### 3.61 $$\int \frac{x^5}{(a x+b x^2)^{5/2}} \, dx$$

Optimal. Leaf size=94 $-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \sqrt{a x+b x^2}}{b^3}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{b^{7/2}}-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}$

[Out]

(-2*x^4)/(3*b*(a*x + b*x^2)^(3/2)) - (10*x^2)/(3*b^2*Sqrt[a*x + b*x^2]) + (5*Sqrt[a*x + b*x^2])/b^3 - (5*a*Arc
Tanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/b^(7/2)

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Rubi [A]  time = 0.0414738, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {668, 640, 620, 206} $-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \sqrt{a x+b x^2}}{b^3}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{b^{7/2}}-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^5/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*x^4)/(3*b*(a*x + b*x^2)^(3/2)) - (10*x^2)/(3*b^2*Sqrt[a*x + b*x^2]) + (5*Sqrt[a*x + b*x^2])/b^3 - (5*a*Arc
Tanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/b^(7/2)

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 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[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^5}{\left (a x+b x^2\right )^{5/2}} \, dx &=-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}+\frac{5 \int \frac{x^3}{\left (a x+b x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \int \frac{x}{\sqrt{a x+b x^2}} \, dx}{b^2}\\ &=-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \sqrt{a x+b x^2}}{b^3}-\frac{(5 a) \int \frac{1}{\sqrt{a x+b x^2}} \, dx}{2 b^3}\\ &=-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \sqrt{a x+b x^2}}{b^3}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a x+b x^2}}\right )}{b^3}\\ &=-\frac{2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}-\frac{10 x^2}{3 b^2 \sqrt{a x+b x^2}}+\frac{5 \sqrt{a x+b x^2}}{b^3}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a x+b x^2}}\right )}{b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0131385, size = 50, normalized size = 0.53 $\frac{2 x^4 \sqrt{\frac{b x}{a}+1} \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )}{7 a^2 \sqrt{x (a+b x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^5/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x^4*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[5/2, 7/2, 9/2, -((b*x)/a)])/(7*a^2*Sqrt[x*(a + b*x)])

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Maple [A]  time = 0.047, size = 149, normalized size = 1.6 \begin{align*}{\frac{{x}^{4}}{b} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,a{x}^{3}}{6\,{b}^{2}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,{a}^{2}{x}^{2}}{4\,{b}^{3}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}-{\frac{5\,x{a}^{3}}{12\,{b}^{4}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{3}{2}}}}+{\frac{35\,ax}{6\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}+{\frac{5\,{a}^{2}}{12\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+ax}}}}-{\frac{5\,a}{2}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{7}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(b*x^2+a*x)^(5/2),x)

[Out]

x^4/b/(b*x^2+a*x)^(3/2)+5/6/b^2*a*x^3/(b*x^2+a*x)^(3/2)-5/4/b^3*a^2*x^2/(b*x^2+a*x)^(3/2)-5/12/b^4*a^3/(b*x^2+
a*x)^(3/2)*x+35/6/b^3*a/(b*x^2+a*x)^(1/2)*x+5/12/b^4*a^2/(b*x^2+a*x)^(1/2)-5/2/b^(7/2)*a*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^5/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92178, size = 493, normalized size = 5.24 \begin{align*} \left [\frac{15 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{b} \log \left (2 \, b x + a - 2 \, \sqrt{b x^{2} + a x} \sqrt{b}\right ) + 2 \,{\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x^{2} + a x}}{6 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{15 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x^{2} + a x} \sqrt{-b}}{b x}\right ) +{\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x^{2} + a x}}{3 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(3*b^3*x^2 + 2
0*a*b^2*x + 15*a^2*b)*sqrt(b*x^2 + a*x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3
)*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (3*b^3*x^2 + 20*a*b^2*x + 15*a^2*b)*sqrt(b*x^2 + a*x))/(
b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\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**5/(b*x**2+a*x)**(5/2),x)

[Out]

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

[Out]

Exception raised: TypeError