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

Optimal. Leaf size=113 \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{2 a^{7/2}}-\frac{15 b \sqrt{a x+b \sqrt{x}}}{2 a^3}+\frac{5 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{a^2}-\frac{4 x^{3/2}}{a \sqrt{a x+b \sqrt{x}}} \]

[Out]

(-4*x^(3/2))/(a*Sqrt[b*Sqrt[x] + a*x]) - (15*b*Sqrt[b*Sqrt[x] + a*x])/(2*a^3) + (5*Sqrt[x]*Sqrt[b*Sqrt[x] + a*
x])/a^2 + (15*b^2*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(2*a^(7/2))

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Rubi [A]  time = 0.102718, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2018, 668, 670, 640, 620, 206} \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{a x+b \sqrt{x}}}\right )}{2 a^{7/2}}-\frac{15 b \sqrt{a x+b \sqrt{x}}}{2 a^3}+\frac{5 \sqrt{x} \sqrt{a x+b \sqrt{x}}}{a^2}-\frac{4 x^{3/2}}{a \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x^(3/2))/(a*Sqrt[b*Sqrt[x] + a*x]) - (15*b*Sqrt[b*Sqrt[x] + a*x])/(2*a^3) + (5*Sqrt[x]*Sqrt[b*Sqrt[x] + a*
x])/a^2 + (15*b^2*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(2*a^(7/2))

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 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 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}}{\left (b \sqrt{x}+a x\right )^{3/2}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x^4}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{4 x^{3/2}}{a \sqrt{b \sqrt{x}+a x}}+\frac{10 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{a}\\ &=-\frac{4 x^{3/2}}{a \sqrt{b \sqrt{x}+a x}}+\frac{5 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{a^2}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{x}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{2 a^2}\\ &=-\frac{4 x^{3/2}}{a \sqrt{b \sqrt{x}+a x}}-\frac{15 b \sqrt{b \sqrt{x}+a x}}{2 a^3}+\frac{5 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{a^2}+\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+a x^2}} \, dx,x,\sqrt{x}\right )}{4 a^3}\\ &=-\frac{4 x^{3/2}}{a \sqrt{b \sqrt{x}+a x}}-\frac{15 b \sqrt{b \sqrt{x}+a x}}{2 a^3}+\frac{5 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{a^2}+\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-a x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{2 a^3}\\ &=-\frac{4 x^{3/2}}{a \sqrt{b \sqrt{x}+a x}}-\frac{15 b \sqrt{b \sqrt{x}+a x}}{2 a^3}+\frac{5 \sqrt{x} \sqrt{b \sqrt{x}+a x}}{a^2}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b \sqrt{x}+a x}}\right )}{2 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.049401, size = 62, normalized size = 0.55 \[ \frac{4 x^2 \sqrt{\frac{a \sqrt{x}}{b}+1} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{a \sqrt{x}}{b}\right )}{7 b \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[1 + (a*Sqrt[x])/b]*x^2*Hypergeometric2F1[3/2, 7/2, 9/2, -((a*Sqrt[x])/b)])/(7*b*Sqrt[b*Sqrt[x] + a*x])

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Maple [B]  time = 0.008, size = 440, normalized size = 3.9 \begin{align*}{\frac{1}{4}\sqrt{b\sqrt{x}+ax} \left ( 4\,\sqrt{b\sqrt{x}+ax}{a}^{9/2}{x}^{3/2}+10\,\sqrt{b\sqrt{x}+ax}{a}^{7/2}xb-32\,{a}^{7/2}x\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b+16\,{a}^{3}\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 ) x{b}^{2}+8\,\sqrt{b\sqrt{x}+ax}{a}^{5/2}\sqrt{x}{b}^{2}-64\,{a}^{5/2}\sqrt{x}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2}+16\,{a}^{5/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}b+32\,{a}^{2}\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 ) \sqrt{x}{b}^{3}-\ln \left ({\frac{1}{2} \left ( 2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b \right ){\frac{1}{\sqrt{a}}}} \right ) x{a}^{3}{b}^{2}+2\,\sqrt{b\sqrt{x}+ax}{a}^{3/2}{b}^{3}-32\,{a}^{3/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{3}+16\,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}-2\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ) \sqrt{x}{a}^{2}{b}^{3}-\ln \left ({\frac{1}{2} \left ( 2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b \right ){\frac{1}{\sqrt{a}}}} \right ) a{b}^{4} \right ){a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}} \left ( b+a\sqrt{x} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^(1/2)+a*x)^(1/2)/a^(9/2)*(4*(b*x^(1/2)+a*x)^(1/2)*a^(9/2)*x^(3/2)+10*(b*x^(1/2)+a*x)^(1/2)*a^(7/2)*x*
b-32*a^(7/2)*x*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b+16*a^3*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))*x*b^2+8*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*x^(1/2)*b^2-64*a^(5/2)*x^(1/2)*(x^(1/2)*(b+a*x^(1/2)))^(
1/2)*b^2+16*a^(5/2)*(x^(1/2)*(b+a*x^(1/2)))^(3/2)*b+32*a^2*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))*x^(1/2)*b^3-ln(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a^(1/2))*x*a^3*b^2+2*(
b*x^(1/2)+a*x)^(1/2)*a^(3/2)*b^3-32*a^(3/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b^3+16*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^4-2*ln(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a^(1
/2))*x^(1/2)*a^2*b^3-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)/(b+a*x^(1/2))^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{3}{2}}}{{\left (a x + b \sqrt{x}\right )}^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError