Optimal. Leaf size=278 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\sqrt {b} \sqrt [4]{b^2 c+a^2 d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\sqrt {b} \sqrt [4]{b^2 c+a^2 d}}-\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x} \]
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Rubi [A]
time = 0.20, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {760, 408, 504,
1232, 455, 65, 304, 211, 214} \begin {gather*} -\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {d a^2+b^2 c}};\left .\text {ArcSin}\left (\frac {\sqrt [4]{d x^2+c}}{\sqrt [4]{c}}\right )\right |-1\right )}{b x \sqrt {a^2 d+b^2 c}}+\frac {\text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{a^2 d+b^2 c}}\right )}{\sqrt {b} \sqrt [4]{a^2 d+b^2 c}} \end {gather*}
Antiderivative was successfully verified.
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Rule 65
Rule 211
Rule 214
Rule 304
Rule 408
Rule 455
Rule 504
Rule 760
Rule 1232
Rubi steps
\begin {align*} \int \frac {1}{(a+b x) \sqrt [4]{c+d x^2}} \, dx &=a \int \frac {1}{\left (a^2-b^2 x^2\right ) \sqrt [4]{c+d x^2}} \, dx-b \int \frac {x}{\left (a^2-b^2 x^2\right ) \sqrt [4]{c+d x^2}} \, dx\\ &=-\left (\frac {1}{2} b \text {Subst}\left (\int \frac {1}{\left (a^2-b^2 x\right ) \sqrt [4]{c+d x}} \, dx,x,x^2\right )\right )+\frac {\left (2 a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {x^2}{\left (b^2 c+a^2 d-b^2 x^4\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{x}\\ &=-\frac {(2 b) \text {Subst}\left (\int \frac {x^2}{a^2+\frac {b^2 c}{d}-\frac {b^2 x^4}{d}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{d}+\frac {\left (a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {b^2 c+a^2 d}-b x^2\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{b x}-\frac {\left (a \sqrt {-\frac {d x^2}{c}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {b^2 c+a^2 d}+b x^2\right ) \sqrt {1-\frac {x^4}{c}}} \, dx,x,\sqrt [4]{c+d x^2}\right )}{b x}\\ &=-\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x}-\text {Subst}\left (\int \frac {1}{\sqrt {b^2 c+a^2 d}-b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )+\text {Subst}\left (\int \frac {1}{\sqrt {b^2 c+a^2 d}+b x^2} \, dx,x,\sqrt [4]{c+d x^2}\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\sqrt {b} \sqrt [4]{b^2 c+a^2 d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{c+d x^2}}{\sqrt [4]{b^2 c+a^2 d}}\right )}{\sqrt {b} \sqrt [4]{b^2 c+a^2 d}}-\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (-\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x}+\frac {a \sqrt [4]{c} \sqrt {-\frac {d x^2}{c}} \Pi \left (\frac {b \sqrt {c}}{\sqrt {b^2 c+a^2 d}};\left .\sin ^{-1}\left (\frac {\sqrt [4]{c+d x^2}}{\sqrt [4]{c}}\right )\right |-1\right )}{b \sqrt {b^2 c+a^2 d} x}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in
optimal.
time = 6.35, size = 126, normalized size = 0.45 \begin {gather*} -\frac {2 \sqrt [4]{\frac {b \left (-\sqrt {-\frac {c}{d}}+x\right )}{a+b x}} \sqrt [4]{\frac {b \left (\sqrt {-\frac {c}{d}}+x\right )}{a+b x}} F_1\left (\frac {1}{2};\frac {1}{4},\frac {1}{4};\frac {3}{2};\frac {a-b \sqrt {-\frac {c}{d}}}{a+b x},\frac {a+b \sqrt {-\frac {c}{d}}}{a+b x}\right )}{b \sqrt [4]{c+d x^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (b x +a \right ) \left (d \,x^{2}+c \right )^{\frac {1}{4}}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right ) \sqrt [4]{c + d x^{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {1}{{\left (d\,x^2+c\right )}^{1/4}\,\left (a+b\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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