Optimal. Leaf size=270 \[ \frac{\sqrt{2 x+1} (10 x+3)}{31 \left (5 x^2+3 x+2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]
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Rubi [A] time = 0.34043, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {736, 826, 1169, 634, 618, 204, 628} \[ \frac{\sqrt{2 x+1} (10 x+3)}{31 \left (5 x^2+3 x+2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (47 \sqrt{35}-218\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )+\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]
Antiderivative was successfully verified.
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Rule 736
Rule 826
Rule 1169
Rule 634
Rule 618
Rule 204
Rule 628
Rubi steps
\begin{align*} \int \frac{\sqrt{1+2 x}}{\left (2+3 x+5 x^2\right )^2} \, dx &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{1}{31} \int \frac{-7-10 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{2}{31} \operatorname{Subst}\left (\int \frac{-4-10 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{-4 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-4+2 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{31 \sqrt{14 \left (2+\sqrt{35}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{-4 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-4+2 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{31 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}+\frac{\left (35+2 \sqrt{35}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{1085}+\frac{\left (35+2 \sqrt{35}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{1085}+\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \operatorname{Subst}\left (\int \frac{-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 x}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 x}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}+\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{\left (2 \left (35+2 \sqrt{35}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )}{1085}-\frac{\left (2 \left (35+2 \sqrt{35}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )}{1085}\\ &=\frac{\sqrt{1+2 x} (3+10 x)}{31 \left (2+3 x+5 x^2\right )}-\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-2 \sqrt{1+2 x}\right )\right )+\frac{1}{31} \sqrt{\frac{2}{217} \left (218+47 \sqrt{35}\right )} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )\right )+\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )-\frac{1}{31} \sqrt{\frac{1}{434} \left (-218+47 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )\\ \end{align*}
Mathematica [C] time = 0.369282, size = 145, normalized size = 0.54 \[ \frac{1}{31} \left (\frac{\sqrt{2 x+1} (10 x+3)}{5 x^2+3 x+2}+\frac{2 \sqrt{10-5 i \sqrt{31}} \left (62-39 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )+2 \sqrt{10+5 i \sqrt{31}} \left (62+39 i \sqrt{31}\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )}{1085}\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.243, size = 972, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x + 1}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.09446, size = 2176, normalized size = 8.06 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 5.22867, size = 83, normalized size = 0.31 \begin{align*} \frac{80 \left (2 x + 1\right )^{\frac{3}{2}}}{- 992 x + 620 \left (2 x + 1\right )^{2} + 372} - \frac{32 \sqrt{2 x + 1}}{- 992 x + 620 \left (2 x + 1\right )^{2} + 372} + 16 \operatorname{RootSum}{\left (407144088666112 t^{4} + 3325152256 t^{2} + 11045, \left ( t \mapsto t \log{\left (\frac{33312534528 t^{3}}{235} + \frac{166784 t}{235} + \sqrt{2 x + 1} \right )} \right )\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 \, x + 1}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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