Optimal. Leaf size=494 \[ -\frac{\left (-8 i a^2+4 a+3 i\right ) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{8 b^3}-\frac{\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt{2} b^3}+\frac{\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt{2} b^3}+\frac{\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt{2} b^3}-\frac{\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt{2} b^3}+\frac{x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac{(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{12 b^3} \]
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Rubi [A] time = 0.405493, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5095, 90, 80, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\left (-8 i a^2+4 a+3 i\right ) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{8 b^3}-\frac{\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt{2} b^3}+\frac{\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac{\sqrt{-i a-i b x+1}}{\sqrt{i a+i b x+1}}+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt{2} b^3}+\frac{\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt{2} b^3}-\frac{\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt{2} b^3}+\frac{x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2}-\frac{(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{12 b^3} \]
Antiderivative was successfully verified.
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Rule 5095
Rule 90
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^2 \, dx &=\int \frac{x^2 \sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx\\ &=\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac{\int \frac{\sqrt [4]{1+i a+i b x} \left (-1-a^2-\frac{1}{2} (i+8 a) b x\right )}{\sqrt [4]{1-i a-i b x}} \, dx}{3 b^2}\\ &=-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3-4 i a-8 a^2\right ) \int \frac{\sqrt [4]{1+i a+i b x}}{\sqrt [4]{1-i a-i b x}} \, dx}{8 b^2}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3-4 i a-8 a^2\right ) \int \frac{1}{\sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}} \, dx}{16 b^2}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{4 b^3}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{4 b^3}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{8 b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{8 b^3}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}+\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{8 \sqrt{2} b^3}+\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{8 \sqrt{2} b^3}\\ &=-\frac{\left (3 i+4 a-8 i a^2\right ) (1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}}{8 b^3}-\frac{(i+8 a) (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{12 b^3}+\frac{x (1-i a-i b x)^{3/4} (1+i a+i b x)^{5/4}}{3 b^2}+\frac{\left (3 i+4 a-8 i a^2\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{1-i a-i b x}}{\sqrt [4]{1+i a+i b x}}\right )}{8 \sqrt{2} b^3}-\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}+\frac{\left (3 i+4 a-8 i a^2\right ) \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 )}{16 \sqrt{2} b^3}\\ \end{align*}
Mathematica [C] time = 0.0852281, size = 121, normalized size = 0.24 \[ \frac{(-i (a+b x+i))^{3/4} \left (2 i \sqrt [4]{2} \left (8 a^2+4 i a-3\right ) \, _2F_1\left (-\frac{1}{4},\frac{3}{4};\frac{7}{4};-\frac{1}{2} i (a+b x+i)\right )-i \sqrt [4]{i a+i b x+1} \left (8 a^2+a (4 b x-7 i)-4 b^2 x^2+5 i b x+1\right )\right )}{12 b^3} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.312, 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}^{2}\, 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^{2} \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 = 2.46128, size = 1432, normalized size = 2.9 \begin{align*} \frac{3 \, b^{3} \sqrt{\frac{64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac{i \, b^{3} \sqrt{\frac{64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} +{\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt{\frac{64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac{-i \, b^{3} \sqrt{\frac{64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} +{\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 3 \, b^{3} \sqrt{\frac{-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac{i \, b^{3} \sqrt{\frac{-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} +{\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt{\frac{-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac{-i \, b^{3} \sqrt{\frac{-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} +{\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 2 \,{\left (8 \, b^{3} x^{3} - 2 i \, b^{2} x^{2} + 8 \, a^{3} +{\left (8 i \, a - 1\right )} b x + 34 i \, a^{2} - 37 \, a - 11 i\right )} \sqrt{\frac{i \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{48 \, b^{3}} \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^{2} \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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