3.103 \(\int \frac{x^2}{\sqrt{b \sqrt{x}+a x}} \, dx\)

Optimal. Leaf size=174 \[ \frac{63 b^4 \sqrt{a x+b \sqrt{x}}}{64 a^5}-\frac{21 b^3 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{32 a^4}+\frac{21 b^2 x \sqrt{a x+b \sqrt{x}}}{40 a^3}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{64 a^{11/2}}-\frac{9 b x^{3/2} \sqrt{a x+b \sqrt{x}}}{20 a^2}+\frac{2 x^2 \sqrt{a x+b \sqrt{x}}}{5 a} \]

[Out]

(63*b^4*Sqrt[b*Sqrt[x] + a*x])/(64*a^5) - (21*b^3*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (21*b^2*x*Sqrt[b*S
qrt[x] + a*x])/(40*a^3) - (9*b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(20*a^2) + (2*x^2*Sqrt[b*Sqrt[x] + a*x])/(5*a) -
(63*b^5*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(64*a^(11/2))

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Rubi [A]  time = 0.151043, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2018, 670, 640, 620, 206} \[ \frac{63 b^4 \sqrt{a x+b \sqrt{x}}}{64 a^5}-\frac{21 b^3 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{32 a^4}+\frac{21 b^2 x \sqrt{a x+b \sqrt{x}}}{40 a^3}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{64 a^{11/2}}-\frac{9 b x^{3/2} \sqrt{a x+b \sqrt{x}}}{20 a^2}+\frac{2 x^2 \sqrt{a x+b \sqrt{x}}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(63*b^4*Sqrt[b*Sqrt[x] + a*x])/(64*a^5) - (21*b^3*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (21*b^2*x*Sqrt[b*S
qrt[x] + a*x])/(40*a^3) - (9*b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(20*a^2) + (2*x^2*Sqrt[b*Sqrt[x] + a*x])/(5*a) -
(63*b^5*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(64*a^(11/2))

Rule 2018

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]

Rule 670

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 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^2}{\sqrt{b \sqrt{x}+a x}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^5}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}-\frac{(9 b) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{5 a}\\ &=-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}+\frac{\left (63 b^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{40 a^2}\\ &=\frac{21 b^2 x \sqrt{b \sqrt{x}+a x}}{40 a^3}-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}-\frac{\left (21 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{16 a^3}\\ &=-\frac{21 b^3 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{21 b^2 x \sqrt{b \sqrt{x}+a x}}{40 a^3}-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}+\frac{\left (63 b^4\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{64 a^4}\\ &=\frac{63 b^4 \sqrt{b \sqrt{x}+a x}}{64 a^5}-\frac{21 b^3 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{21 b^2 x \sqrt{b \sqrt{x}+a x}}{40 a^3}-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}-\frac{\left (63 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{128 a^5}\\ &=\frac{63 b^4 \sqrt{b \sqrt{x}+a x}}{64 a^5}-\frac{21 b^3 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{21 b^2 x \sqrt{b \sqrt{x}+a x}}{40 a^3}-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}-\frac{\left (63 b^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{64 a^5}\\ &=\frac{63 b^4 \sqrt{b \sqrt{x}+a x}}{64 a^5}-\frac{21 b^3 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{21 b^2 x \sqrt{b \sqrt{x}+a x}}{40 a^3}-\frac{9 b x^{3/2} \sqrt{b \sqrt{x}+a x}}{20 a^2}+\frac{2 x^2 \sqrt{b \sqrt{x}+a x}}{5 a}-\frac{63 b^5 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{64 a^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.170274, size = 151, normalized size = 0.87 \[ \frac{\left (a \sqrt{x}+b\right ) \left (\sqrt{a} \sqrt{x} \sqrt{\frac{a \sqrt{x}}{b}+1} \left (168 a^2 b^2 x-144 a^3 b x^{3/2}+128 a^4 x^2-210 a b^3 \sqrt{x}+315 b^4\right )-315 b^{9/2} \sqrt [4]{x} \sinh ^{-1}\left (\frac{\sqrt{a} \sqrt [4]{x}}{\sqrt{b}}\right )\right )}{320 a^{11/2} \sqrt{\frac{a \sqrt{x}}{b}+1} \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Sqrt[x])*(Sqrt[a]*Sqrt[1 + (a*Sqrt[x])/b]*Sqrt[x]*(315*b^4 - 210*a*b^3*Sqrt[x] + 168*a^2*b^2*x - 144*a
^3*b*x^(3/2) + 128*a^4*x^2) - 315*b^(9/2)*x^(1/4)*ArcSinh[(Sqrt[a]*x^(1/4))/Sqrt[b]]))/(320*a^(11/2)*Sqrt[1 +
(a*Sqrt[x])/b]*Sqrt[b*Sqrt[x] + a*x])

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Maple [A]  time = 0.06, size = 223, normalized size = 1.3 \begin{align*}{\frac{1}{640}\sqrt{b\sqrt{x}+ax} \left ( -544\,\sqrt{x}{a}^{7/2} \left ( b\sqrt{x}+ax \right ) ^{3/2}b+256\,x \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{9/2}-1300\,\sqrt{x}{a}^{5/2}\sqrt{b\sqrt{x}+ax}{b}^{3}+880\,{a}^{5/2} \left ( b\sqrt{x}+ax \right ) ^{3/2}{b}^{2}+1280\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{a}^{3/2}{b}^{4}-640\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ) a{b}^{5}-650\,{a}^{3/2}\sqrt{b\sqrt{x}+ax}{b}^{4}+325\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ) a{b}^{5} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{a}^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^(1/2)+a*x)^(1/2),x)

[Out]

1/640*(b*x^(1/2)+a*x)^(1/2)*(-544*x^(1/2)*a^(7/2)*(b*x^(1/2)+a*x)^(3/2)*b+256*x*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)-
1300*x^(1/2)*a^(5/2)*(b*x^(1/2)+a*x)^(1/2)*b^3+880*a^(5/2)*(b*x^(1/2)+a*x)^(3/2)*b^2+1280*(x^(1/2)*(b+a*x^(1/2
)))^(1/2)*a^(3/2)*b^4-640*ln(1/2*(2*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*a^(1/2)+2*a*x^(1/2)+b)/a^(1/2))*a*b^5-650*a^
(3/2)*(b*x^(1/2)+a*x)^(1/2)*b^4+325*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)*(b+a*x^(1/2)))^(1/2)/a^(13/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a x + b \sqrt{x}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\sqrt{a x + b \sqrt{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.39757, size = 150, normalized size = 0.86 \begin{align*} \frac{1}{320} \, \sqrt{a x + b \sqrt{x}}{\left (2 \,{\left (4 \,{\left (2 \, \sqrt{x}{\left (\frac{8 \, \sqrt{x}}{a} - \frac{9 \, b}{a^{2}}\right )} + \frac{21 \, b^{2}}{a^{3}}\right )} \sqrt{x} - \frac{105 \, b^{3}}{a^{4}}\right )} \sqrt{x} + \frac{315 \, b^{4}}{a^{5}}\right )} + \frac{63 \, b^{5} \log \left ({\left | -2 \, \sqrt{a}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )} - b \right |}\right )}{128 \, a^{\frac{11}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*sqrt(a*x + b*sqrt(x))*(2*(4*(2*sqrt(x)*(8*sqrt(x)/a - 9*b/a^2) + 21*b^2/a^3)*sqrt(x) - 105*b^3/a^4)*sqrt
(x) + 315*b^4/a^5) + 63/128*b^5*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(11/2)