Optimal. Leaf size=156 \[ -\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (18 a^2+2 (1-6 a) b x-10 a+7\right )}{24 b^4}-\frac {\left (-8 a^3+12 a^2-12 a+3\right ) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{8 b^4}+\frac {\left (-8 a^3+12 a^2-12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac {x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b^2} \]
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Rubi [A] time = 0.16, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6163, 100, 147, 50, 53, 619, 216} \[ -\frac {\sqrt {-a-b x+1} (a+b x+1)^{3/2} \left (18 a^2+2 (1-6 a) b x-10 a+7\right )}{24 b^4}-\frac {\left (-8 a^3+12 a^2-12 a+3\right ) \sqrt {-a-b x+1} \sqrt {a+b x+1}}{8 b^4}+\frac {\left (-8 a^3+12 a^2-12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac {x^2 \sqrt {-a-b x+1} (a+b x+1)^{3/2}}{4 b^2} \]
Antiderivative was successfully verified.
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Rule 50
Rule 53
Rule 100
Rule 147
Rule 216
Rule 619
Rule 6163
Rubi steps
\begin {align*} \int e^{\tanh ^{-1}(a+b x)} x^3 \, dx &=\int \frac {x^3 \sqrt {1+a+b x}}{\sqrt {1-a-b x}} \, dx\\ &=-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\int \frac {x \sqrt {1+a+b x} \left (-2 \left (1-a^2\right )-(1-6 a) b x\right )}{\sqrt {1-a-b x}} \, dx}{4 b^2}\\ &=-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac {\left (3-12 a+12 a^2-8 a^3\right ) \int \frac {\sqrt {1+a+b x}}{\sqrt {1-a-b x}} \, dx}{8 b^3}\\ &=-\frac {\left (3-12 a+12 a^2-8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac {\left (3-12 a+12 a^2-8 a^3\right ) \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{8 b^3}\\ &=-\frac {\left (3-12 a+12 a^2-8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac {\left (3-12 a+12 a^2-8 a^3\right ) \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{8 b^3}\\ &=-\frac {\left (3-12 a+12 a^2-8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}-\frac {\left (3-12 a+12 a^2-8 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{16 b^5}\\ &=-\frac {\left (3-12 a+12 a^2-8 a^3\right ) \sqrt {1-a-b x} \sqrt {1+a+b x}}{8 b^4}-\frac {x^2 \sqrt {1-a-b x} (1+a+b x)^{3/2}}{4 b^2}-\frac {\sqrt {1-a-b x} (1+a+b x)^{3/2} \left (7-10 a+18 a^2+2 (1-6 a) b x\right )}{24 b^4}+\frac {\left (3-12 a+12 a^2-8 a^3\right ) \sin ^{-1}(a+b x)}{8 b^4}\\ \end {align*}
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Mathematica [A] time = 0.39, size = 149, normalized size = 0.96 \[ -\frac {\left (8 a^3-12 a^2+12 a-3\right ) \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {b}}\right )}{4 b^{9/2}}-\frac {\sqrt {-a^2-2 a b x-b^2 x^2+1} \left (-6 a^3+a^2 (6 b x+44)-a \left (6 b^2 x^2+20 b x+39\right )+6 b^3 x^3+8 b^2 x^2+9 b x+16\right )}{24 b^4} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.60, size = 144, normalized size = 0.92 \[ \frac {3 \, {\left (8 \, a^{3} - 12 \, a^{2} + 12 \, a - 3\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - {\left (6 \, b^{3} x^{3} - 2 \, {\left (3 \, a - 4\right )} b^{2} x^{2} - 6 \, a^{3} + {\left (6 \, a^{2} - 20 \, a + 9\right )} b x + 44 \, a^{2} - 39 \, a + 16\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{24 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.23, size = 139, normalized size = 0.89 \[ -\frac {1}{24} \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left ({\left (2 \, x {\left (\frac {3 \, x}{b} - \frac {3 \, a b^{5} - 4 \, b^{5}}{b^{7}}\right )} + \frac {6 \, a^{2} b^{4} - 20 \, a b^{4} + 9 \, b^{4}}{b^{7}}\right )} x - \frac {6 \, a^{3} b^{3} - 44 \, a^{2} b^{3} + 39 \, a b^{3} - 16 \, b^{3}}{b^{7}}\right )} + \frac {{\left (8 \, a^{3} - 12 \, a^{2} + 12 \, a - 3\right )} \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{8 \, b^{3} {\left | b \right |}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.04, size = 487, normalized size = 3.12 \[ -\frac {x^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b}+\frac {a \,x^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{2}}-\frac {a^{2} x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{3}}+\frac {a^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{4 b^{4}}+\frac {3 a^{2} \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{3} \sqrt {b^{2}}}+\frac {13 a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{8 b^{4}}-\frac {3 x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{8 b^{3}}+\frac {3 \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{8 b^{3} \sqrt {b^{2}}}-\frac {x^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 b^{2}}+\frac {5 a x \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 b^{3}}-\frac {11 a^{2} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{6 b^{4}}-\frac {a^{3} \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{b^{3} \sqrt {b^{2}}}-\frac {3 a \arctan \left (\frac {\sqrt {b^{2}}\, \left (x +\frac {a}{b}\right )}{\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}\right )}{2 b^{3} \sqrt {b^{2}}}-\frac {2 \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{3 b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.43, size = 540, normalized size = 3.46 \[ -\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x^{3}}{4 \, b} - \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a + 1\right )} x^{2}}{3 \, b^{2}} + \frac {7 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x^{2}}{12 \, b^{2}} + \frac {5 \, {\left (a + 1\right )} a^{3} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{4}} - \frac {35 \, a^{4} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{8 \, b^{4}} + \frac {5 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a + 1\right )} a x}{6 \, b^{3}} - \frac {35 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2} x}{24 \, b^{3}} - \frac {3 \, {\left (a^{2} - 1\right )} {\left (a + 1\right )} a \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{2 \, b^{4}} + \frac {15 \, {\left (a^{2} - 1\right )} a^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{4 \, b^{4}} - \frac {5 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a + 1\right )} a^{2}}{2 \, b^{4}} + \frac {35 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{8 \, b^{4}} + \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} x}{8 \, b^{3}} - \frac {3 \, {\left (a^{2} - 1\right )}^{2} \arcsin \left (-\frac {b^{2} x + a b}{\sqrt {a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}}}\right )}{8 \, b^{4}} + \frac {2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} {\left (a + 1\right )}}{3 \, b^{4}} - \frac {55 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )} a}{24 \, b^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,x+1\right )}{\sqrt {1-{\left (a+b\,x\right )}^2}} \,d x \]
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 \[ \int \frac {x^{3} \left (a + b x + 1\right )}{\sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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