3.1.62 \(\int \frac {x^4}{(a x+b x^2)^{5/2}} \, dx\) [62]

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

[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 = 0.02, 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 = {682, 666, 634, 212} \begin {gather*} \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}} \end {gather*}

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 212

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

Rule 634

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 666

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 682

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 \text {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 = 0.12, size = 76, normalized size = 1.07 \begin {gather*} -\frac {2 x \left (\sqrt {b} x (3 a+4 b x)+3 \sqrt {x} (a+b x)^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{3 b^{5/2} (x (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a*x + b*x^2)^(5/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(242\) vs. \(2(59)=118\).
time = 0.39, size = 243, normalized size = 3.42

method	result	size
default	\(-\frac {x^{3}}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\right )}{2 b}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}}{b}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3/b/(b*x^2+a*x)^(3/2)-1/2*a/b*(-x^2/b/(b*x^2+a*x)^(3/2)+1/2*a/b*(-1/2*x/b/(b*x^2+a*x)^(3/2)-1/4*a/b*(-1
/3/b/(b*x^2+a*x)^(3/2)-1/2*a/b*(-2/3*(2*b*x+a)/a^2/(b*x^2+a*x)^(3/2)+16/3*b*(2*b*x+a)/a^4/(b*x^2+a*x)^(1/2))))
)+1/b*(-x/b/(b*x^2+a*x)^(1/2)-1/2*a/b*(-1/b/(b*x^2+a*x)^(1/2)+1/a/b*(2*b*x+a)/(b*x^2+a*x)^(1/2))+1/b^(3/2)*ln(
(1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2)))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (59) = 118\).
time = 0.28, size = 140, normalized size = 1.97 \begin {gather*} -\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}} \end {gather*}

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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Fricas [A]
time = 1.87, size = 193, normalized size = 2.72 \begin {gather*} \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 ] \end {gather*}

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

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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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{1,[2]%%%},[4,4]%%%}+%%%{%%{[%%%{-4,[1]%%%},0]:[1,0,%
%%{-1,[1]%%

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

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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