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

Optimal. Leaf size=146 \[ -\frac{35 b^3 \sqrt{a x+b \sqrt{x}}}{32 a^4}+\frac{35 b^2 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{48 a^3}+\frac{35 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{32 a^{9/2}}-\frac{7 b x \sqrt{a x+b \sqrt{x}}}{12 a^2}+\frac{x^{3/2} \sqrt{a x+b \sqrt{x}}}{2 a} \]

[Out]

(-35*b^3*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (35*b^2*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(48*a^3) - (7*b*x*Sqrt[b*Sqr
t[x] + a*x])/(12*a^2) + (x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(2*a) + (35*b^4*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[
x] + a*x]])/(32*a^(9/2))

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Rubi [A]  time = 0.119912, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2018, 670, 640, 620, 206} \[ -\frac{35 b^3 \sqrt{a x+b \sqrt{x}}}{32 a^4}+\frac{35 b^2 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{48 a^3}+\frac{35 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{32 a^{9/2}}-\frac{7 b x \sqrt{a x+b \sqrt{x}}}{12 a^2}+\frac{x^{3/2} \sqrt{a x+b \sqrt{x}}}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(-35*b^3*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (35*b^2*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(48*a^3) - (7*b*x*Sqrt[b*Sqr
t[x] + a*x])/(12*a^2) + (x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(2*a) + (35*b^4*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[
x] + a*x]])/(32*a^(9/2))

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^{3/2}}{\sqrt{b \sqrt{x}+a x}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}-\frac{(7 b) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{4 a}\\ &=-\frac{7 b x \sqrt{b \sqrt{x}+a x}}{12 a^2}+\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}+\frac{\left (35 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{24 a^2}\\ &=\frac{35 b^2 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{48 a^3}-\frac{7 b x \sqrt{b \sqrt{x}+a x}}{12 a^2}+\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}-\frac{\left (35 b^3\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{32 a^3}\\ &=-\frac{35 b^3 \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{35 b^2 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{48 a^3}-\frac{7 b x \sqrt{b \sqrt{x}+a x}}{12 a^2}+\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}+\frac{\left (35 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{64 a^4}\\ &=-\frac{35 b^3 \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{35 b^2 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{48 a^3}-\frac{7 b x \sqrt{b \sqrt{x}+a x}}{12 a^2}+\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}+\frac{\left (35 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{32 a^4}\\ &=-\frac{35 b^3 \sqrt{b \sqrt{x}+a x}}{32 a^4}+\frac{35 b^2 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{48 a^3}-\frac{7 b x \sqrt{b \sqrt{x}+a x}}{12 a^2}+\frac{x^{3/2} \sqrt{b \sqrt{x}+a x}}{2 a}+\frac{35 b^4 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{32 a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.137471, size = 142, normalized size = 0.97 \[ -\frac{35 b^5 \left (\frac{a \sqrt{x}}{b}+1\right ) \left (-\frac{32 a^4 x^2}{35 b^4}+\frac{16 a^3 x^{3/2}}{15 b^3}-\frac{4 a^2 x}{3 b^2}+\frac{2 a \sqrt{x}}{b}-\frac{2 \sqrt{a} \sqrt [4]{x} \sinh ^{-1}\left (\frac{\sqrt{a} \sqrt [4]{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{\frac{a \sqrt{x}}{b}+1}}\right )}{64 a^5 \sqrt{\sqrt{x} \left (a \sqrt{x}+b\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*b^5*(1 + (a*Sqrt[x])/b)*((2*a*Sqrt[x])/b - (4*a^2*x)/(3*b^2) + (16*a^3*x^(3/2))/(15*b^3) - (32*a^4*x^2)/(
35*b^4) - (2*Sqrt[a]*x^(1/4)*ArcSinh[(Sqrt[a]*x^(1/4))/Sqrt[b]])/(Sqrt[b]*Sqrt[1 + (a*Sqrt[x])/b])))/(64*a^5*S
qrt[(b + a*Sqrt[x])*Sqrt[x]])

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Maple [A]  time = 0.006, size = 203, normalized size = 1.4 \begin{align*}{\frac{1}{192}\sqrt{b\sqrt{x}+ax} \left ( 96\,\sqrt{x} \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{7/2}-208\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}b+348\,\sqrt{b\sqrt{x}+ax}{a}^{5/2}\sqrt{x}{b}^{2}+174\,\sqrt{b\sqrt{x}+ax}{a}^{3/2}{b}^{3}-384\,{a}^{3/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{3}+192\,a\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 ){b}^{4}-87\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ) a{b}^{4} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{a}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(b*x^(1/2)+a*x)^(1/2)*(96*x^(1/2)*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)-208*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*b+348*
(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*x^(1/2)*b^2+174*(b*x^(1/2)+a*x)^(1/2)*a^(3/2)*b^3-384*a^(3/2)*(x^(1/2)*(b+a*x^(1
/2)))^(1/2)*b^3+192*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^4-87*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^4)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/a^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{\sqrt{a x + b \sqrt{x}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(3/2)/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^(3/2)/(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^{\frac{3}{2}}}{\sqrt{a x + b \sqrt{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.31092, size = 131, normalized size = 0.9 \begin{align*} \frac{1}{96} \, \sqrt{a x + b \sqrt{x}}{\left (2 \,{\left (4 \, \sqrt{x}{\left (\frac{6 \, \sqrt{x}}{a} - \frac{7 \, b}{a^{2}}\right )} + \frac{35 \, b^{2}}{a^{3}}\right )} \sqrt{x} - \frac{105 \, b^{3}}{a^{4}}\right )} - \frac{35 \, b^{4} \log \left ({\left | -2 \, \sqrt{a}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )} - b \right |}\right )}{64 \, a^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*sqrt(a*x + b*sqrt(x))*(2*(4*sqrt(x)*(6*sqrt(x)/a - 7*b/a^2) + 35*b^2/a^3)*sqrt(x) - 105*b^3/a^4) - 35/64*
b^4*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(9/2)