Optimal. Leaf size=348 \[ \frac {5 i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 i c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {a^2 c x^2+c}}-\frac {5 c^2 \sqrt {a^2 c x^2+c}}{16 a}+\frac {5}{16} c^2 x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {5 c \left (a^2 c x^2+c\right )^{3/2}}{72 a}-\frac {\left (a^2 c x^2+c\right )^{5/2}}{30 a}+\frac {5}{24} c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x) \]
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Rubi [A] time = 0.19, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4878, 4890, 4886} \[ \frac {5 i c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 i c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {a^2 c x^2+c}}-\frac {5 c^2 \sqrt {a^2 c x^2+c}}{16 a}-\frac {5 i c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {a^2 c x^2+c}}+\frac {5}{16} c^2 x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {5 c \left (a^2 c x^2+c\right )^{3/2}}{72 a}-\frac {\left (a^2 c x^2+c\right )^{5/2}}{30 a}+\frac {5}{24} c x \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x) \]
Antiderivative was successfully verified.
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Rule 4878
Rule 4886
Rule 4890
Rubi steps
\begin {align*} \int \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x) \, dx &=-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)+\frac {1}{6} (5 c) \int \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx\\ &=-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)+\frac {1}{8} \left (5 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=-\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)+\frac {1}{16} \left (5 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)+\frac {\left (5 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{16 \sqrt {c+a^2 c x^2}}\\ &=-\frac {5 c^2 \sqrt {c+a^2 c x^2}}{16 a}-\frac {5 c \left (c+a^2 c x^2\right )^{3/2}}{72 a}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a}+\frac {5}{16} c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {5}{24} c x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{6} x \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {5 i c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a \sqrt {c+a^2 c x^2}}+\frac {5 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}}-\frac {5 i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a \sqrt {c+a^2 c x^2}}\\ \end {align*}
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Mathematica [A] time = 6.50, size = 643, normalized size = 1.85 \[ \frac {c^2 \sqrt {a^2 c x^2+c} \left (\frac {3}{4} \left (a^2 x^2+1\right )^{5/2}+720 \sqrt {a^2 x^2+1} \left (a x \tan ^{-1}(a x)-1\right )+\frac {55}{8} \left (a^2 x^2+1\right )^3 \cos \left (3 \tan ^{-1}(a x)\right )-\frac {45}{8} \left (a^2 x^2+1\right )^3 \cos \left (5 \tan ^{-1}(a x)\right )+\frac {15}{16} \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x) \left (\frac {156 a x}{\sqrt {a^2 x^2+1}}+30 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-30 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-94 \sin \left (3 \tan ^{-1}(a x)\right )+6 \sin \left (5 \tan ^{-1}(a x)\right )+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right ) \cos \left (6 \tan ^{-1}(a x)\right )+45 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+18 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right ) \cos \left (6 \tan ^{-1}(a x)\right )\right )-15 \left (a^2 x^2+1\right )^2 \left (-\frac {2}{\sqrt {a^2 x^2+1}}+3 \tan ^{-1}(a x) \left (-\frac {14 a x}{\sqrt {a^2 x^2+1}}+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+2 \sin \left (3 \tan ^{-1}(a x)\right )+4 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+\left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )\right )-6 \cos \left (3 \tan ^{-1}(a x)\right )\right )+450 i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-450 i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )+720 \tan ^{-1}(a x) \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right )\right )}{1440 a \sqrt {a^2 x^2+1}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right ), x\right ) \]
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 [A] time = 0.61, size = 225, normalized size = 0.65 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (120 \arctan \left (a x \right ) x^{5} a^{5}-24 a^{4} x^{4}+390 \arctan \left (a x \right ) x^{3} a^{3}-98 a^{2} x^{2}+495 \arctan \left (a x \right ) x a -299\right )}{720 a}-\frac {5 c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{16 a \sqrt {a^{2} x^{2}+1}} \]
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 \[ \int {\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} \arctan \left (a x\right )\,{d x} \]
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 \[ \int \mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2} \,d x \]
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 \[ \int \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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