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 = 0.31, antiderivative size = 410, normalized size of antiderivative = 1.00, 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, 240, 211, 1165, 628, 1162, 617, 204} \[ \frac {(i a+i b x+1)^{3/4} (-i a-i b x+1)^{5/4}}{2 b^2}+\frac {(1+4 i a) (i a+i b x+1)^{3/4} \sqrt [4]{-i a-i b x+1}}{4 b^2}+\frac {(1+4 i a) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt {2} b^2}-\frac {(1+4 i a) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{8 \sqrt {2} b^2}+\frac {(1+4 i a) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt {2} b^2}-\frac {(1+4 i a) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{4 \sqrt {2} b^2} \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 80
Rule 204
Rule 211
Rule 240
Rule 617
Rule 628
Rule 1162
Rule 1165
Rule 5095
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)^{5/4} (1+i a+i b x)^{3/4}}{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) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{2 b^2}+\frac {(i-4 a) \int \frac {1}{(1-i a-i b x)^{3/4} \sqrt [4]{1+i a+i b x}} \, dx}{8 b}\\ &=\frac {(1+4 i a) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{2 b^2}-\frac {(1+4 i a) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2-x^4}} \, dx,x,\sqrt [4]{1-i a-i b x}\right )}{2 b^2}\\ &=\frac {(1+4 i a) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{2 b^2}-\frac {(1+4 i a) \operatorname {Subst}\left (\int \frac {1}{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) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{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) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{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) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{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) \sqrt [4]{1-i a-i b x} (1+i a+i b x)^{3/4}}{4 b^2}+\frac {(1-i a-i b x)^{5/4} (1+i a+i b x)^{3/4}}{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*}
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Mathematica [C] time = 0.06, size = 84, normalized size = 0.20 \[ -\frac {i (-i (a+b x+i))^{5/4} \left (2^{3/4} (4 a-i) \, _2F_1\left (\frac {1}{4},\frac {5}{4};\frac {9}{4};-\frac {1}{2} i (a+b x+i)\right )+5 i (i a+i b x+1)^{3/4}\right )}{10 b^2} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.45, size = 421, normalized size = 1.03 \[ -\frac {b^{2} \sqrt {\frac {16 i \, a^{2} + 8 \, a - i}{b^{4}}} \log \left (\frac {b^{2} \sqrt {\frac {16 i \, a^{2} + 8 \, a - i}{b^{4}}} + {\left (4 \, a - i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a - i}\right ) - b^{2} \sqrt {\frac {16 i \, a^{2} + 8 \, a - i}{b^{4}}} \log \left (-\frac {b^{2} \sqrt {\frac {16 i \, a^{2} + 8 \, a - i}{b^{4}}} - {\left (4 \, a - i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a - i}\right ) - b^{2} \sqrt {\frac {-16 i \, a^{2} - 8 \, a + i}{b^{4}}} \log \left (\frac {b^{2} \sqrt {\frac {-16 i \, a^{2} - 8 \, a + i}{b^{4}}} + {\left (4 \, a - i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a - i}\right ) + b^{2} \sqrt {\frac {-16 i \, a^{2} - 8 \, a + i}{b^{4}}} \log \left (-\frac {b^{2} \sqrt {\frac {-16 i \, a^{2} - 8 \, a + i}{b^{4}}} - {\left (4 \, a - i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{4 \, a - i}\right ) - 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (-2 i \, b x + 2 i \, a + 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, b^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}}\, dx \]
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 \[ \int \frac {x}{\sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x}{\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}}} \,d x \]
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 \[ \int \frac {x}{\sqrt {\frac {i \left (a + b x - i\right )}{\sqrt {a^{2} + 2 a b x + b^{2} x^{2} + 1}}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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