3.2.11 \(\int \frac {x^2}{(b \sqrt {x}+a x)^{3/2}} \, dx\) [111]

Optimal. Leaf size=139 \[ -\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {35 b^2 \sqrt {b \sqrt {x}+a x}}{4 a^4}-\frac {35 b \sqrt {x} \sqrt {b \sqrt {x}+a x}}{6 a^3}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}-\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{4 a^{9/2}} \]

[Out]

-35/4*b^3*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(9/2)-4*x^2/a/(b*x^(1/2)+a*x)^(1/2)+35/4*b^2*(b*x^(
1/2)+a*x)^(1/2)/a^4+14/3*x*(b*x^(1/2)+a*x)^(1/2)/a^2-35/6*b*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)/a^3

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Rubi [A]
time = 0.09, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2043, 682, 684, 654, 634, 212} \begin {gather*} -\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{4 a^{9/2}}+\frac {35 b^2 \sqrt {a x+b \sqrt {x}}}{4 a^4}-\frac {35 b \sqrt {x} \sqrt {a x+b \sqrt {x}}}{6 a^3}+\frac {14 x \sqrt {a x+b \sqrt {x}}}{3 a^2}-\frac {4 x^2}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(-4*x^2)/(a*Sqrt[b*Sqrt[x] + a*x]) + (35*b^2*Sqrt[b*Sqrt[x] + a*x])/(4*a^4) - (35*b*Sqrt[x]*Sqrt[b*Sqrt[x] + a
*x])/(6*a^3) + (14*x*Sqrt[b*Sqrt[x] + a*x])/(3*a^2) - (35*b^3*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]]
)/(4*a^(9/2))

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]

Rule 684

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*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \text {Subst}\left (\int \frac {x^5}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {14 \text {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}-\frac {(35 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{3 a^2}\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}-\frac {35 b \sqrt {x} \sqrt {b \sqrt {x}+a x}}{6 a^3}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}+\frac {\left (35 b^2\right ) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{4 a^3}\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {35 b^2 \sqrt {b \sqrt {x}+a x}}{4 a^4}-\frac {35 b \sqrt {x} \sqrt {b \sqrt {x}+a x}}{6 a^3}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}-\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{8 a^4}\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {35 b^2 \sqrt {b \sqrt {x}+a x}}{4 a^4}-\frac {35 b \sqrt {x} \sqrt {b \sqrt {x}+a x}}{6 a^3}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}-\frac {\left (35 b^3\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{4 a^4}\\ &=-\frac {4 x^2}{a \sqrt {b \sqrt {x}+a x}}+\frac {35 b^2 \sqrt {b \sqrt {x}+a x}}{4 a^4}-\frac {35 b \sqrt {x} \sqrt {b \sqrt {x}+a x}}{6 a^3}+\frac {14 x \sqrt {b \sqrt {x}+a x}}{3 a^2}-\frac {35 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{4 a^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 119, normalized size = 0.86 \begin {gather*} \frac {\sqrt {b \sqrt {x}+a x} \left (105 b^3+35 a b^2 \sqrt {x}-14 a^2 b x+8 a^3 x^{3/2}\right )}{12 a^4 \left (b+a \sqrt {x}\right )}+\frac {35 b^3 \log \left (a^4 b+2 a^5 \sqrt {x}-2 a^{9/2} \sqrt {b \sqrt {x}+a x}\right )}{8 a^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(b*Sqrt[x] + a*x)^(3/2),x]

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(105*b^3 + 35*a*b^2*Sqrt[x] - 14*a^2*b*x + 8*a^3*x^(3/2)))/(12*a^4*(b + a*Sqrt[x])) + (
35*b^3*Log[a^4*b + 2*a^5*Sqrt[x] - 2*a^(9/2)*Sqrt[b*Sqrt[x] + a*x]])/(8*a^(9/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(502\) vs. \(2(103)=206\).
time = 0.39, size = 503, normalized size = 3.62

method	result	size
derivativedivides	\(\frac {2 x^{2}}{3 a \sqrt {b \sqrt {x}+a x}}-\frac {7 b \left (\frac {x^{\frac {3}{2}}}{2 a \sqrt {b \sqrt {x}+a x}}-\frac {5 b \left (\frac {x}{a \sqrt {b \sqrt {x}+a x}}-\frac {3 b \left (-\frac {\sqrt {x}}{a \sqrt {b \sqrt {x}+a x}}-\frac {b \left (-\frac {1}{a \sqrt {b \sqrt {x}+a x}}+\frac {b +2 a \sqrt {x}}{b a \sqrt {b \sqrt {x}+a x}}\right )}{2 a}+\frac {\ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )}{4 a}\right )}{3 a}\)	\(175\)
default	\(\frac {\sqrt {b \sqrt {x}+a x}\, \left (16 x \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {9}{2}}-60 \sqrt {b \sqrt {x}+a x}\, x^{\frac {3}{2}} a^{\frac {9}{2}} b +32 \sqrt {x}\, \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b -150 \sqrt {b \sqrt {x}+a x}\, x \,a^{\frac {7}{2}} b^{2}+240 x \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {7}{2}} b^{2}-120 x \,a^{3} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{3}+16 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}-120 \sqrt {x}\, \sqrt {b \sqrt {x}+a x}\, a^{\frac {5}{2}} b^{3}+15 x \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{3}+480 \sqrt {x}\, \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {5}{2}} b^{3}-240 \sqrt {x}\, a^{2} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{4}-96 \left (\sqrt {x}\, \left (a \sqrt {x}+b \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}-30 \sqrt {b \sqrt {x}+a x}\, a^{\frac {3}{2}} b^{4}+30 \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{4}+240 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}} b^{4}-120 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{5}+15 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{5}\right )}{24 a^{\frac {11}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \left (a \sqrt {x}+b \right )^{2}}\)	\(503\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^(1/2)+a*x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/24*(b*x^(1/2)+a*x)^(1/2)/a^(11/2)*(16*x*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)-60*(b*x^(1/2)+a*x)^(1/2)*x^(3/2)*a^(9/
2)*b+32*x^(1/2)*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*b-150*(b*x^(1/2)+a*x)^(1/2)*x*a^(7/2)*b^2+240*x*(x^(1/2)*(a*x^(1
/2)+b))^(1/2)*a^(7/2)*b^2-120*x*a^3*ln(1/2*(2*a*x^(1/2)+2*(x^(1/2)*(a*x^(1/2)+b))^(1/2)*a^(1/2)+b)/a^(1/2))*b^
3+16*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*b^2-120*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*b^3+15*x*ln(1/2*(2*a*x^(1/2)+
2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a^(1/2))*a^3*b^3+480*x^(1/2)*(x^(1/2)*(a*x^(1/2)+b))^(1/2)*a^(5/2)*b^3-240*
x^(1/2)*a^2*ln(1/2*(2*a*x^(1/2)+2*(x^(1/2)*(a*x^(1/2)+b))^(1/2)*a^(1/2)+b)/a^(1/2))*b^4-96*(x^(1/2)*(a*x^(1/2)
+b))^(3/2)*a^(5/2)*b^2-30*(b*x^(1/2)+a*x)^(1/2)*a^(3/2)*b^4+30*x^(1/2)*ln(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(
1/2)*a^(1/2)+b)/a^(1/2))*a^2*b^4+240*(x^(1/2)*(a*x^(1/2)+b))^(1/2)*a^(3/2)*b^4-120*ln(1/2*(2*a*x^(1/2)+2*(x^(1
/2)*(a*x^(1/2)+b))^(1/2)*a^(1/2)+b)/a^(1/2))*a*b^5+15*ln(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a
^(1/2))*a*b^5)/(x^(1/2)*(a*x^(1/2)+b))^(1/2)/(a*x^(1/2)+b)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(a*x + b*sqrt(x))^(3/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (a x + b \sqrt {x}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2/(a*x + b*sqrt(x))**(3/2), x)

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Giac [A]
time = 0.70, size = 122, normalized size = 0.88 \begin {gather*} \frac {1}{12} \, \sqrt {a x + b \sqrt {x}} {\left (2 \, \sqrt {x} {\left (\frac {4 \, \sqrt {x}}{a^{2}} - \frac {11 \, b}{a^{3}}\right )} + \frac {57 \, b^{2}}{a^{4}}\right )} + \frac {35 \, b^{3} \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{8 \, a^{\frac {9}{2}}} + \frac {4 \, b^{4}}{{\left (a {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + \sqrt {a} b\right )} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(a*x + b*sqrt(x))*(2*sqrt(x)*(4*sqrt(x)/a^2 - 11*b/a^3) + 57*b^2/a^4) + 35/8*b^3*log(abs(-2*sqrt(a)*(
sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(9/2) + 4*b^4/((a*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) +
sqrt(a)*b)*a^4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\left (a\,x+b\,\sqrt {x}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x + b*x^(1/2))^(3/2),x)

[Out]

int(x^2/(a*x + b*x^(1/2))^(3/2), x)

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