3.585 \(\int \frac{x^{3/2}}{(a+b x)^{5/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{a+b x}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b (a+b x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.021607, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 63, 217, 206} \[ -\frac{2 \sqrt{x}}{b^2 \sqrt{a+b x}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{5/2}}-\frac{2 x^{3/2}}{3 b (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b*x)^(5/2),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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

Mathematica [A]  time = 0.130184, size = 80, normalized size = 1.16 \[ \frac{6 \sqrt{a} (a+b x) \sqrt{\frac{b x}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )-2 \sqrt{b} \sqrt{x} (3 a+4 b x)}{3 b^{5/2} (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b*x)^(5/2),x]

[Out]

(-2*Sqrt[b]*Sqrt[x]*(3*a + 4*b*x) + 6*Sqrt[a]*(a + b*x)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/
(3*b^(5/2)*(a + b*x)^(3/2))

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{{x}^{{\frac{3}{2}}} \left ( bx+a \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x+a)^(5/2),x)

[Out]

int(x^(3/2)/(b*x+a)^(5/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08751, size = 451, normalized size = 6.54 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac{2 \,{\left (3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (4 \, b^{2} x + 3 \, a b\right )} \sqrt{b x + a} \sqrt{x}\right )}}{3 \,{\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(4*b^2*x + 3*a*
b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -2/3*(3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-b)*arctan(s
qrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (4*b^2*x + 3*a*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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Sympy [B]  time = 6.02493, size = 328, normalized size = 4.75 \begin{align*} \frac{6 a^{\frac{39}{2}} b^{11} x^{\frac{27}{2}} \sqrt{1 + \frac{b x}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{\frac{29}{2}} \sqrt{1 + \frac{b x}{a}}} + \frac{6 a^{\frac{37}{2}} b^{12} x^{\frac{29}{2}} \sqrt{1 + \frac{b x}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{\frac{29}{2}} \sqrt{1 + \frac{b x}{a}}} - \frac{6 a^{19} b^{\frac{23}{2}} x^{14}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{\frac{29}{2}} \sqrt{1 + \frac{b x}{a}}} - \frac{8 a^{18} b^{\frac{25}{2}} x^{15}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} x^{\frac{27}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{\frac{29}{2}} \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(b*x+a)**(5/2),x)

[Out]

6*a**(39/2)*b**11*x**(27/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sq
rt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) + 6*a**(37/2)*b**12*x**(29/2)*sqrt(1 + b*x/a)
*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)*b**(29/2)*x**(2
9/2)*sqrt(1 + b*x/a)) - 6*a**19*b**(23/2)*x**14/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/a) + 3*a**(37/2)
*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a)) - 8*a**18*b**(25/2)*x**15/(3*a**(39/2)*b**(27/2)*x**(27/2)*sqrt(1 + b*x/
a) + 3*a**(37/2)*b**(29/2)*x**(29/2)*sqrt(1 + b*x/a))

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Giac [B]  time = 58.8763, size = 223, normalized size = 3.23 \begin{align*} -\frac{{\left (\frac{3 \, \log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{\sqrt{b}} + \frac{8 \,{\left (3 \, a{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt{b} + 3 \, a^{2}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{3}{2}} + 2 \, a^{3} b^{\frac{5}{2}}\right )}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )}^{3}}\right )}{\left | b \right |}}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(b*x+a)^(5/2),x, algorithm="giac")

[Out]

-1/3*(3*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/sqrt(b) + 8*(3*a*(sqrt(b*x + a)*sqrt(b) - sqr
t((b*x + a)*b - a*b))^4*sqrt(b) + 3*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2) + 2*a^3*b^
(5/2))/((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2 + a*b)^3)*abs(b)/b^3