Optimal. Leaf size=106 \[ -\frac {a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}+\frac {\left (1-6 a^2\right ) x}{4 b^3}-\frac {\left (a^4-6 a^2+1\right ) \tan ^{-1}(a+b x)}{4 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {a (a+b x)^2}{2 b^4}+\frac {1}{4} x^4 \tan ^{-1}(a+b x) \]
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Rubi [A] time = 0.11, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5047, 4862, 702, 635, 203, 260} \[ \frac {\left (1-6 a^2\right ) x}{4 b^3}-\frac {a \left (1-a^2\right ) \log \left ((a+b x)^2+1\right )}{2 b^4}-\frac {\left (a^4-6 a^2+1\right ) \tan ^{-1}(a+b x)}{4 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {a (a+b x)^2}{2 b^4}+\frac {1}{4} x^4 \tan ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 203
Rule 260
Rule 635
Rule 702
Rule 4862
Rule 5047
Rubi steps
\begin {align*} \int x^3 \tan ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \tan ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \tan ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{1+x^2} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \tan ^{-1}(a+b x)-\frac {1}{4} \operatorname {Subst}\left (\int \left (-\frac {1-6 a^2}{b^4}-\frac {4 a x}{b^4}+\frac {x^2}{b^4}+\frac {1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{b^4 \left (1+x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \tan ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int \frac {1-6 a^2+a^4+4 a \left (1-a^2\right ) x}{1+x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \tan ^{-1}(a+b x)-\frac {\left (a \left (1-a^2\right )\right ) \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,a+b x\right )}{b^4}-\frac {\left (1-6 a^2+a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {\left (1-6 a^2\right ) x}{4 b^3}+\frac {a (a+b x)^2}{2 b^4}-\frac {(a+b x)^3}{12 b^4}-\frac {\left (1-6 a^2+a^4\right ) \tan ^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \tan ^{-1}(a+b x)-\frac {a \left (1-a^2\right ) \log \left (1+(a+b x)^2\right )}{2 b^4}\\ \end {align*}
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Mathematica [C] time = 0.08, size = 95, normalized size = 0.90 \[ \frac {6 \left (1-6 a^2\right ) b x+6 b^4 x^4 \tan ^{-1}(a+b x)-2 (a+b x)^3+12 a (a+b x)^2+3 i (a-i)^4 \log (-a-b x+i)-3 i (a+i)^4 \log (a+b x+i)}{24 b^4} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.41, size = 87, normalized size = 0.82 \[ -\frac {b^{3} x^{3} - 3 \, a b^{2} x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} b x - 3 \, {\left (b^{4} x^{4} - a^{4} + 6 \, a^{2} - 1\right )} \arctan \left (b x + a\right ) - 6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{12 \, b^{4}} \]
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 = 0.04, size = 132, normalized size = 1.25 \[ \frac {x^{4} \arctan \left (b x +a \right )}{4}-\frac {\arctan \left (b x +a \right ) a^{4}}{4 b^{4}}-\frac {x^{3}}{12 b}+\frac {x^{2} a}{4 b^{2}}-\frac {3 x \,a^{2}}{4 b^{3}}-\frac {13 a^{3}}{12 b^{4}}+\frac {x}{4 b^{3}}+\frac {a}{4 b^{4}}+\frac {\ln \left (1+\left (b x +a \right )^{2}\right ) a^{3}}{2 b^{4}}-\frac {\ln \left (1+\left (b x +a \right )^{2}\right ) a}{2 b^{4}}+\frac {3 \arctan \left (b x +a \right ) a^{2}}{2 b^{4}}-\frac {\arctan \left (b x +a \right )}{4 b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 104, normalized size = 0.98 \[ \frac {1}{4} \, x^{4} \arctan \left (b x + a\right ) - \frac {1}{12} \, b {\left (\frac {b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} - 1\right )} x}{b^{4}} + \frac {3 \, {\left (a^{4} - 6 \, a^{2} + 1\right )} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{b^{5}} - \frac {6 \, {\left (a^{3} - a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{b^{5}}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.59, size = 133, normalized size = 1.25 \[ \frac {x^4\,\mathrm {atan}\left (a+b\,x\right )}{4}-\frac {\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {x}{4\,b^3}-\frac {x^3}{12\,b}+\frac {a^3\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4}+\frac {3\,a^2\,\mathrm {atan}\left (a+b\,x\right )}{2\,b^4}-\frac {a^4\,\mathrm {atan}\left (a+b\,x\right )}{4\,b^4}+\frac {a\,x^2}{4\,b^2}-\frac {3\,a^2\,x}{4\,b^3}-\frac {a\,\ln \left (a^2+2\,a\,b\,x+b^2\,x^2+1\right )}{2\,b^4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.58, size = 155, normalized size = 1.46 \[ \begin {cases} - \frac {a^{4} \operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} + \frac {a^{3} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} - \frac {3 a^{2} x}{4 b^{3}} + \frac {3 a^{2} \operatorname {atan}{\left (a + b x \right )}}{2 b^{4}} + \frac {a x^{2}}{4 b^{2}} - \frac {a \log {\left (a^{2} + 2 a b x + b^{2} x^{2} + 1 \right )}}{2 b^{4}} + \frac {x^{4} \operatorname {atan}{\left (a + b x \right )}}{4} - \frac {x^{3}}{12 b} + \frac {x}{4 b^{3}} - \frac {\operatorname {atan}{\left (a + b x \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atan}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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