Optimal. Leaf size=494 \[ -\frac {\left (-8 i a^2+4 a+3 i\right ) (-i a-i b x+1)^{3/4} \sqrt [4]{i a+i b x+1}}{8 b^3}-\frac {\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}+\frac {\left (-8 i a^2+4 a+3 i\right ) \log \left (\frac {\sqrt {-i a-i b x+1}}{\sqrt {i a+i b x+1}}+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}+1\right )}{16 \sqrt {2} b^3}+\frac {\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {\left (-8 i a^2+4 a+3 i\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-i a-i b x+1}}{\sqrt [4]{i a+i b x+1}}\right )}{8 \sqrt {2} b^3}-\frac {(8 a+i) (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{12 b^3}+\frac {x (-i a-i b x+1)^{3/4} (i a+i b x+1)^{5/4}}{3 b^2} \]
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Rubi [A] time = 0.41, antiderivative size = 494, normalized size of antiderivative = 1.00, 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 50
Rule 63
Rule 80
Rule 90
Rule 204
Rule 297
Rule 331
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^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*}
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Mathematica [C] time = 0.09, 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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fricas [A] time = 0.51, size = 554, normalized size = 1.12 \[ \frac {3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {64 i \, a^{4} - 64 \, a^{3} - 64 i \, a^{2} + 24 \, a + 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) - 3 \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} \log \left (\frac {-i \, b^{3} \sqrt {\frac {-64 i \, a^{4} + 64 \, a^{3} + 64 i \, a^{2} - 24 \, a - 9 i}{b^{6}}} + {\left (8 \, a^{2} + 4 i \, a - 3\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{8 \, a^{2} + 4 i \, a - 3}\right ) + 2 \, {\left (8 \, b^{3} x^{3} - 2 i \, b^{2} x^{2} + 8 \, a^{3} + {\left (8 i \, a - 1\right )} b x + 34 i \, a^{2} - 37 \, a - 11 i\right )} \sqrt {\frac {i \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{b x + a + i}}}{48 \, b^{3}} \]
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 = 0.34, size = 0, normalized size = 0.00 \[ \int \sqrt {\frac {1+i \left (b x +a \right )}{\sqrt {1+\left (b x +a \right )^{2}}}}\, x^{2}\, dx \]
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 \[ \int x^{2} \sqrt {\frac {i \, b x + i \, a + 1}{\sqrt {{\left (b x + a\right )}^{2} + 1}}}\,{d x} \]
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 \[ \int x^2\,\sqrt {\frac {1+a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}{\sqrt {{\left (a+b\,x\right )}^2+1}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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