Optimal. Leaf size=103 \[ -\frac {2 (1-a)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(a+1) \sqrt {1-a^2}}+\frac {4 \sqrt {-a-b x+1}}{(a+1) \sqrt {a+b x+1}}+\sin ^{-1}(a+b x) \]
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Rubi [A] time = 0.09, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6163, 98, 157, 53, 619, 216, 93, 208} \[ -\frac {2 (1-a)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(a+1) \sqrt {1-a^2}}+\frac {4 \sqrt {-a-b x+1}}{(a+1) \sqrt {a+b x+1}}+\sin ^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 53
Rule 93
Rule 98
Rule 157
Rule 208
Rule 216
Rule 619
Rule 6163
Rubi steps
\begin {align*} \int \frac {e^{-3 \tanh ^{-1}(a+b x)}}{x} \, dx &=\int \frac {(1-a-b x)^{3/2}}{x (1+a+b x)^{3/2}} \, dx\\ &=\frac {4 \sqrt {1-a-b x}}{(1+a) \sqrt {1+a+b x}}+\frac {2 \int \frac {\frac {1}{2} (1-a)^2 b+\frac {1}{2} (1+a) b^2 x}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{(1+a) b}\\ &=\frac {4 \sqrt {1-a-b x}}{(1+a) \sqrt {1+a+b x}}+\frac {(1-a)^2 \int \frac {1}{x \sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx}{1+a}+b \int \frac {1}{\sqrt {1-a-b x} \sqrt {1+a+b x}} \, dx\\ &=\frac {4 \sqrt {1-a-b x}}{(1+a) \sqrt {1+a+b x}}+\frac {\left (2 (1-a)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1-a-(-1+a) x^2} \, dx,x,\frac {\sqrt {1+a+b x}}{\sqrt {1-a-b x}}\right )}{1+a}+b \int \frac {1}{\sqrt {(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx\\ &=\frac {4 \sqrt {1-a-b x}}{(1+a) \sqrt {1+a+b x}}-\frac {2 (1-a)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1+a) \sqrt {1-a^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{2 b}\\ &=\frac {4 \sqrt {1-a-b x}}{(1+a) \sqrt {1+a+b x}}+\sin ^{-1}(a+b x)-\frac {2 (1-a)^2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {1+a+b x}}{\sqrt {1+a} \sqrt {1-a-b x}}\right )}{(1+a) \sqrt {1-a^2}}\\ \end {align*}
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Mathematica [A] time = 0.44, size = 137, normalized size = 1.33 \[ -\frac {4 (a+b x-1)}{(a+1) \sqrt {-((a+b x-1) (a+b x+1))}}+\frac {2 \sqrt {-b} \sinh ^{-1}\left (\frac {\sqrt {-b} \sqrt {-a-b x+1}}{\sqrt {2} \sqrt {b}}\right )}{\sqrt {b}}-\frac {2 (a-1)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )}{(-a-1)^{3/2}} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.46, size = 438, normalized size = 4.25 \[ \left [\frac {{\left ({\left (a - 1\right )} b x + a^{2} - 1\right )} \sqrt {-\frac {a - 1}{a + 1}} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{3} + {\left (a^{2} + a\right )} b x + a^{2} - a - 1\right )} \sqrt {-\frac {a - 1}{a + 1}} + 2}{x^{2}}\right ) - 2 \, {\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 8 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{2 \, {\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )}}, \frac {{\left ({\left (a - 1\right )} b x + a^{2} - 1\right )} \sqrt {\frac {a - 1}{a + 1}} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {\frac {a - 1}{a + 1}}}{{\left (a - 1\right )} b^{2} x^{2} + a^{3} + 2 \, {\left (a^{2} - a\right )} b x - a^{2} - a + 1}\right ) - {\left ({\left (a + 1\right )} b x + a^{2} + 2 \, a + 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 4 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{{\left (a + 1\right )} b x + a^{2} + 2 \, a + 1}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 154, normalized size = 1.50 \[ -\frac {b \arcsin \left (-b x - a\right ) \mathrm {sgn}\relax (b)}{{\left | b \right |}} + \frac {2 \, {\left (a^{2} b - 2 \, a b + b\right )} \arctan \left (\frac {\frac {{\left (\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} + {\left | b \right |}\right )}} - \frac {8 \, b}{{\left (a {\left | b \right |} + {\left | b \right |}\right )} {\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.05, size = 1062, normalized size = 10.31 \[ \text {result too large to display} \]
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 \[ \int \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}}}{{\left (b x + a + 1\right )}^{3} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-{\left (a+b\,x\right )}^2\right )}^{3/2}}{x\,{\left (a+b\,x+1\right )}^3} \,d x \]
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 \[ \int \frac {\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}{x \left (a + b x + 1\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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