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

Optimal result

Integrand size = 17, antiderivative size = 71 \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {682, 666, 634, 212} \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{5/2}}-\frac {2 x}{b^2 \sqrt {a x+b x^2}}-\frac {2 x^3}{3 b \left (a x+b x^2\right )^{3/2}} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.18 \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 x \left (\sqrt {b} x (3 a+4 b x)+6 \sqrt {x} (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )\right )}{3 b^{5/2} (x (a+b x))^{3/2}} \]

[In]

Integrate[x^4/(a*x + b*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03

method	result	size
pseudoelliptic	\(\frac {6 \sqrt {x \left (b x +a \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right ) \left (b x +a \right )-6 x a \sqrt {b}-8 b^{\frac {3}{2}} x^{2}}{b^{\frac {5}{2}} \sqrt {x \left (b x +a \right )}\, \left (3 b x +3 a \right )}\)	\(73\)
default	\(-\frac {x^{3}}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\right )}{2 b}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}}{b}\)	\(243\)

[In]

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

[Out]

6/b^(5/2)*((x*(b*x+a))^(1/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))*(b*x+a)-x*a*b^(1/2)-4/3*b^(3/2)*x^2)/(x*(b*x
+a))^(1/2)/(3*b*x+3*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 193, normalized size of antiderivative = 2.72 \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) - 2 \, {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {2 \, {\left (3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt {b x^{2} + a x}\right )}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]

[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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (59) = 118\).

Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.97 \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {a x}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, x}{\sqrt {b x^{2} + a x} a b} - \frac {1}{\sqrt {b x^{2} + a x} b^{2}}\right )} - \frac {4 \, x}{3 \, \sqrt {b x^{2} + a x} b^{2}} + \frac {\log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sqrt {b x^{2} + a x}}{3 \, a b^{2}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (59) = 118\).

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.75 \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {\log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{b^{\frac {5}{2}}} - \frac {2 \, {\left (6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a b + 9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{2} \sqrt {b} + 4 \, a^{3}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )}^{3} b^{\frac {5}{2}}} \]

[In]

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

[Out]

-log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(5/2) - 2/3*(6*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a*
b + 9*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^2*sqrt(b) + 4*a^3)/(((sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a)^3*b^
(5/2))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \]

[In]

int(x^4/(a*x + b*x^2)^(5/2),x)

[Out]

int(x^4/(a*x + b*x^2)^(5/2), x)