^{2} \log \left ({\left (g x + f\right )}^{m} h\right )}{\sqrt {-c^{2} x^{2} + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 9.97, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arccos \left (c x \right )\right )^{2} \ln \left (h \left (g x +f \right )^{m}\right )}{\sqrt {-c^{2} x^{2}+1}}\, dx \]
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 \[ \frac {-\frac {1}{3} \, b^{2} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )^{3} \log \relax (h) - a b \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )^{2} \log \relax (h) + b^{2} c \int \frac {\arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right )^{2} \log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + 2 \, a b c \int \frac {\arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right ) \log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + a^{2} c \int \frac {\log \left ({\left (g x + f\right )}^{m}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1}}\,{d x} + a^{2} \arctan \left (c x, \sqrt {-c^{2} x^{2} + 1}\right ) \log \relax (h)}{c} \]
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 \[ \int \frac {\ln \left (h\,{\left (f+g\,x\right )}^m\right )\,{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2}{\sqrt {1-c^2\,x^2}} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2} \log {\left (h \left (f + g x\right )^{m} \right )}}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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