Optimal. Leaf size=166 \[ \frac {\sin ^{-1}(d x) \left (8 a^2 d^4+8 a c d^2+4 b^2 d^2+3 c^2\right )}{8 d^5}-\frac {x \sqrt {1-d^2 x^2} \left (c \left (8 a+\frac {3 c}{d^2}\right )+4 b^2\right )}{8 d^2}-\frac {2 b \sqrt {1-d^2 x^2} \left (3 a d^2+2 c\right )}{3 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2} \]
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Rubi [A] time = 0.32, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {899, 1815, 641, 216} \[ \frac {\sin ^{-1}(d x) \left (8 a^2 d^4+8 a c d^2+4 b^2 d^2+3 c^2\right )}{8 d^5}-\frac {x \sqrt {1-d^2 x^2} \left (c \left (8 a+\frac {3 c}{d^2}\right )+4 b^2\right )}{8 d^2}-\frac {2 b \sqrt {1-d^2 x^2} \left (3 a d^2+2 c\right )}{3 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2} \]
Antiderivative was successfully verified.
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Rule 216
Rule 641
Rule 899
Rule 1815
Rubi steps
\begin {align*} \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx &=\int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d^2 x^2}} \, dx\\ &=-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}-\frac {\int \frac {-4 a^2 d^2-8 a b d^2 x-\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x^2-8 b c d^2 x^3}{\sqrt {1-d^2 x^2}} \, dx}{4 d^2}\\ &=-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\int \frac {12 a^2 d^4+8 b d^2 \left (2 c+3 a d^2\right ) x+3 d^2 \left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x^2}{\sqrt {1-d^2 x^2}} \, dx}{12 d^4}\\ &=-\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}-\frac {\int \frac {-3 d^2 \left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right )-16 b d^4 \left (2 c+3 a d^2\right ) x}{\sqrt {1-d^2 x^2}} \, dx}{24 d^6}\\ &=-\frac {2 b \left (2 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{3 d^4}-\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{8 d^4}\\ &=-\frac {2 b \left (2 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{3 d^4}-\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \sin ^{-1}(d x)}{8 d^5}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 114, normalized size = 0.69 \[ \frac {3 \sin ^{-1}(d x) \left (8 a^2 d^4+8 a c d^2+4 b^2 d^2+3 c^2\right )-d \sqrt {1-d^2 x^2} \left (16 b \left (3 a d^2+c d^2 x^2+2 c\right )+3 c x \left (8 a d^2+2 c d^2 x^2+3 c\right )+12 b^2 d^2 x\right )}{24 d^5} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.68, size = 134, normalized size = 0.81 \[ -\frac {{\left (6 \, c^{2} d^{3} x^{3} + 16 \, b c d^{3} x^{2} + 48 \, a b d^{3} + 32 \, b c d + 3 \, {\left (4 \, {\left (b^{2} + 2 \, a c\right )} d^{3} + 3 \, c^{2} d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (8 \, a^{2} d^{4} + 4 \, {\left (b^{2} + 2 \, a c\right )} d^{2} + 3 \, c^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.42, size = 196, normalized size = 1.18 \[ -\frac {{\left ({\left (d x + 1\right )} {\left (2 \, {\left (d x + 1\right )} {\left (\frac {3 \, {\left (d x + 1\right )} c^{2}}{d^{4}} + \frac {8 \, b c d^{17} - 9 \, c^{2} d^{16}}{d^{20}}\right )} + \frac {12 \, b^{2} d^{18} + 24 \, a c d^{18} - 32 \, b c d^{17} + 27 \, c^{2} d^{16}}{d^{20}}\right )} + \frac {3 \, {\left (16 \, a b d^{19} - 4 \, b^{2} d^{18} - 8 \, a c d^{18} + 16 \, b c d^{17} - 5 \, c^{2} d^{16}\right )}}{d^{20}}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - \frac {6 \, {\left (8 \, a^{2} d^{4} + 4 \, b^{2} d^{2} + 8 \, a c d^{2} + 3 \, c^{2}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{d^{4}}}{24 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.03, size = 291, normalized size = 1.75 \[ -\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (6 \sqrt {-d^{2} x^{2}+1}\, c^{2} d^{3} x^{3} \mathrm {csgn}\relax (d )+16 \sqrt {-d^{2} x^{2}+1}\, b c \,d^{3} x^{2} \mathrm {csgn}\relax (d )-24 a^{2} d^{4} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+24 \sqrt {-d^{2} x^{2}+1}\, a c \,d^{3} x \,\mathrm {csgn}\relax (d )+12 \sqrt {-d^{2} x^{2}+1}\, b^{2} d^{3} x \,\mathrm {csgn}\relax (d )+48 \sqrt {-d^{2} x^{2}+1}\, a b \,d^{3} \mathrm {csgn}\relax (d )-24 a c \,d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )-12 b^{2} d^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )+9 \sqrt {-d^{2} x^{2}+1}\, c^{2} d x \,\mathrm {csgn}\relax (d )+32 \sqrt {-d^{2} x^{2}+1}\, b c d \,\mathrm {csgn}\relax (d )-9 c^{2} \arctan \left (\frac {d x \,\mathrm {csgn}\relax (d )}{\sqrt {-d^{2} x^{2}+1}}\right )\right ) \mathrm {csgn}\relax (d )}{24 \sqrt {-d^{2} x^{2}+1}\, d^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.97, size = 171, normalized size = 1.03 \[ -\frac {\sqrt {-d^{2} x^{2} + 1} c^{2} x^{3}}{4 \, d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} b c x^{2}}{3 \, d^{2}} + \frac {a^{2} \arcsin \left (d x\right )}{d} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} a b}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (b^{2} + 2 \, a c\right )} x}{2 \, d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} c^{2} x}{8 \, d^{4}} + \frac {{\left (b^{2} + 2 \, a c\right )} \arcsin \left (d x\right )}{2 \, d^{3}} - \frac {4 \, \sqrt {-d^{2} x^{2} + 1} b c}{3 \, d^{4}} + \frac {3 \, c^{2} \arcsin \left (d x\right )}{8 \, d^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 13.85, size = 897, normalized size = 5.40 \[ -\frac {\frac {{\left (\sqrt {1-d\,x}-1\right )}^{15}\,\left (2\,b^2\,d^2+\frac {3\,c^2}{2}+4\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{15}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^3\,\left (6\,b^2\,d^2-\frac {23\,c^2}{2}+12\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^{13}\,\left (6\,b^2\,d^2-\frac {23\,c^2}{2}+12\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{13}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^5\,\left (30\,b^2\,d^2+\frac {333\,c^2}{2}+60\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^{11}\,\left (30\,b^2\,d^2+\frac {333\,c^2}{2}+60\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{11}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^7\,\left (22\,b^2\,d^2-\frac {671\,c^2}{2}+44\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^9\,\left (22\,b^2\,d^2-\frac {671\,c^2}{2}+44\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^9}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^4\,\left (96\,a\,b\,d^3+128\,b\,c\,d\right )}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^{12}\,\left (96\,a\,b\,d^3+128\,b\,c\,d\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{12}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^8\,\left (320\,a\,b\,d^3+\frac {256\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^8}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^6\,\left (240\,a\,b\,d^3+\frac {512\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^{10}\,\left (240\,a\,b\,d^3+\frac {512\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{10}}-\frac {\left (\sqrt {1-d\,x}-1\right )\,\left (2\,b^2\,d^2+\frac {3\,c^2}{2}+4\,a\,c\,d^2\right )}{\sqrt {d\,x+1}-1}+\frac {16\,a\,b\,d^3\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {16\,a\,b\,d^3\,{\left (\sqrt {1-d\,x}-1\right )}^{14}}{{\left (\sqrt {d\,x+1}-1\right )}^{14}}}{d^5+\frac {8\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {28\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {56\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {70\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}+\frac {56\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{10}}{{\left (\sqrt {d\,x+1}-1\right )}^{10}}+\frac {28\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{12}}{{\left (\sqrt {d\,x+1}-1\right )}^{12}}+\frac {8\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{14}}{{\left (\sqrt {d\,x+1}-1\right )}^{14}}+\frac {d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{16}}{{\left (\sqrt {d\,x+1}-1\right )}^{16}}}-\frac {\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\,\left (8\,a^2\,d^4+8\,a\,c\,d^2+4\,b^2\,d^2+3\,c^2\right )}{2\,d^5} \]
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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