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

Optimal. Leaf size=112 \[ \frac{5 a^2 \sqrt{a x^3+b x^4}}{8 b^3 x}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x^3+b x^4}}\right )}{8 b^{7/2}}-\frac{5 a \sqrt{a x^3+b x^4}}{12 b^2}+\frac{x \sqrt{a x^3+b x^4}}{3 b} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.178822, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2024, 2029, 206} \[ \frac{5 a^2 \sqrt{a x^3+b x^4}}{8 b^3 x}-\frac{5 a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x^3+b x^4}}\right )}{8 b^{7/2}}-\frac{5 a \sqrt{a x^3+b x^4}}{12 b^2}+\frac{x \sqrt{a x^3+b x^4}}{3 b} \]

Antiderivative was successfully verified.

[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 2024

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 2029

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]

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])

Rubi steps

\begin{align*} \int \frac{x^4}{\sqrt{a x^3+b x^4}} \, dx &=\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 ) \operatorname{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]  time = 0.145038, size = 94, normalized size = 0.84 \[ \frac{\sqrt{x^3 (a+b x)} \left (\sqrt{b} \sqrt{x} \left (15 a^2-10 a b x+8 b^2 x^2\right )-\frac{15 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{24 b^{7/2} x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a*x^3 + b*x^4],x]

[Out]

(Sqrt[x^3*(a + b*x)]*(Sqrt[b]*Sqrt[x]*(15*a^2 - 10*a*b*x + 8*b^2*x^2) - (15*a^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[x])/
Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(24*b^(7/2)*x^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 120, normalized size = 1.1 \begin{align*}{\frac{x}{48}\sqrt{x \left ( bx+a \right ) } \left ( 16\,{x}^{2}\sqrt{b{x}^{2}+ax}{b}^{7/2}-20\,\sqrt{b{x}^{2}+ax}{b}^{5/2}xa+30\,\sqrt{b{x}^{2}+ax}{b}^{3/2}{a}^{2}-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{b{x}^{2}+ax}\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b \right ){\frac{1}{\sqrt{b{x}^{4}+a{x}^{3}}}}{b}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*x*(x*(b*x+a))^(1/2)*(16*x^2*(b*x^2+a*x)^(1/2)*b^(7/2)-20*(b*x^2+a*x)^(1/2)*b^(5/2)*x*a+30*(b*x^2+a*x)^(1/
2)*b^(3/2)*a^2-15*ln(1/2*(2*(b*x^2+a*x)^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b)/(b*x^4+a*x^3)^(1/2)/b^(9/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{b x^{4} + a x^{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 0.91539, size = 390, normalized size = 3.48 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{x^{3} \left (a + b x\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{b x^{4} + a x^{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/sqrt(b*x^4 + a*x^3), x)