Optimal. Leaf size=156 \[ -\frac{(-a-b x+1)^{3/2} \sqrt{a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right )}{24 b^4}-\frac{\left (8 a^3+12 a^2+12 a+3\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}-\frac{\left (8 a^3+12 a^2+12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac{x^2 (-a-b x+1)^{3/2} \sqrt{a+b x+1}}{4 b^2} \]
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Rubi [A] time = 0.166045, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6163, 100, 147, 50, 53, 619, 216} \[ -\frac{(-a-b x+1)^{3/2} \sqrt{a+b x+1} \left (18 a^2-2 (6 a+1) b x+10 a+7\right )}{24 b^4}-\frac{\left (8 a^3+12 a^2+12 a+3\right ) \sqrt{-a-b x+1} \sqrt{a+b x+1}}{8 b^4}-\frac{\left (8 a^3+12 a^2+12 a+3\right ) \sin ^{-1}(a+b x)}{8 b^4}-\frac{x^2 (-a-b x+1)^{3/2} \sqrt{a+b x+1}}{4 b^2} \]
Antiderivative was successfully verified.
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Rule 6163
Rule 100
Rule 147
Rule 50
Rule 53
Rule 619
Rule 216
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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{4 b^2}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x} \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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{4 b^2}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x} \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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{4 b^2}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x} \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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{4 b^2}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x} \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 (1-a-b x)^{3/2} \sqrt{1+a+b x}}{4 b^2}-\frac{(1-a-b x)^{3/2} \sqrt{1+a+b x} \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*}
Mathematica [A] time = 0.438158, size = 160, normalized size = 1.03 \[ \frac{\frac{\sqrt{a+b x+1} \left (5 a^2 (6 b x-1)+6 a^4+38 a^3+a \left (-18 b^2 x^2+50 b x-23\right )-6 b^4 x^4+14 b^3 x^3-17 b^2 x^2+25 b x-16\right )}{\sqrt{-a-b x+1}}+\frac{6 \left (8 a^3+12 a^2+12 a+3\right ) \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{\sqrt{-b}}}{24 b^4} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.043, size = 809, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.47801, size = 456, normalized size = 2.92 \begin{align*} \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2} x}{2 \, b^{3}} - \frac{a^{3} \arcsin \left (b x + a\right )}{b^{4}} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{3}}{2 \, b^{4}} - \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}} x}{4 \, b^{3}} + \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a x}{2 \, b^{3}} - \frac{3 \, a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac{3 \,{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}} a}{4 \, b^{4}} - \frac{3 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a^{2}}{2 \, b^{4}} + \frac{5 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} x}{8 \, b^{3}} - \frac{3 \, a \arcsin \left (b x + a\right )}{2 \, b^{4}} + \frac{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}{3 \, b^{4}} - \frac{19 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{8 \, b^{4}} - \frac{3 \, \arcsin \left (b x + a\right )}{8 \, b^{4}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.03707, size = 338, normalized size = 2.17 \begin{align*} \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}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \sqrt{- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}{a + b x + 1}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.22074, size = 200, normalized size = 1.28 \begin{align*} \frac{1}{24} \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{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}\left (b\right )}{8 \, b^{3}{\left | b \right |}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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