3.1.68 \(\int \frac {x^2}{(b \sqrt {x}+a x)^{3/2}} \, dx\)

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

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Rubi [A]  time = 0.13, 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 = {2018, 668, 670, 640, 620, 206} \begin {gather*} \frac {35 b^2 \sqrt {a x+b \sqrt {x}}}{4 a^4}-\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 \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 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]

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}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \operatorname {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 \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 [C]  time = 0.08, size = 64, normalized size = 0.46 \begin {gather*} \frac {4 x^{5/2} \sqrt {\frac {a \sqrt {x}}{b}+1} \, _2F_1\left (\frac {3}{2},\frac {9}{2};\frac {11}{2};-\frac {a \sqrt {x}}{b}\right )}{9 b \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[1 + (a*Sqrt[x])/b]*x^(5/2)*Hypergeometric2F1[3/2, 9/2, 11/2, -((a*Sqrt[x])/b)])/(9*b*Sqrt[b*Sqrt[x] +
a*x])

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IntegrateAlgebraic [A]  time = 0.52, size = 119, normalized size = 0.86 \begin {gather*} \frac {35 b^3 \log \left (-2 a^{9/2} \sqrt {a x+b \sqrt {x}}+2 a^5 \sqrt {x}+a^4 b\right )}{8 a^{9/2}}+\frac {\sqrt {a x+b \sqrt {x}} \left (8 a^3 x^{3/2}-14 a^2 b x+35 a b^2 \sqrt {x}+105 b^3\right )}{12 a^4 \left (a \sqrt {x}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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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)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueDone

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maple [B]  time = 0.06, size = 503, normalized size = 3.62 \begin {gather*} \frac {\sqrt {a x +b \sqrt {x}}\, \left (-120 a^{3} b^{3} x \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+15 a^{3} b^{3} x \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-240 a^{2} b^{4} \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+30 a^{2} b^{4} \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-60 \sqrt {a x +b \sqrt {x}}\, a^{\frac {9}{2}} b \,x^{\frac {3}{2}}-120 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 )+15 a \,b^{5} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-150 \sqrt {a x +b \sqrt {x}}\, a^{\frac {7}{2}} b^{2} x +240 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {7}{2}} b^{2} x -120 \sqrt {a x +b \sqrt {x}}\, a^{\frac {5}{2}} b^{3} \sqrt {x}+480 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {5}{2}} b^{3} \sqrt {x}+16 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} x -30 \sqrt {a x +b \sqrt {x}}\, a^{\frac {3}{2}} b^{4}+240 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {3}{2}} b^{4}+32 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b \sqrt {x}+16 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}-96 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}\right )}{24 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} a^{\frac {11}{2}}} \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]

1/24*(a*x+b*x^(1/2))^(1/2)/a^(11/2)*(16*(a*x+b*x^(1/2))^(3/2)*a^(9/2)*x-60*(a*x+b*x^(1/2))^(1/2)*a^(9/2)*x^(3/
2)*b-120*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*a^3*x*b^3+32*(a*x+b*x^(1/2))^
(3/2)*a^(7/2)*b*x^(1/2)-150*(a*x+b*x^(1/2))^(1/2)*a^(7/2)*x*b^2+240*a^(7/2)*x*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b^
2+15*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x*a^3*b^3-240*ln(1/2*(2*a*x^(1/2)+b+2*((a
*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*a^2*x^(1/2)*b^4+16*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*b^2-120*(a*x+b*x
^(1/2))^(1/2)*a^(5/2)*b^3*x^(1/2)+480*a^(5/2)*x^(1/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b^3-96*a^(5/2)*((a*x^(1/2)
+b)*x^(1/2))^(3/2)*b^2+30*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^(1/2)*a^2*b^4-120*
a*b^5*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))-30*(a*x+b*x^(1/2))^(1/2)*a^(3/2)
*b^4+240*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(3/2)*b^4+15*a*b^5*ln(1/2*(2*a*x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1
/2))/a^(1/2)))/((a*x^(1/2)+b)*x^(1/2))^(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*} \int \frac {x^{2}}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}}}\,{d x} \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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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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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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