Optimal. Leaf size=202 \[ \frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) \sqrt{d+e x}}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)^{3/2}}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^4 (a+b x)} \]
[Out]
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Rubi [A] time = 0.202229, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067 \[ \frac{6 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}{e^4 (a+b x) \sqrt{d+e x}}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^4 (a+b x) (d+e x)^{3/2}}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}{5 e^4 (a+b x) (d+e x)^{5/2}}+\frac{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x}}{e^4 (a+b x)} \]
Antiderivative was successfully verified.
[In] Int[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(7/2),x]
[Out]
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Rubi in Sympy [A] time = 21.5415, size = 168, normalized size = 0.83 \[ \frac{16 b^{2} \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5 e^{3} \sqrt{d + e x}} - \frac{32 b^{2} \left (a e - b d\right ) \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{5 e^{4} \left (a + b x\right ) \sqrt{d + e x}} - \frac{4 b \left (3 a + 3 b x\right ) \sqrt{a^{2} + 2 a b x + b^{2} x^{2}}}{15 e^{2} \left (d + e x\right )^{\frac{3}{2}}} - \frac{2 \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac{3}{2}}}{5 e \left (d + e x\right )^{\frac{5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(7/2),x)
[Out]
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Mathematica [A] time = 0.186338, size = 98, normalized size = 0.49 \[ \frac{2 \left ((a+b x)^2\right )^{3/2} \sqrt{d+e x} \left (\frac{15 b^2 (b d-a e)}{d+e x}-\frac{5 b (b d-a e)^2}{(d+e x)^2}+\frac{(b d-a e)^3}{(d+e x)^3}+5 b^3\right )}{5 e^4 (a+b x)^3} \]
Antiderivative was successfully verified.
[In] Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(d + e*x)^(7/2),x]
[Out]
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Maple [A] time = 0.01, size = 131, normalized size = 0.7 \[ -{\frac{-10\,{x}^{3}{b}^{3}{e}^{3}+30\,{x}^{2}a{b}^{2}{e}^{3}-60\,{x}^{2}{b}^{3}d{e}^{2}+10\,x{a}^{2}b{e}^{3}+40\,xa{b}^{2}d{e}^{2}-80\,x{b}^{3}{d}^{2}e+2\,{a}^{3}{e}^{3}+4\,{a}^{2}bd{e}^{2}+16\,a{b}^{2}{d}^{2}e-32\,{b}^{3}{d}^{3}}{5\, \left ( bx+a \right ) ^{3}{e}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int((b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(7/2),x)
[Out]
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Maxima [A] time = 0.740461, size = 185, normalized size = 0.92 \[ \frac{2 \,{\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \,{\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \,{\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )}}{5 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt{e x + d}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((b^2*x^2 + 2*a*b*x + a^2)^(3/2)/(e*x + d)^(7/2),x, algorithm="maxima")
[Out]
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Fricas [A] time = 0.207344, size = 185, normalized size = 0.92 \[ \frac{2 \,{\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \,{\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \,{\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )}}{5 \,{\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )} \sqrt{e x + d}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((b^2*x^2 + 2*a*b*x + a^2)^(3/2)/(e*x + d)^(7/2),x, algorithm="fricas")
[Out]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((b**2*x**2+2*a*b*x+a**2)**(3/2)/(e*x+d)**(7/2),x)
[Out]
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GIAC/XCAS [A] time = 0.21894, size = 265, normalized size = 1.31 \[ 2 \, \sqrt{x e + d} b^{3} e^{\left (-4\right )}{\rm sign}\left (b x + a\right ) + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} b^{3} d{\rm sign}\left (b x + a\right ) - 5 \,{\left (x e + d\right )} b^{3} d^{2}{\rm sign}\left (b x + a\right ) + b^{3} d^{3}{\rm sign}\left (b x + a\right ) - 15 \,{\left (x e + d\right )}^{2} a b^{2} e{\rm sign}\left (b x + a\right ) + 10 \,{\left (x e + d\right )} a b^{2} d e{\rm sign}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e{\rm sign}\left (b x + a\right ) - 5 \,{\left (x e + d\right )} a^{2} b e^{2}{\rm sign}\left (b x + a\right ) + 3 \, a^{2} b d e^{2}{\rm sign}\left (b x + a\right ) - a^{3} e^{3}{\rm sign}\left (b x + a\right )\right )} e^{\left (-4\right )}}{5 \,{\left (x e + d\right )}^{\frac{5}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate((b^2*x^2 + 2*a*b*x + a^2)^(3/2)/(e*x + d)^(7/2),x, algorithm="giac")
[Out]