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

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

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Rubi [A]  time = 0.15, antiderivative size = 174, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.19, size = 151, normalized size = 0.87 \begin {gather*} \frac {\left (a \sqrt {x}+b\right ) \left (\sqrt {a} \sqrt {x} \sqrt {\frac {a \sqrt {x}}{b}+1} \left (128 a^4 x^2-144 a^3 b x^{3/2}+168 a^2 b^2 x-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}}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.31, size = 113, normalized size = 0.65 \begin {gather*} \frac {63 b^5 \log \left (-2 \sqrt {a} \sqrt {a x+b \sqrt {x}}+2 a \sqrt {x}+b\right )}{128 a^{11/2}}+\frac {\sqrt {a x+b \sqrt {x}} \left (128 a^4 x^2-144 a^3 b x^{3/2}+168 a^2 b^2 x-210 a b^3 \sqrt {x}+315 b^4\right )}{320 a^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sqrt[x] + a*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))/(320*a
^5) + (63*b^5*Log[b + 2*a*Sqrt[x] - 2*Sqrt[a]*Sqrt[b*Sqrt[x] + a*x]])/(128*a^(11/2))

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fricas [F(-1)]  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)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.29, size = 111, normalized size = 0.64 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.09, size = 223, normalized size = 1.28 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (-640 a \,b^{5} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+325 a \,b^{5} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-1300 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{3} \sqrt {x}+256 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} x +1280 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{4}-650 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{4}-544 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b \sqrt {x}+880 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}\right )}{640 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {13}{2}}} \end {gather*}

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {a x + b \sqrt {x}}}\,{d x} \end {gather*}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x + b*x^(1/2))^(1/2),x)

[Out]

int(x^2/(a*x + b*x^(1/2))^(1/2), x)

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

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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