3.62 \(\int \frac {x^4}{(a x+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=71 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}} \]

[Out]

-2/3*x^3/b/(b*x^2+a*x)^(3/2)+2*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(5/2)-2*x/b^2/(b*x^2+a*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {668, 652, 620, 206} \[ -\frac {2 x}{b^2 \sqrt {a x+b x^2}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}}-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a*x + b*x^2)^(5/2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 652

Int[((d_.) + (e_.)*(x_))^2*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + b*x +
c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fr
eeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p,
-1]

Rule 668

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] - Dist[(e^2*(m + p))/(c*(p + 1)), Int[(d + e*x)^(m - 2)*(a + b*x +
c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx &=-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}}+\frac {\int \frac {x^2}{\left (a x+b x^2\right )^{3/2}} \, dx}{b}\\ &=-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}+\frac {\int \frac {1}{\sqrt {a x+b x^2}} \, dx}{b^2}\\ &=-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a x+b x^2}}\right )}{b^2}\\ &=-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 84, normalized size = 1.18 \[ \frac {x \left (6 \sqrt {a} \sqrt {x} (a+b x) \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-2 \sqrt {b} x (3 a+4 b x)\right )}{3 b^{5/2} (x (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a*x + b*x^2)^(5/2),x]

[Out]

(x*(-2*Sqrt[b]*x*(3*a + 4*b*x) + 6*Sqrt[a]*Sqrt[x]*(a + b*x)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[
a]]))/(3*b^(5/2)*(x*(a + b*x))^(3/2))

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fricas [A]  time = 0.98, size = 193, normalized size = 2.72 \[ \left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x^{2} + a x}\right )}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a*x)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) - 2*(4*b^2*x + 3*a*b)*s
qrt(b*x^2 + a*x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -2/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-b)*arctan(sqrt(b*x^
2 + a*x)*sqrt(-b)/(b*x)) + (4*b^2*x + 3*a*b)*sqrt(b*x^2 + a*x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to divide, perhaps due to rounding error%%%{%%%{1,[2]%%%},[4,4]%%%}+%%%{%%{[%%%{-4,[1]%%%},0]:[1,0,%%%{-1,[
1]%%%}]%%},[3,5]%%%}+%%%{%%%{6,[1]%%%},[2,6]%%%}+%%%{%%{[-4,0]:[1,0,%%%{-1,[1]%%%}]%%},[1,7]%%%}+%%%{1,[0,8]%%
%} / %%%{%%%{1,[4]%%%},[4,0]%%%}+%%%{%%{[%%%{-4,[3]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[3,1]%%%}+%%%{%%%{6,[3]%%%}
,[2,2]%%%}+%%%{%%{[%%%{-4,[2]%%%},0]:[1,0,%%%{-1,[1]%%%}]%%},[1,3]%%%}+%%%{%%%{1,[2]%%%},[0,4]%%%} Error: Bad
Argument Value

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maple [B]  time = 0.05, size = 123, normalized size = 1.73 \[ -\frac {x^{3}}{3 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b}+\frac {a \,x^{2}}{2 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{2}}+\frac {a^{2} x}{6 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{3}}-\frac {7 x}{3 \sqrt {b \,x^{2}+a x}\, b^{2}}+\frac {\ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {5}{2}}}-\frac {a}{6 \sqrt {b \,x^{2}+a x}\, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a*x)^(5/2),x)

[Out]

-1/3*x^3/b/(b*x^2+a*x)^(3/2)+1/2*a/b^2*x^2/(b*x^2+a*x)^(3/2)+1/6*a^2/b^3/(b*x^2+a*x)^(3/2)*x-7/3*x/b^2/(b*x^2+
a*x)^(1/2)-1/6*a/b^3/(b*x^2+a*x)^(1/2)+1/b^(5/2)*ln((b*x+1/2*a)/b^(1/2)+(b*x^2+a*x)^(1/2))

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maxima [B]  time = 1.50, size = 140, normalized size = 1.97 \[ -\frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {a x}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, x}{\sqrt {b x^{2} + a x} a b} - \frac {1}{\sqrt {b x^{2} + a x} b^{2}}\right )} - \frac {4 \, x}{3 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {\log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sqrt {b x^{2} + a x}}{3 \, a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

-1/3*x*(3*x^2/((b*x^2 + a*x)^(3/2)*b) + a*x/((b*x^2 + a*x)^(3/2)*b^2) - 2*x/(sqrt(b*x^2 + a*x)*a*b) - 1/(sqrt(
b*x^2 + a*x)*b^2)) - 4/3*x/(sqrt(b*x^2 + a*x)*b^2) + log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 2/
3*sqrt(b*x^2 + a*x)/(a*b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x + b*x^2)^(5/2),x)

[Out]

int(x^4/(a*x + b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a*x)**(5/2),x)

[Out]

Integral(x**4/(x*(a + b*x))**(5/2), x)

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