Optimal. Leaf size=236 \[ -\frac {x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{15 e}-\frac {d x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{8 e^2}-\frac {d^2 \sqrt {d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{15 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (20 A e^2+15 B d e+13 C d^2\right )}{8 e^3}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2} \]
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Rubi [A] time = 0.66, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1815, 641, 217, 203} \[ -\frac {x^2 \sqrt {d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{15 e}-\frac {d x \sqrt {d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{8 e^2}-\frac {d^2 \sqrt {d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{15 e^3}+\frac {d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (20 A e^2+15 B d e+13 C d^2\right )}{8 e^3}-\frac {1}{4} x^3 \sqrt {d^2-e^2 x^2} (B e+3 C d)-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2} \]
Antiderivative was successfully verified.
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Rule 203
Rule 217
Rule 641
Rule 1815
Rubi steps
\begin {align*} \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-5 A d^3 e^2-5 d^2 e^2 (B d+3 A e) x-5 d e^2 \left (C d^2+3 e (B d+A e)\right ) x^2-e^3 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3-5 e^4 (3 C d+B e) x^4}{\sqrt {d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {20 A d^3 e^4+20 d^2 e^4 (B d+3 A e) x+5 d e^4 \left (13 C d^2+15 B d e+12 A e^2\right ) x^2+4 e^5 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3}{\sqrt {d^2-e^2 x^2}} \, dx}{20 e^4}\\ &=-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}-\frac {\int \frac {-60 A d^3 e^6-4 d^2 e^5 \left (38 C d^2+45 B d e+55 A e^2\right ) x-15 d e^6 \left (13 C d^2+15 B d e+12 A e^2\right ) x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{60 e^6}\\ &=-\frac {d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {\int \frac {15 d^3 e^6 \left (13 C d^2+15 B d e+20 A e^2\right )+8 d^2 e^7 \left (38 C d^2+45 B d e+55 A e^2\right ) x}{\sqrt {d^2-e^2 x^2}} \, dx}{120 e^8}\\ &=-\frac {d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (d^3 \left (13 C d^2+15 B d e+20 A e^2\right )\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac {d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {\left (d^3 \left (13 C d^2+15 B d e+20 A e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^2}\\ &=-\frac {d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt {d^2-e^2 x^2}}{15 e^3}-\frac {d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt {d^2-e^2 x^2}}{8 e^2}-\frac {\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt {d^2-e^2 x^2}}{15 e}-\frac {1}{4} (3 C d+B e) x^3 \sqrt {d^2-e^2 x^2}-\frac {1}{5} C e x^4 \sqrt {d^2-e^2 x^2}+\frac {d^3 \left (13 C d^2+15 B d e+20 A e^2\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e^3}\\ \end {align*}
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Mathematica [A] time = 0.45, size = 174, normalized size = 0.74 \[ \frac {15 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (5 e (4 A e+3 B d)+13 C d^2\right )-\sqrt {d^2-e^2 x^2} \left (5 e \left (4 A e \left (22 d^2+9 d e x+2 e^2 x^2\right )+3 B \left (24 d^3+15 d^2 e x+8 d e^2 x^2+2 e^3 x^3\right )\right )+C \left (304 d^4+195 d^3 e x+152 d^2 e^2 x^2+90 d e^3 x^3+24 e^4 x^4\right )\right )}{120 e^3} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 178, normalized size = 0.75 \[ -\frac {30 \, {\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (24 \, C e^{4} x^{4} + 304 \, C d^{4} + 360 \, B d^{3} e + 440 \, A d^{2} e^{2} + 30 \, {\left (3 \, C d e^{3} + B e^{4}\right )} x^{3} + 8 \, {\left (19 \, C d^{2} e^{2} + 15 \, B d e^{3} + 5 \, A e^{4}\right )} x^{2} + 15 \, {\left (13 \, C d^{3} e + 15 \, B d^{2} e^{2} + 12 \, A d e^{3}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.37, size = 166, normalized size = 0.70 \[ \frac {1}{8} \, {\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{120} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, C x e + 5 \, {\left (3 \, C d e^{6} + B e^{7}\right )} e^{\left (-6\right )}\right )} x + 4 \, {\left (19 \, C d^{2} e^{5} + 15 \, B d e^{6} + 5 \, A e^{7}\right )} e^{\left (-6\right )}\right )} x + 15 \, {\left (13 \, C d^{3} e^{4} + 15 \, B d^{2} e^{5} + 12 \, A d e^{6}\right )} e^{\left (-6\right )}\right )} x + 8 \, {\left (38 \, C d^{4} e^{3} + 45 \, B d^{3} e^{4} + 55 \, A d^{2} e^{5}\right )} e^{\left (-6\right )}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 374, normalized size = 1.58 \[ -\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C e \,x^{4}}{5}+\frac {5 A \,d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}+\frac {15 B \,d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, B e \,x^{3}}{4}+\frac {13 C \,d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}\, e^{2}}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, C d \,x^{3}}{4}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, A e \,x^{2}}{3}-\sqrt {-e^{2} x^{2}+d^{2}}\, B d \,x^{2}-\frac {19 \sqrt {-e^{2} x^{2}+d^{2}}\, C \,d^{2} x^{2}}{15 e}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, A d x}{2}-\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, B \,d^{2} x}{8 e}-\frac {13 \sqrt {-e^{2} x^{2}+d^{2}}\, C \,d^{3} x}{8 e^{2}}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, A \,d^{2}}{3 e}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, B \,d^{3}}{e^{2}}-\frac {38 \sqrt {-e^{2} x^{2}+d^{2}}\, C \,d^{4}}{15 e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.98, size = 390, normalized size = 1.65 \[ -\frac {1}{5} \, \sqrt {-e^{2} x^{2} + d^{2}} C e x^{4} - \frac {4 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2} x^{2}}{15 \, e} + \frac {A d^{3} \arcsin \left (\frac {e x}{d}\right )}{e} - \frac {8 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{4}}{15 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d^{3}}{e^{2}} - \frac {3 \, \sqrt {-e^{2} x^{2} + d^{2}} A d^{2}}{e} - \frac {{\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x^{3}}{4 \, e^{2}} - \frac {{\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e^{2}} + \frac {3 \, {\left (3 \, C d e^{2} + B e^{3}\right )} d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e^{5}} + \frac {{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {3 \, {\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{4}} - \frac {{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}} x}{2 \, e^{2}} - \frac {2 \, {\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^3\,\left (C\,x^2+B\,x+A\right )}{\sqrt {d^2-e^2\,x^2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 24.44, size = 1268, normalized size = 5.37 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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