Optimal. Leaf size=134 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {a} e+\sqrt {c} d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}} \]
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Rubi [A] time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {708, 1093, 208} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {a} e+\sqrt {c} d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}} \end {gather*}
Antiderivative was successfully verified.
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Rule 208
Rule 708
Rule 1093
Rubi steps
\begin {align*} \int \frac {1}{\sqrt {d+e x} \left (a-c x^2\right )} \, dx &=(2 e) \operatorname {Subst}\left (\int \frac {1}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt {d+e x}\right )\\ &=-\frac {\sqrt {c} \operatorname {Subst}\left (\int \frac {1}{c d-\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {a}}+\frac {\sqrt {c} \operatorname {Subst}\left (\int \frac {1}{c d+\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{\sqrt {a}}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d-\sqrt {a} e}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{\sqrt {a} \sqrt [4]{c} \sqrt {\sqrt {c} d+\sqrt {a} e}}\\ \end {align*}
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Mathematica [A] time = 0.08, size = 125, normalized size = 0.93 \begin {gather*} \frac {\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{\sqrt {\sqrt {a} e+\sqrt {c} d}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{\sqrt {\sqrt {c} d-\sqrt {a} e}}}{\sqrt {a} \sqrt [4]{c}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.33, size = 167, normalized size = 1.25 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-\sqrt {a} \sqrt {c} e-c d}}{\sqrt {a} e+\sqrt {c} d}\right )}{\sqrt {a} \sqrt {-\sqrt {c} \left (\sqrt {a} e+\sqrt {c} d\right )}}-\frac {\tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {a} \sqrt {c} e-c d}}{\sqrt {c} d-\sqrt {a} e}\right )}{\sqrt {a} \sqrt {-\sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.43, size = 949, normalized size = 7.08 \begin {gather*} \frac {1}{2} \, \sqrt {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} \log \left (\sqrt {e x + d} e + {\left (a e^{2} - {\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}}\right ) - \frac {1}{2} \, \sqrt {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}} \log \left (\sqrt {e x + d} e - {\left (a e^{2} - {\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} + d}{a c d^{2} - a^{2} e^{2}}}\right ) + \frac {1}{2} \, \sqrt {-\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} \log \left (\sqrt {e x + d} e + {\left (a e^{2} + {\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {-\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}}\right ) - \frac {1}{2} \, \sqrt {-\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}} \log \left (\sqrt {e x + d} e - {\left (a e^{2} + {\left (a c^{2} d^{3} - a^{2} c d e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}}\right )} \sqrt {-\frac {{\left (a c d^{2} - a^{2} e^{2}\right )} \sqrt {\frac {e^{2}}{a c^{3} d^{4} - 2 \, a^{2} c^{2} d^{2} e^{2} + a^{3} c e^{4}}} - d}{a c d^{2} - a^{2} e^{2}}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 159, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {a c} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d + \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d - \sqrt {a c} c e} a c} + \frac {\sqrt {a c} {\left | c \right |} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {c d - \sqrt {c^{2} d^{2} - {\left (c d^{2} - a e^{2}\right )} c}}{c}}}\right )}{\sqrt {-c^{2} d + \sqrt {a c} c e} a c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 110, normalized size = 0.82 \begin {gather*} \frac {c e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {c e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{\sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}} \end {gather*}
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*} -\int \frac {1}{{\left (c x^{2} - a\right )} \sqrt {e x + d}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.65, size = 1366, normalized size = 10.19 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {32\,a^2\,c^5\,d^2\,e^2\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}-\frac {32\,c^3\,e^2\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^2\,c^4\,d\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a\,c^3\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}+\frac {32\,a\,c^4\,d\,e^3\,\sqrt {a^3\,c}\,\sqrt {-\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}\right )\,\sqrt {-\frac {e\,\sqrt {a^3\,c}+a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}-2\,\mathrm {atanh}\left (\frac {32\,c^3\,e^2\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^2\,c^4\,d\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a\,c^3\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}-\frac {32\,a^2\,c^5\,d^2\,e^2\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}+\frac {32\,a\,c^4\,d\,e^3\,\sqrt {a^3\,c}\,\sqrt {\frac {e\,\sqrt {a^3\,c}}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}-\frac {a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}}\,\sqrt {d+e\,x}}{\frac {16\,a^4\,c^6\,d^3\,e^3}{a^3\,c\,e^2-a^2\,c^2\,d^2}+\frac {16\,a^4\,c^4\,e^6\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^5\,c^5\,d\,e^5}{a^3\,c\,e^2-a^2\,c^2\,d^2}-\frac {16\,a^3\,c^5\,d^2\,e^4\,\sqrt {a^3\,c}}{a^3\,c\,e^2-a^2\,c^2\,d^2}}\right )\,\sqrt {\frac {e\,\sqrt {a^3\,c}-a\,c\,d}{4\,\left (a^3\,c\,e^2-a^2\,c^2\,d^2\right )}} \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}{- a \sqrt {d + e x} + c x^{2} \sqrt {d + e x}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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