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

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

[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)

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Rubi [A]  time = 0.0300053, antiderivative size = 71, normalized size of antiderivative = 1., 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 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]

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 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 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}{\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*}

Mathematica [A]  time = 0.160041, 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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Maple [B]  time = 0.044, size = 123, normalized size = 1.7 \begin{align*} -{\frac{{x}^{3}}{3\,b} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{a{x}^{2}}{2\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{2}x}{6\,{b}^{3}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{7\,x}{3\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}-{\frac{a}{6\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}+{\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.02773, size = 432, normalized size = 6.08 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\left (x \left (a + b x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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