Optimal. Leaf size=225 \[ \frac{x \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right )}{16 a^3 c^2 \left (a+c x^2\right )}-\frac{x \left (3 a e^2 (a C+A c)-c d (2 a B e+a C d+5 A c d)\right )+2 a e (a B e+2 a C d+4 A c d)}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right )}{16 a^{7/2} c^{5/2}}-\frac{(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]
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Rubi [A] time = 0.397508, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1645, 778, 199, 205} \[ \frac{x \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right )}{16 a^3 c^2 \left (a+c x^2\right )}-\frac{x \left (3 a e^2 (a C+A c)-c d (2 a B e+a C d+5 A c d)\right )+2 a e (a B e+2 a C d+4 A c d)}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (c d (2 a B e+a C d+5 A c d)+a e^2 (a C+A c)\right )}{16 a^{7/2} c^{5/2}}-\frac{(d+e x)^2 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 1645
Rule 778
Rule 199
Rule 205
Rubi steps
\begin{align*} \int \frac{(d+e x)^2 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx &=-\frac{(a B-(A c-a C) x) (d+e x)^2}{6 a c \left (a+c x^2\right )^3}-\frac{\int \frac{(d+e x) (-5 A c d-a C d-2 a B e-3 (A c+a C) e x)}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^2}{6 a c \left (a+c x^2\right )^3}-\frac{2 a e (4 A c d+2 a C d+a B e)+\left (3 a (A c+a C) e^2-c d (5 A c d+a C d+2 a B e)\right ) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) \int \frac{1}{\left (a+c x^2\right )^2} \, dx}{8 a^2 c^2}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^2}{6 a c \left (a+c x^2\right )^3}-\frac{2 a e (4 A c d+2 a C d+a B e)+\left (3 a (A c+a C) e^2-c d (5 A c d+a C d+2 a B e)\right ) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) x}{16 a^3 c^2 \left (a+c x^2\right )}+\frac{\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) \int \frac{1}{a+c x^2} \, dx}{16 a^3 c^2}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)^2}{6 a c \left (a+c x^2\right )^3}-\frac{2 a e (4 A c d+2 a C d+a B e)+\left (3 a (A c+a C) e^2-c d (5 A c d+a C d+2 a B e)\right ) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) x}{16 a^3 c^2 \left (a+c x^2\right )}+\frac{\left (a (A c+a C) e^2+c d (5 A c d+a C d+2 a B e)\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.169364, size = 266, normalized size = 1.18 \[ \frac{x \left (A c \left (a e^2+5 c d^2\right )+a \left (a C e^2+c d (2 B e+C d)\right )\right )}{16 a^3 c^2 \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c \left (a e^2+5 c d^2\right )+a \left (a C e^2+c d (2 B e+C d)\right )\right )}{16 a^{7/2} c^{5/2}}+\frac{a^2 (-e) (6 B e+12 C d+7 C e x)+a c x \left (e (A e+2 B d)+C d^2\right )+5 A c^2 d^2 x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{a^2 e (B e+2 C d+C e x)-a c \left (A e (2 d+e x)+B d (d+2 e x)+C d^2 x\right )+A c^2 d^2 x}{6 a c^2 \left (a+c x^2\right )^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.053, size = 333, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{3}} \left ({\frac{ \left ( aA{e}^{2}c+5\,A{c}^{2}{d}^{2}+2\,Bacde+{a}^{2}C{e}^{2}+Cac{d}^{2} \right ){x}^{5}}{16\,{a}^{3}}}+{\frac{ \left ( aA{e}^{2}c+5\,A{c}^{2}{d}^{2}+2\,Bacde-{a}^{2}C{e}^{2}+Cac{d}^{2} \right ){x}^{3}}{6\,{a}^{2}c}}-{\frac{e \left ( Be+2\,Cd \right ){x}^{2}}{4\,c}}-{\frac{ \left ( aA{e}^{2}c-11\,A{c}^{2}{d}^{2}+2\,Bacde+{a}^{2}C{e}^{2}+Cac{d}^{2} \right ) x}{16\,a{c}^{2}}}-{\frac{4\,Acde+aB{e}^{2}+2\,Bc{d}^{2}+2\,Cade}{12\,{c}^{2}}} \right ) }+{\frac{A{e}^{2}}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{5\,A{d}^{2}}{16\,{a}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{Bde}{8\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{C{e}^{2}}{16\,a{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{C{d}^{2}}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.55657, size = 2163, normalized size = 9.61 \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 [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 [A] time = 1.15398, size = 443, normalized size = 1.97 \begin{align*} \frac{{\left (C a c d^{2} + 5 \, A c^{2} d^{2} + 2 \, B a c d e + C a^{2} e^{2} + A a c e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3} c^{2}} + \frac{3 \, C a c^{3} d^{2} x^{5} + 15 \, A c^{4} d^{2} x^{5} + 6 \, B a c^{3} d x^{5} e + 3 \, C a^{2} c^{2} x^{5} e^{2} + 3 \, A a c^{3} x^{5} e^{2} + 8 \, C a^{2} c^{2} d^{2} x^{3} + 40 \, A a c^{3} d^{2} x^{3} + 16 \, B a^{2} c^{2} d x^{3} e - 8 \, C a^{3} c x^{3} e^{2} + 8 \, A a^{2} c^{2} x^{3} e^{2} - 24 \, C a^{3} c d x^{2} e - 3 \, C a^{3} c d^{2} x + 33 \, A a^{2} c^{2} d^{2} x - 12 \, B a^{3} c x^{2} e^{2} - 6 \, B a^{3} c d x e - 8 \, B a^{3} c d^{2} - 3 \, C a^{4} x e^{2} - 3 \, A a^{3} c x e^{2} - 8 \, C a^{4} d e - 16 \, A a^{3} c d e - 4 \, B a^{4} e^{2}}{48 \,{\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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