Optimal. Leaf size=192 \[ -\frac {21 c^2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}-\frac {21 c^2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}+\frac {7 c \sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \]
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Rubi [A] time = 0.14, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {687, 694, 329, 212, 206, 203} \begin {gather*} -\frac {21 c^2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}-\frac {21 c^2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt {d} \sqrt [4]{b^2-4 a c}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}+\frac {7 c \sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {\sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 212
Rule 329
Rule 687
Rule 694
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^3} \, dx &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}-\frac {(7 c) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {\left (21 c^2\right ) \int \frac {1}{\sqrt {b d+2 c d x} \left (a+b x+c x^2\right )} \, dx}{2 \left (b^2-4 a c\right )^2}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(21 c) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (a-\frac {b^2}{4 c}+\frac {x^2}{4 c d^2}\right )} \, dx,x,b d+2 c d x\right )}{4 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}+\frac {(21 c) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b^2}{4 c}+\frac {x^4}{4 c d^2}} \, dx,x,\sqrt {d (b+2 c x)}\right )}{2 \left (b^2-4 a c\right )^2 d}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {\left (21 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d-x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{5/2}}-\frac {\left (21 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c} d+x^2} \, dx,x,\sqrt {d (b+2 c x)}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ &=-\frac {\sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right ) d \left (a+b x+c x^2\right )^2}+\frac {7 c \sqrt {b d+2 c d x}}{2 \left (b^2-4 a c\right )^2 d \left (a+b x+c x^2\right )}-\frac {21 c^2 \tan ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{11/4} \sqrt {d}}-\frac {21 c^2 \tanh ^{-1}\left (\frac {\sqrt {d (b+2 c x)}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )}{\left (b^2-4 a c\right )^{11/4} \sqrt {d}}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 179, normalized size = 0.93 \begin {gather*} \frac {c^2 \left (\frac {\left (b^2-4 a c\right ) (b+2 c x) \left (-c \left (11 a+7 c x^2\right )+b^2-7 b c x\right )}{2 c^2 (a+x (b+c x))^2}+21 \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x} \tan ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )+21 \sqrt [4]{b^2-4 a c} \sqrt {b+2 c x} \tanh ^{-1}\left (\frac {\sqrt {b+2 c x}}{\sqrt [4]{b^2-4 a c}}\right )\right )}{\left (4 a c-b^2\right )^3 \sqrt {d (b+2 c x)}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [C] time = 1.04, size = 273, normalized size = 1.42 \begin {gather*} \frac {\left (11 a c-b^2+7 b c x+7 c^2 x^2\right ) \sqrt {b d+2 c d x}}{2 d \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}+\frac {\left (\frac {21}{2}-\frac {21 i}{2}\right ) c^2 \tan ^{-1}\left (\frac {-\frac {(1+i) c \sqrt {d} x}{\sqrt [4]{b^2-4 a c}}-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {d}}{\sqrt [4]{b^2-4 a c}}+\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {d} \sqrt [4]{b^2-4 a c}}{\sqrt {b d+2 c d x}}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}}-\frac {\left (\frac {21}{2}-\frac {21 i}{2}\right ) c^2 \tanh ^{-1}\left (\frac {(1+i) \sqrt [4]{b^2-4 a c} \sqrt {b d+2 c d x}}{\sqrt {d} \left (\sqrt {b^2-4 a c}+i b+2 i c x\right )}\right )}{\sqrt {d} \left (b^2-4 a c\right )^{11/4}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.48, size = 2021, normalized size = 10.53
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.27, size = 645, normalized size = 3.36 \begin {gather*} -\frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} + 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} - 2 \, \sqrt {2 \, c d x + b d}\right )}}{2 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}}}\right )}{\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d} - \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \log \left (2 \, c d x + b d + \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} + \frac {21 \, {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} c^{2} \log \left (2 \, c d x + b d - \sqrt {2} {\left (-b^{2} d^{2} + 4 \, a c d^{2}\right )}^{\frac {1}{4}} \sqrt {2 \, c d x + b d} + \sqrt {-b^{2} d^{2} + 4 \, a c d^{2}}\right )}{2 \, {\left (\sqrt {2} b^{6} d - 12 \, \sqrt {2} a b^{4} c d + 48 \, \sqrt {2} a^{2} b^{2} c^{2} d - 64 \, \sqrt {2} a^{3} c^{3} d\right )}} - \frac {2 \, {\left (11 \, \sqrt {2 \, c d x + b d} b^{2} c^{2} d^{3} - 44 \, \sqrt {2 \, c d x + b d} a c^{3} d^{3} - 7 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d\right )}}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (b^{2} d^{2} - 4 \, a c d^{2} - {\left (2 \, c d x + b d\right )}^{2}\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.06, size = 419, normalized size = 2.18 \begin {gather*} -\frac {21 \sqrt {2}\, c^{2} d^{5} \arctan \left (-\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {21 \sqrt {2}\, c^{2} d^{5} \arctan \left (\frac {\sqrt {2}\, \sqrt {2 c d x +b d}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}}}+1\right )}{2 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {21 \sqrt {2}\, c^{2} d^{5} \ln \left (\frac {2 c d x +b d +\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}{2 c d x +b d -\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {1}{4}} \sqrt {2 c d x +b d}\, \sqrt {2}+\sqrt {4 a c \,d^{2}-b^{2} d^{2}}}\right )}{4 \left (4 a c \,d^{2}-b^{2} d^{2}\right )^{\frac {11}{4}}}+\frac {8 \sqrt {2 c d x +b d}\, c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right ) \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )^{2}}+\frac {14 \sqrt {2 c d x +b d}\, c^{2} d^{5}}{\left (4 a c \,d^{2}-b^{2} d^{2}\right )^{2} \left (4 c^{2} d^{2} x^{2}+4 b c \,d^{2} x +4 a c \,d^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.18, size = 317, normalized size = 1.65 \begin {gather*} \frac {\frac {14\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {22\,c^2\,d^3\,\sqrt {b\,d+2\,c\,d\,x}}{4\,a\,c-b^2}}{{\left (b\,d+2\,c\,d\,x\right )}^4-{\left (b\,d+2\,c\,d\,x\right )}^2\,\left (2\,b^2\,d^2-8\,a\,c\,d^2\right )+b^4\,d^4+16\,a^2\,c^2\,d^4-8\,a\,b^2\,c\,d^4}-\frac {21\,c^2\,\mathrm {atan}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{15/4}}{\sqrt {d}\,\left (256\,a^4\,c^4-256\,a^3\,b^2\,c^3+96\,a^2\,b^4\,c^2-16\,a\,b^6\,c+b^8\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{11/4}}-\frac {21\,c^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,d+2\,c\,d\,x}\,{\left (b^2-4\,a\,c\right )}^{15/4}}{\sqrt {d}\,\left (256\,a^4\,c^4-256\,a^3\,b^2\,c^3+96\,a^2\,b^4\,c^2-16\,a\,b^6\,c+b^8\right )}\right )}{\sqrt {d}\,{\left (b^2-4\,a\,c\right )}^{11/4}} \end {gather*}
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 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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