Optimal. Leaf size=169 \[ -\frac {15 x}{64 c^3 \left (a^2 x^2+1\right )}-\frac {x}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}-\frac {15 \tan ^{-1}(a x)}{64 a c^3} \]
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Rubi [A] time = 0.12, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {4900, 4892, 4930, 199, 205} \[ -\frac {15 x}{64 c^3 \left (a^2 x^2+1\right )}-\frac {x}{32 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (a^2 x^2+1\right )}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (a^2 x^2+1\right )^2}+\frac {3 \tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (a^2 x^2+1\right )^2}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}-\frac {15 \tan ^{-1}(a x)}{64 a c^3} \]
Antiderivative was successfully verified.
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Rule 199
Rule 205
Rule 4892
Rule 4900
Rule 4930
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^3} \, dx &=\frac {\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}-\frac {1}{8} \int \frac {1}{\left (c+a^2 c x^2\right )^3} \, dx+\frac {3 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=-\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{32 c}-\frac {(3 a) \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{4 c}\\ &=-\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {3 x}{64 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{c+a^2 c x^2} \, dx}{64 c^2}-\frac {3 \int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{8 c}\\ &=-\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {3 \tan ^{-1}(a x)}{64 a c^3}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}-\frac {3 \int \frac {1}{c+a^2 c x^2} \, dx}{16 c^2}\\ &=-\frac {x}{32 c^3 \left (1+a^2 x^2\right )^2}-\frac {15 x}{64 c^3 \left (1+a^2 x^2\right )}-\frac {15 \tan ^{-1}(a x)}{64 a c^3}+\frac {\tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )^2}+\frac {3 \tan ^{-1}(a x)}{8 a c^3 \left (1+a^2 x^2\right )}+\frac {x \tan ^{-1}(a x)^2}{4 c^3 \left (1+a^2 x^2\right )^2}+\frac {3 x \tan ^{-1}(a x)^2}{8 c^3 \left (1+a^2 x^2\right )}+\frac {\tan ^{-1}(a x)^3}{8 a c^3}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 98, normalized size = 0.58 \[ \frac {-a x \left (15 a^2 x^2+17\right )+8 \left (a^2 x^2+1\right )^2 \tan ^{-1}(a x)^3+8 a x \left (3 a^2 x^2+5\right ) \tan ^{-1}(a x)^2+\left (-15 a^4 x^4-6 a^2 x^2+17\right ) \tan ^{-1}(a x)}{64 a c^3 \left (a^2 x^2+1\right )^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.69, size = 113, normalized size = 0.67 \[ -\frac {15 \, a^{3} x^{3} - 8 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} - 8 \, {\left (3 \, a^{3} x^{3} + 5 \, a x\right )} \arctan \left (a x\right )^{2} + 17 \, a x + {\left (15 \, a^{4} x^{4} + 6 \, a^{2} x^{2} - 17\right )} \arctan \left (a x\right )}{64 \, {\left (a^{5} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \]
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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 159, normalized size = 0.94 \[ \frac {x \arctan \left (a x \right )^{2}}{4 c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 x \arctan \left (a x \right )^{2}}{8 c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\arctan \left (a x \right )^{3}}{8 a \,c^{3}}+\frac {\arctan \left (a x \right )}{8 a \,c^{3} \left (a^{2} x^{2}+1\right )^{2}}+\frac {3 \arctan \left (a x \right )}{8 a \,c^{3} \left (a^{2} x^{2}+1\right )}-\frac {15 a^{2} x^{3}}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {17 x}{64 c^{3} \left (a^{2} x^{2}+1\right )^{2}}-\frac {15 \arctan \left (a x \right )}{64 a \,c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 232, normalized size = 1.37 \[ \frac {1}{8} \, {\left (\frac {3 \, a^{2} x^{3} + 5 \, x}{a^{4} c^{3} x^{4} + 2 \, a^{2} c^{3} x^{2} + c^{3}} + \frac {3 \, \arctan \left (a x\right )}{a c^{3}}\right )} \arctan \left (a x\right )^{2} - \frac {{\left (15 \, a^{3} x^{3} - 8 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{3} + 17 \, a x + 15 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )\right )} a^{2}}{64 \, {\left (a^{7} c^{3} x^{4} + 2 \, a^{5} c^{3} x^{2} + a^{3} c^{3}\right )}} + \frac {{\left (3 \, a^{2} x^{2} - 3 \, {\left (a^{4} x^{4} + 2 \, a^{2} x^{2} + 1\right )} \arctan \left (a x\right )^{2} + 4\right )} a \arctan \left (a x\right )}{8 \, {\left (a^{6} c^{3} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{2} c^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.53, size = 157, normalized size = 0.93 \[ \frac {\mathrm {atan}\left (a\,x\right )\,\left (\frac {1}{2\,a^3\,c^3}+\frac {3\,x^2}{8\,a\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}-\frac {15\,\mathrm {atan}\left (a\,x\right )}{64\,a\,c^3}-\frac {\frac {15\,a^2\,x^3}{8}+\frac {17\,x}{8}}{8\,a^4\,c^3\,x^4+16\,a^2\,c^3\,x^2+8\,c^3}+\frac {{\mathrm {atan}\left (a\,x\right )}^2\,\left (\frac {3\,x^3}{8\,c^3}+\frac {5\,x}{8\,a^2\,c^3}\right )}{\frac {1}{a^2}+2\,x^2+a^2\,x^4}+\frac {{\mathrm {atan}\left (a\,x\right )}^3}{8\,a\,c^3} \]
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 \[ \frac {\int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{a^{6} x^{6} + 3 a^{4} x^{4} + 3 a^{2} x^{2} + 1}\, dx}{c^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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