\(\int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx\) [310]

Optimal result

Integrand size = 19, antiderivative size = 112 \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=-\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {5 a^2 \sqrt {a x^3+b x^4}}{8 b^3 x}+\frac {x \sqrt {a x^3+b x^4}}{3 b}-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{8 b^{7/2}} \]

[Out]

-5/8*a^3*arctanh(x^2*b^(1/2)/(b*x^4+a*x^3)^(1/2))/b^(7/2)-5/12*a*(b*x^4+a*x^3)^(1/2)/b^2+5/8*a^2*(b*x^4+a*x^3)
^(1/2)/b^3/x+1/3*x*(b*x^4+a*x^3)^(1/2)/b

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2049, 2054, 212} \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{8 b^{7/2}}+\frac {5 a^2 \sqrt {a x^3+b x^4}}{8 b^3 x}-\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {x \sqrt {a x^3+b x^4}}{3 b} \]

[In]

Int[x^4/Sqrt[a*x^3 + b*x^4],x]

[Out]

(-5*a*Sqrt[a*x^3 + b*x^4])/(12*b^2) + (5*a^2*Sqrt[a*x^3 + b*x^4])/(8*b^3*x) + (x*Sqrt[a*x^3 + b*x^4])/(3*b) -
(5*a^3*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x^3 + b*x^4]])/(8*b^(7/2))

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 2049

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n +
1)*((a*x^j + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^(n - j)*((m + j*p - n + j + 1)/(b*(m + n*p + 1))
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \frac {x \sqrt {a x^3+b x^4}}{3 b}-\frac {(5 a) \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx}{6 b} \\ & = -\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {x \sqrt {a x^3+b x^4}}{3 b}+\frac {\left (5 a^2\right ) \int \frac {x^2}{\sqrt {a x^3+b x^4}} \, dx}{8 b^2} \\ & = -\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {5 a^2 \sqrt {a x^3+b x^4}}{8 b^3 x}+\frac {x \sqrt {a x^3+b x^4}}{3 b}-\frac {\left (5 a^3\right ) \int \frac {x}{\sqrt {a x^3+b x^4}} \, dx}{16 b^3} \\ & = -\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {5 a^2 \sqrt {a x^3+b x^4}}{8 b^3 x}+\frac {x \sqrt {a x^3+b x^4}}{3 b}-\frac {\left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a x^3+b x^4}}\right )}{8 b^3} \\ & = -\frac {5 a \sqrt {a x^3+b x^4}}{12 b^2}+\frac {5 a^2 \sqrt {a x^3+b x^4}}{8 b^3 x}+\frac {x \sqrt {a x^3+b x^4}}{3 b}-\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{8 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\frac {\sqrt {b} x^2 \left (15 a^3+5 a^2 b x-2 a b^2 x^2+8 b^3 x^3\right )+30 a^3 x^{3/2} \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )}{24 b^{7/2} \sqrt {x^3 (a+b x)}} \]

[In]

Integrate[x^4/Sqrt[a*x^3 + b*x^4],x]

[Out]

(Sqrt[b]*x^2*(15*a^3 + 5*a^2*b*x - 2*a*b^2*x^2 + 8*b^3*x^3) + 30*a^3*x^(3/2)*Sqrt[a + b*x]*ArcTanh[(Sqrt[b]*Sq
rt[x])/(Sqrt[a] - Sqrt[a + b*x])])/(24*b^(7/2)*Sqrt[x^3*(a + b*x)])

Maple [A] (verified)

Time = 2.43 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {8 b^{\frac {5}{2}} x^{2} \sqrt {x^{3} \left (b x +a \right )}-10 a \,b^{\frac {3}{2}} x \sqrt {x^{3} \left (b x +a \right )}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{3} \left (b x +a \right )}}{x^{2} \sqrt {b}}\right ) a^{3} x +15 a^{2} \sqrt {x^{3} \left (b x +a \right )}\, \sqrt {b}}{24 b^{\frac {7}{2}} x}\)	\(91\)
risch	\(\frac {\left (8 b^{2} x^{2}-10 a b x +15 a^{2}\right ) x^{2} \left (b x +a \right )}{24 b^{3} \sqrt {x^{3} \left (b x +a \right )}}-\frac {5 a^{3} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) x \sqrt {x \left (b x +a \right )}}{16 b^{\frac {7}{2}} \sqrt {x^{3} \left (b x +a \right )}}\)	\(98\)
default	\(\frac {x \sqrt {x \left (b x +a \right )}\, \left (16 x^{2} \sqrt {b \,x^{2}+a x}\, b^{\frac {7}{2}}-20 \sqrt {b \,x^{2}+a x}\, b^{\frac {5}{2}} a x +30 \sqrt {b \,x^{2}+a x}\, b^{\frac {3}{2}} a^{2}-15 \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{3} b \right )}{48 \sqrt {b \,x^{4}+a \,x^{3}}\, b^{\frac {9}{2}}}\)	\(120\)

[In]

int(x^4/(b*x^4+a*x^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(8*b^(5/2)*x^2*(x^3*(b*x+a))^(1/2)-10*a*b^(3/2)*x*(x^3*(b*x+a))^(1/2)-15*arctanh((x^3*(b*x+a))^(1/2)/x^2/
b^(1/2))*a^3*x+15*a^2*(x^3*(b*x+a))^(1/2)*b^(1/2))/b^(7/2)/x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.53 \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\left [\frac {15 \, a^{3} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{4} + a x^{3}} \sqrt {b}}{x}\right ) + 2 \, {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{4} + a x^{3}}}{48 \, b^{4} x}, \frac {15 \, a^{3} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{4} + a x^{3}} \sqrt {-b}}{b x^{2}}\right ) + {\left (8 \, b^{3} x^{2} - 10 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{4} + a x^{3}}}{24 \, b^{4} x}\right ] \]

[In]

integrate(x^4/(b*x^4+a*x^3)^(1/2),x, algorithm="fricas")

[Out]

[1/48*(15*a^3*sqrt(b)*x*log((2*b*x^2 + a*x - 2*sqrt(b*x^4 + a*x^3)*sqrt(b))/x) + 2*(8*b^3*x^2 - 10*a*b^2*x + 1
5*a^2*b)*sqrt(b*x^4 + a*x^3))/(b^4*x), 1/24*(15*a^3*sqrt(-b)*x*arctan(sqrt(b*x^4 + a*x^3)*sqrt(-b)/(b*x^2)) +
(8*b^3*x^2 - 10*a*b^2*x + 15*a^2*b)*sqrt(b*x^4 + a*x^3))/(b^4*x)]

Sympy [F]

\[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\int \frac {x^{4}}{\sqrt {x^{3} \left (a + b x\right )}}\, dx \]

[In]

integrate(x**4/(b*x**4+a*x**3)**(1/2),x)

[Out]

Integral(x**4/sqrt(x**3*(a + b*x)), x)

Maxima [F]

\[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {b x^{4} + a x^{3}}} \,d x } \]

[In]

integrate(x^4/(b*x^4+a*x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(b*x^4 + a*x^3), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93 \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\frac {1}{24} \, \sqrt {b x^{2} + a x} {\left (2 \, x {\left (\frac {4 \, x}{b \mathrm {sgn}\left (x\right )} - \frac {5 \, a}{b^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {15 \, a^{2}}{b^{3} \mathrm {sgn}\left (x\right )}\right )} - \frac {5 \, a^{3} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{16 \, b^{\frac {7}{2}}} + \frac {5 \, a^{3} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{16 \, b^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^4/(b*x^4+a*x^3)^(1/2),x, algorithm="giac")

[Out]

1/24*sqrt(b*x^2 + a*x)*(2*x*(4*x/(b*sgn(x)) - 5*a/(b^2*sgn(x))) + 15*a^2/(b^3*sgn(x))) - 5/16*a^3*log(abs(a))*
sgn(x)/b^(7/2) + 5/16*a^3*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/(b^(7/2)*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {a x^3+b x^4}} \, dx=\int \frac {x^4}{\sqrt {b\,x^4+a\,x^3}} \,d x \]

[In]

int(x^4/(a*x^3 + b*x^4)^(1/2),x)

[Out]

int(x^4/(a*x^3 + b*x^4)^(1/2), x)