Optimal. Leaf size=146 \[ -\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{5/2}}+\frac {e^2 \sqrt {d+e x}}{8 b (a+b x) (b d-a e)^2}-\frac {e \sqrt {d+e x}}{12 b (a+b x)^2 (b d-a e)}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3} \]
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Rubi [A] time = 0.07, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 47, 51, 63, 208} \begin {gather*} -\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{5/2}}+\frac {e^2 \sqrt {d+e x}}{8 b (a+b x) (b d-a e)^2}-\frac {e \sqrt {d+e x}}{12 b (a+b x)^2 (b d-a e)}-\frac {\sqrt {d+e x}}{3 b (a+b x)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 27
Rule 47
Rule 51
Rule 63
Rule 208
Rubi steps
\begin {align*} \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {\sqrt {d+e x}}{(a+b x)^4} \, dx\\ &=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}+\frac {e \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{6 b}\\ &=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x)^2}-\frac {e^2 \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{8 b (b d-a e)}\\ &=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x)^2}+\frac {e^2 \sqrt {d+e x}}{8 b (b d-a e)^2 (a+b x)}+\frac {e^3 \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b (b d-a e)^2}\\ &=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x)^2}+\frac {e^2 \sqrt {d+e x}}{8 b (b d-a e)^2 (a+b x)}+\frac {e^2 \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 b (b d-a e)^2}\\ &=-\frac {\sqrt {d+e x}}{3 b (a+b x)^3}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x)^2}+\frac {e^2 \sqrt {d+e x}}{8 b (b d-a e)^2 (a+b x)}-\frac {e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{5/2}}\\ \end {align*}
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Mathematica [C] time = 0.01, size = 52, normalized size = 0.36 \begin {gather*} \frac {2 e^3 (d+e x)^{3/2} \, _2F_1\left (\frac {3}{2},4;\frac {5}{2};-\frac {b (d+e x)}{a e-b d}\right )}{3 (a e-b d)^4} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.75, size = 176, normalized size = 1.21 \begin {gather*} \frac {e^3 \sqrt {d+e x} \left (3 a^2 e^2-8 a b e (d+e x)-6 a b d e+3 b^2 d^2-3 b^2 (d+e x)^2+8 b^2 d (d+e x)\right )}{24 b (b d-a e)^2 (-a e-b (d+e x)+b d)^3}-\frac {e^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{8 b^{3/2} (b d-a e)^2 \sqrt {a e-b d}} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.43, size = 785, normalized size = 5.38 \begin {gather*} \left [\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{3} - 22 \, a b^{3} d^{2} e + 17 \, a^{2} b^{2} d e^{2} - 3 \, a^{3} b e^{3} - 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\right )} x\right )}}, \frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (8 \, b^{4} d^{3} - 22 \, a b^{3} d^{2} e + 17 \, a^{2} b^{2} d e^{2} - 3 \, a^{3} b e^{3} - 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (b^{4} d^{2} e - 5 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\right )} x\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 211, normalized size = 1.45 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {x e + d} b^{2} d^{2} e^{3} + 8 \, {\left (x e + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {x e + d} a b d e^{4} - 3 \, \sqrt {x e + d} a^{2} e^{5}}{24 \, {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 170, normalized size = 1.16 \begin {gather*} \frac {\left (e x +d \right )^{\frac {5}{2}} b \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, b}+\frac {\left (e x +d \right )^{\frac {3}{2}} e^{3}}{3 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {\sqrt {e x +d}\, e^{3}}{8 \left (b e x +a e \right )^{3} b} \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 = 0.65, size = 207, normalized size = 1.42 \begin {gather*} \frac {\frac {e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,\left (a\,e-b\,d\right )}-\frac {e^3\,\sqrt {d+e\,x}}{8\,b}+\frac {b\,e^3\,{\left (d+e\,x\right )}^{5/2}}{8\,{\left (a\,e-b\,d\right )}^2}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2}+\frac {e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{8\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{5/2}} \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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