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

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

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Rubi [A]  time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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])

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 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 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]

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*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.53 \begin {gather*} \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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.47, size = 90, normalized size = 0.96 \begin {gather*} \frac {\sqrt {a x+b x^2} \left (15 a^2+20 a b x+3 b^2 x^2\right )}{3 b^3 (a+b x)^2}+\frac {5 a \log \left (-2 b^{7/2} \sqrt {a x+b x^2}+a b^3+2 b^4 x\right )}{2 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/(a*x + b*x^2)^(5/2),x]

[Out]

(Sqrt[a*x + b*x^2]*(15*a^2 + 20*a*b*x + 3*b^2*x^2))/(3*b^3*(a + b*x)^2) + (5*a*Log[a*b^3 + 2*b^4*x - 2*b^(7/2)
*Sqrt[a*x + b*x^2]])/(2*b^(7/2))

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fricas [A]  time = 0.46, size = 221, normalized size = 2.35 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.27, size = 146, normalized size = 1.55 \begin {gather*} \frac {5 \, a \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{2 \, b^{\frac {7}{2}}} + \frac {\sqrt {b x^{2} + a x}}{b^{3}} + \frac {2 \, {\left (9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{2} b + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{3} \sqrt {b} + 7 \, a^{4}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} b + a \sqrt {b}\right )}^{3} b^{2}} \end {gather*}

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]

5/2*a*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(7/2) + sqrt(b*x^2 + a*x)/b^3 + 2/3*(9*(sqrt(
b)*x - sqrt(b*x^2 + a*x))^2*a^2*b + 15*(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b) + 7*a^4)/(((sqrt(b)*x - sqr
t(b*x^2 + a*x))*b + a*sqrt(b))^3*b^2)

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maple [A]  time = 0.05, size = 149, normalized size = 1.59 \begin {gather*} \frac {x^{4}}{\left (b \,x^{2}+a x \right )^{\frac {3}{2}} b}+\frac {5 a \,x^{3}}{6 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{2}}-\frac {5 a^{2} x^{2}}{4 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{3}}-\frac {5 a^{3} x}{12 \left (b \,x^{2}+a x \right )^{\frac {3}{2}} b^{4}}+\frac {35 a x}{6 \sqrt {b \,x^{2}+a x}\, b^{3}}-\frac {5 a \ln \left (\frac {b x +\frac {a}{2}}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {7}{2}}}+\frac {5 a^{2}}{12 \sqrt {b \,x^{2}+a x}\, b^{4}} \end {gather*}

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*a/b^2*x^3/(b*x^2+a*x)^(3/2)-5/4*a^2/b^3*x^2/(b*x^2+a*x)^(3/2)-5/12*a^3/b^4/(b*x^2+
a*x)^(3/2)*x+35/6*a/b^3/(b*x^2+a*x)^(1/2)*x+5/12*a^2/b^4/(b*x^2+a*x)^(1/2)-5/2*a/b^(7/2)*ln((b*x+1/2*a)/b^(1/2
)+(b*x^2+a*x)^(1/2))

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maxima [B]  time = 1.50, size = 162, normalized size = 1.72 \begin {gather*} \frac {5 \, a 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 )}}{6 \, b} + \frac {x^{4}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {10 \, a x}{3 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {5 \, a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {7}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x}}{3 \, b^{3}} \end {gather*}

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]

5/6*a*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))/b + x^4/((b*x^2 + a*x)^(3/2)*b) + 10/3*a*x/(sqrt(b*x^2 + a*x)*b^3) - 5/2*a*log(2*b*x + a +
2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 5/3*sqrt(b*x^2 + a*x)/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^5}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a*x + b*x^2)^(5/2),x)

[Out]

int(x^5/(a*x + b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

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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