Optimal. Leaf size=410 \[ \frac{(-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b^2}+\frac{(1-4 i a) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{4 b^2}+\frac{(1-4 i a) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt{2} b^2}+\frac{(1-4 i a) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt{2} b^2} \]
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Rubi [A] time = 0.307194, antiderivative size = 410, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 11, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.688, Rules used = {5095, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ \frac{(-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{2 b^2}+\frac{(1-4 i a) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{4 b^2}+\frac{(1-4 i a) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt{2} b^2}+\frac{(1-4 i a) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt{2} b^2} \]
Antiderivative was successfully verified.
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Rule 5095
Rule 80
Rule 50
Rule 63
Rule 331
Rule 297
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int e^{\frac{1}{2} i \tan ^{-1}(a+b x)} x \, dx &=\int \frac{x \sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx\\ &=\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac{(i+4 a) \int \frac{\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx}{4 b}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac{(i+4 a) \int \frac{1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx}{8 b}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{x^2}{\left (2-x^4\right )^{3/4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{2 b^2}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{2 b^2}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 b^2}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\frac{\sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}+\frac{(1-4 i a) \log \left (1+\frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}}-\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \log \left (1+\frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}}+\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}+\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt{2} b^2}-\frac{(1-4 i a) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt{2} b^2}\\ &=\frac{(1-4 i a) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{4 b^2}+\frac{(1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{2 b^2}-\frac{(1-4 i a) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt{2} b^2}+\frac{(1-4 i a) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{4 \sqrt{2} b^2}+\frac{(1-4 i a) \log \left (1+\frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}}-\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}-\frac{(1-4 i a) \log \left (1+\frac{\sqrt{1-i a-i b x}}{\sqrt{1+i a+i b x}}+\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^2}\\ \end{align*}
Mathematica [C] time = 0.0460453, size = 81, normalized size = 0.2 \[ \frac{(-i (a+b x+i))^{3/4} \left (2 \sqrt [4]{2} (1-4 i a) \, _2F_1\left (-\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{1}{2} i (a+b x+i)\right )+3 (i a+i b x+1)^{5/4}\right )}{6 b^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.224, size = 0, normalized size = 0. \begin{align*} \int \sqrt{{(1+i \left ( bx+a \right ) ){\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}}}}x\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{\frac{i \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.33364, size = 1015, normalized size = 2.48 \begin{align*} -\frac{b^{2} \sqrt{\frac{16 i \, a^{2} - 8 \, a - i}{b^{4}}} \log \left (\frac{i \, b^{2} \sqrt{\frac{16 i \, a^{2} - 8 \, a - i}{b^{4}}} +{\left (4 \, a + i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - b^{2} \sqrt{\frac{16 i \, a^{2} - 8 \, a - i}{b^{4}}} \log \left (\frac{-i \, b^{2} \sqrt{\frac{16 i \, a^{2} - 8 \, a - i}{b^{4}}} +{\left (4 \, a + i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) + b^{2} \sqrt{\frac{-16 i \, a^{2} + 8 \, a + i}{b^{4}}} \log \left (\frac{i \, b^{2} \sqrt{\frac{-16 i \, a^{2} + 8 \, a + i}{b^{4}}} +{\left (4 \, a + i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - b^{2} \sqrt{\frac{-16 i \, a^{2} + 8 \, a + i}{b^{4}}} \log \left (\frac{-i \, b^{2} \sqrt{\frac{-16 i \, a^{2} + 8 \, a + i}{b^{4}}} +{\left (4 \, a + i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a + i}\right ) - 2 \,{\left (2 \, b^{2} x^{2} - 2 \, a^{2} - i \, b x - 5 i \, a + 3\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sqrt{\frac{i \, b x + i \, a + 1}{\sqrt{{\left (b x + a\right )}^{2} + 1}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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