3.1.61 \(\int \frac {x^5}{(a x+b x^2)^{5/2}} \, dx\) [61]

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

[Out]

-2/3*x^4/b/(b*x^2+a*x)^(3/2)-5*a*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(7/2)-10/3*x^2/b^2/(b*x^2+a*x)^(1/2)+5
*(b*x^2+a*x)^(1/2)/b^3

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Rubi [A]
time = 0.03, 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 = {682, 654, 634, 212} \begin {gather*} -\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}} \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 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 634

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 654

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 682

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) \text {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 [A]
time = 0.13, size = 88, normalized size = 0.94 \begin {gather*} \frac {x \left (\sqrt {b} x \left (15 a^2+20 a b x+3 b^2 x^2\right )+15 a \sqrt {x} (a+b x)^{3/2} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{3 b^{7/2} (x (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a*x + b*x^2)^(5/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(267\) vs. \(2(78)=156\).
time = 0.43, size = 268, normalized size = 2.85

method	result	size
risch	\(\frac {x \left (b x +a \right )}{b^{3} \sqrt {x \left (b x +a \right )}}-\frac {5 a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {7}{2}}}+\frac {14 a \sqrt {\left (x +\frac {a}{b}\right )^{2} b -\left (x +\frac {a}{b}\right ) a}}{3 b^{4} \left (x +\frac {a}{b}\right )}-\frac {2 a^{2} \sqrt {\left (x +\frac {a}{b}\right )^{2} b -\left (x +\frac {a}{b}\right ) a}}{3 b^{5} \left (x +\frac {a}{b}\right )^{2}}\)	\(131\)
default	\(\frac {x^{4}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\right )}{2 b}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\)	\(268\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^4/b/(b*x^2+a*x)^(3/2)-5/2*a/b*(-1/3*x^3/b/(b*x^2+a*x)^(3/2)-1/2*a/b*(-x^2/b/(b*x^2+a*x)^(3/2)+1/2*a/b*(-1/2*
x/b/(b*x^2+a*x)^(3/2)-1/4*a/b*(-1/3/b/(b*x^2+a*x)^(3/2)-1/2*a/b*(-2/3*(2*b*x+a)/a^2/(b*x^2+a*x)^(3/2)+16/3*b*(
2*b*x+a)/a^4/(b*x^2+a*x)^(1/2)))))+1/b*(-x/b/(b*x^2+a*x)^(1/2)-1/2*a/b*(-1/b/(b*x^2+a*x)^(1/2)+1/a/b*(2*b*x+a)
/(b*x^2+a*x)^(1/2))+1/b^(3/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).
time = 0.28, 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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Fricas [A]
time = 1.63, 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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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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Giac [A]
time = 1.15, 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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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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