Optimal. Leaf size=417 \[ -\frac{\sqrt{c} \left (-2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (d \sqrt{b^2-4 a c}-2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-2 a \left (c d^2-e \left (d \sqrt{b^2-4 a c}+a e\right )\right )-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]
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Rubi [A] time = 3.24265, antiderivative size = 416, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {1251, 897, 1287, 199, 206, 1166, 208} \[ -\frac{\sqrt{c} \left (b d \left (d \sqrt{b^2-4 a c}-2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}-a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}+\frac{\sqrt{c} \left (-b d \left (d \sqrt{b^2-4 a c}+2 a e\right )+2 a e \left (d \sqrt{b^2-4 a c}+a e\right )-2 a c d^2+b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a} \]
Antiderivative was successfully verified.
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Rule 1251
Rule 897
Rule 1287
Rule 199
Rule 206
Rule 1166
Rule 208
Rubi steps
\begin{align*} \int \frac{\left (d+e x^2\right )^{3/2}}{x^3 \left (a+b x^2+c x^4\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2}}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (-\frac{d}{e}+\frac{x^2}{e}\right )^2 \left (\frac{c d^2-b d e+a e^2}{e^2}-\frac{(2 c d-b e) x^2}{e^2}+\frac{c x^4}{e^2}\right )} \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{d^2 e^2}{a \left (d-x^2\right )^2}-\frac{d e (-b d+2 a e)}{a^2 \left (d-x^2\right )}+\frac{e \left (-(b d-a e) \left (c d^2-b d e+a e^2\right )+c d (b d-2 a e) x^2\right )}{a^2 \left (c d^2-b d e+a e^2-(2 c d-b e) x^2+c x^4\right )}\right ) \, dx,x,\sqrt{d+e x^2}\right )}{e}\\ &=\frac{\operatorname{Subst}\left (\int \frac{-(b d-a e) \left (c d^2-b d e+a e^2\right )+c d (b d-2 a e) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x^2}\right )}{a^2}+\frac{\left (d^2 e\right ) \operatorname{Subst}\left (\int \frac{1}{\left (d-x^2\right )^2} \, dx,x,\sqrt{d+e x^2}\right )}{a}+\frac{(d (b d-2 a e)) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{a^2}\\ &=-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}+\frac{(d e) \operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a}+\frac{\left (c \left (b^2 d^2-2 a c d^2+b d \left (\sqrt{b^2-4 a c} d-2 a e\right )-2 a e \left (\sqrt{b^2-4 a c} d-a e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a^2 \sqrt{b^2-4 a c}}-\frac{\left (c \left (b^2 d^2-2 a c d^2+2 a e \left (\sqrt{b^2-4 a c} d+a e\right )-b d \left (\sqrt{b^2-4 a c} d+2 a e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x^2}\right )}{2 a^2 \sqrt{b^2-4 a c}}\\ &=-\frac{d \sqrt{d+e x^2}}{2 a x^2}+\frac{\sqrt{d} e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 a}+\frac{\sqrt{d} (b d-2 a e) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{a^2}-\frac{\sqrt{c} \left (b^2 d^2-2 a c d^2+b d \left (\sqrt{b^2-4 a c} d-2 a e\right )-2 a e \left (\sqrt{b^2-4 a c} d-a e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}+\frac{\sqrt{c} \left (b^2 d^2-2 a c d^2+2 a e \left (\sqrt{b^2-4 a c} d+a e\right )-b d \left (\sqrt{b^2-4 a c} d+2 a e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} a^2 \sqrt{b^2-4 a c} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\\ \end{align*}
Mathematica [A] time = 1.62938, size = 380, normalized size = 0.91 \[ \frac{\frac{\sqrt{2} \sqrt{c} \left (\frac{\left (2 a \left (e \left (d \sqrt{b^2-4 a c}-a e\right )+c d^2\right )+b d \left (2 a e-d \sqrt{b^2-4 a c}\right )-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (b d \left (d \sqrt{b^2-4 a c}+2 a e\right )-2 a e \left (d \sqrt{b^2-4 a c}+a e\right )+2 a c d^2-b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x^2}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{b^2-4 a c}}+\sqrt{d} (2 b d-3 a e) \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-\sqrt{d} \log (x) (2 b d-3 a e)-\frac{a d \sqrt{d+e x^2}}{x^2}}{2 a^2} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.03, size = 555, normalized size = 1.3 \begin{align*} -{\frac{{x}^{3}b}{6\,{a}^{2}}{e}^{{\frac{3}{2}}}}+{\frac{{x}^{2}eb}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{3\,bdx}{4\,{a}^{2}}\sqrt{e}}-{\frac{7\,b}{24\,{a}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}+{\frac{x}{2\,a}{e}^{{\frac{3}{2}}}}+{\frac{e}{a}\sqrt{e{x}^{2}+d}}-{\frac{3\,bd}{8\,{a}^{2}}\sqrt{e{x}^{2}+d}}-{\frac{de}{2\,a} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{5\,b{d}^{2}}{8\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-1}}+{\frac{b{d}^{3}}{24\,{a}^{2}} \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x \right ) ^{-3}}+{\frac{1}{4\,{a}^{2}}\sum _{{\it \_R}={\it RootOf} \left ( c{{\it \_Z}}^{8}+ \left ( 4\,be-4\,cd \right ){{\it \_Z}}^{6}+ \left ( 16\,a{e}^{2}-8\,deb+6\,c{d}^{2} \right ){{\it \_Z}}^{4}+ \left ( 4\,b{d}^{2}e-4\,c{d}^{3} \right ){{\it \_Z}}^{2}+c{d}^{4} \right ) }{\frac{cd \left ( -2\,ae+bd \right ){{\it \_R}}^{6}+ \left ( 4\,{a}^{2}{e}^{3}-8\,abd{e}^{2}+2\,e{d}^{2}ca+4\,{b}^{2}{d}^{2}e-3\,bc{d}^{3} \right ){{\it \_R}}^{4}+d \left ( -4\,{a}^{2}{e}^{3}+8\,abd{e}^{2}-2\,e{d}^{2}ca-4\,{b}^{2}{d}^{2}e+3\,bc{d}^{3} \right ){{\it \_R}}^{2}+2\,ac{d}^{4}e-bc{d}^{5}}{{{\it \_R}}^{7}c+3\,{{\it \_R}}^{5}be-3\,{{\it \_R}}^{5}cd+8\,{{\it \_R}}^{3}a{e}^{2}-4\,{{\it \_R}}^{3}bde+3\,{{\it \_R}}^{3}c{d}^{2}+{\it \_R}\,b{d}^{2}e-{\it \_R}\,c{d}^{3}}\ln \left ( \sqrt{e{x}^{2}+d}-\sqrt{e}x-{\it \_R} \right ) }}+{\frac{b}{{a}^{2}}{d}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) }-{\frac{1}{2\,ad{x}^{2}} \left ( e{x}^{2}+d \right ) ^{{\frac{5}{2}}}}+{\frac{e}{2\,ad} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}-{\frac{3\,e}{2\,a}\sqrt{d}\ln \left ({\frac{1}{x} \left ( 2\,d+2\,\sqrt{d}\sqrt{e{x}^{2}+d} \right ) } \right ) } \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{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}{{\left (c x^{4} + b x^{2} + a\right )} x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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