3.32 \(\int \frac{\left (a x+b x^2\right )^{5/2}}{x^6} \, dx\)

Optimal. Leaf size=91 \[ 2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 b^2 \sqrt{a x+b x^2}}{x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{5 x^5}-\frac{2 b \left (a x+b x^2\right )^{3/2}}{3 x^3} \]

[Out]

(-2*b^2*Sqrt[a*x + b*x^2])/x - (2*b*(a*x + b*x^2)^(3/2))/(3*x^3) - (2*(a*x + b*x
^2)^(5/2))/(5*x^5) + 2*b^(5/2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]]

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Rubi [A]  time = 0.120356, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176 \[ 2 b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )-\frac{2 b^2 \sqrt{a x+b x^2}}{x}-\frac{2 \left (a x+b x^2\right )^{5/2}}{5 x^5}-\frac{2 b \left (a x+b x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]  Int[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

(-2*b^2*Sqrt[a*x + b*x^2])/x - (2*b*(a*x + b*x^2)^(3/2))/(3*x^3) - (2*(a*x + b*x
^2)^(5/2))/(5*x^5) + 2*b^(5/2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]]

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Rubi in Sympy [A]  time = 13.0904, size = 83, normalized size = 0.91 \[ 2 b^{\frac{5}{2}} \operatorname{atanh}{\left (\frac{\sqrt{b} x}{\sqrt{a x + b x^{2}}} \right )} - \frac{2 b^{2} \sqrt{a x + b x^{2}}}{x} - \frac{2 b \left (a x + b x^{2}\right )^{\frac{3}{2}}}{3 x^{3}} - \frac{2 \left (a x + b x^{2}\right )^{\frac{5}{2}}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((b*x**2+a*x)**(5/2)/x**6,x)

[Out]

2*b**(5/2)*atanh(sqrt(b)*x/sqrt(a*x + b*x**2)) - 2*b**2*sqrt(a*x + b*x**2)/x - 2
*b*(a*x + b*x**2)**(3/2)/(3*x**3) - 2*(a*x + b*x**2)**(5/2)/(5*x**5)

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Mathematica [A]  time = 0.0723376, size = 93, normalized size = 1.02 \[ -\frac{2 \sqrt{x (a+b x)} \left (\sqrt{a+b x} \left (3 a^2+11 a b x+23 b^2 x^2\right )-15 b^{5/2} x^{5/2} \log \left (\sqrt{b} \sqrt{a+b x}+b \sqrt{x}\right )\right )}{15 x^3 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]  Integrate[(a*x + b*x^2)^(5/2)/x^6,x]

[Out]

(-2*Sqrt[x*(a + b*x)]*(Sqrt[a + b*x]*(3*a^2 + 11*a*b*x + 23*b^2*x^2) - 15*b^(5/2
)*x^(5/2)*Log[b*Sqrt[x] + Sqrt[b]*Sqrt[a + b*x]]))/(15*x^3*Sqrt[a + b*x])

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Maple [B]  time = 0.007, size = 232, normalized size = 2.6 \[ -{\frac{2}{5\,a{x}^{6}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{4\,b}{15\,{a}^{2}{x}^{5}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{16\,{b}^{2}}{15\,{a}^{3}{x}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{32\,{b}^{3}}{5\,{a}^{4}{x}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}-{\frac{256\,{b}^{4}}{15\,{a}^{5}{x}^{2}} \left ( b{x}^{2}+ax \right ) ^{{\frac{7}{2}}}}+{\frac{256\,{b}^{5}}{15\,{a}^{5}} \left ( b{x}^{2}+ax \right ) ^{{\frac{5}{2}}}}+{\frac{32\,{b}^{5}x}{3\,{a}^{4}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}+{\frac{16\,{b}^{4}}{3\,{a}^{3}} \left ( b{x}^{2}+ax \right ) ^{{\frac{3}{2}}}}-4\,{\frac{{b}^{4}\sqrt{b{x}^{2}+ax}x}{{a}^{2}}}-2\,{\frac{{b}^{3}\sqrt{b{x}^{2}+ax}}{a}}+{b}^{{\frac{5}{2}}}\ln \left ({1 \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((b*x^2+a*x)^(5/2)/x^6,x)

[Out]

-2/5/a/x^6*(b*x^2+a*x)^(7/2)-4/15*b/a^2/x^5*(b*x^2+a*x)^(7/2)-16/15*b^2/a^3/x^4*
(b*x^2+a*x)^(7/2)+32/5*b^3/a^4/x^3*(b*x^2+a*x)^(7/2)-256/15*b^4/a^5/x^2*(b*x^2+a
*x)^(7/2)+256/15*b^5/a^5*(b*x^2+a*x)^(5/2)+32/3*b^5/a^4*(b*x^2+a*x)^(3/2)*x+16/3
*b^4/a^3*(b*x^2+a*x)^(3/2)-4*b^4/a^2*(b*x^2+a*x)^(1/2)*x-2*b^3/a*(b*x^2+a*x)^(1/
2)+b^(5/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a*x)^(5/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.233617, size = 1, normalized size = 0.01 \[ \left [\frac{15 \, b^{\frac{5}{2}} x^{3} \log \left (2 \, b x + a + 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) - 2 \,{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{2} + a x}}{15 \, x^{3}}, \frac{2 \,{\left (15 \, \sqrt{-b} b^{2} x^{3} \arctan \left (\frac{\sqrt{b x^{2} + a x}}{\sqrt{-b} x}\right ) -{\left (23 \, b^{2} x^{2} + 11 \, a b x + 3 \, a^{2}\right )} \sqrt{b x^{2} + a x}\right )}}{15 \, x^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a*x)^(5/2)/x^6,x, algorithm="fricas")

[Out]

[1/15*(15*b^(5/2)*x^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(23*b^2*x
^2 + 11*a*b*x + 3*a^2)*sqrt(b*x^2 + a*x))/x^3, 2/15*(15*sqrt(-b)*b^2*x^3*arctan(
sqrt(b*x^2 + a*x)/(sqrt(-b)*x)) - (23*b^2*x^2 + 11*a*b*x + 3*a^2)*sqrt(b*x^2 + a
*x))/x^3]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (x \left (a + b x\right )\right )^{\frac{5}{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x**2+a*x)**(5/2)/x**6,x)

[Out]

Integral((x*(a + b*x))**(5/2)/x**6, x)

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GIAC/XCAS [A]  time = 0.22915, size = 236, normalized size = 2.59 \[ -b^{\frac{5}{2}}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} \sqrt{b} - a \right |}\right ) + \frac{2 \,{\left (45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{4} a b^{2} + 45 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{3} a^{2} b^{\frac{3}{2}} + 35 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{2} a^{3} b + 15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )} a^{4} \sqrt{b} + 3 \, a^{5}\right )}}{15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a x}\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a*x)^(5/2)/x^6,x, algorithm="giac")

[Out]

-b^(5/2)*ln(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a)) + 2/15*(45*(sqr
t(b)*x - sqrt(b*x^2 + a*x))^4*a*b^2 + 45*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*a^2*b
^(3/2) + 35*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^3*b + 15*(sqrt(b)*x - sqrt(b*x^2
+ a*x))*a^4*sqrt(b) + 3*a^5)/(sqrt(b)*x - sqrt(b*x^2 + a*x))^5