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

Optimal. Leaf size=96 \[ \frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{7/2}}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}-\frac{2 x^{5/2}}{b \sqrt{a+b x}} \]

[Out]

(-2*x^(5/2))/(b*Sqrt[a + b*x]) - (15*a*Sqrt[x]*Sqrt[a + b*x])/(4*b^3) + (5*x^(3/2)*Sqrt[a + b*x])/(2*b^2) + (1
5*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(7/2))

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Rubi [A]  time = 0.0293033, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {47, 50, 63, 217, 206} \[ \frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{7/2}}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}-\frac{2 x^{5/2}}{b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x)^(3/2),x]

[Out]

(-2*x^(5/2))/(b*Sqrt[a + b*x]) - (15*a*Sqrt[x]*Sqrt[a + b*x])/(4*b^3) + (5*x^(3/2)*Sqrt[a + b*x])/(2*b^2) + (1
5*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(7/2))

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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^{5/2}}{(a+b x)^{3/2}} \, dx &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}+\frac{5 \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{b}\\ &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}-\frac{(15 a) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{4 b^2}\\ &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}+\frac{\left (15 a^2\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{8 b^3}\\ &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}+\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}+\frac{\left (15 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^3}\\ &=-\frac{2 x^{5/2}}{b \sqrt{a+b x}}-\frac{15 a \sqrt{x} \sqrt{a+b x}}{4 b^3}+\frac{5 x^{3/2} \sqrt{a+b x}}{2 b^2}+\frac{15 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0097587, size = 50, normalized size = 0.52 \[ \frac{2 x^{7/2} \sqrt{\frac{b x}{a}+1} \, _2F_1\left (\frac{3}{2},\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )}{7 a \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.023, size = 119, normalized size = 1.2 \begin{align*} -{\frac{-2\,bx+7\,a}{4\,{b}^{3}}\sqrt{x}\sqrt{bx+a}}+{ \left ({\frac{15\,{a}^{2}}{8}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{7}{2}}}}-2\,{\frac{{a}^{2}}{{b}^{4}}\sqrt{b \left ( x+{\frac{a}{b}} \right ) ^{2}-a \left ( x+{\frac{a}{b}} \right ) } \left ( x+{\frac{a}{b}} \right ) ^{-1}} \right ) \sqrt{x \left ( bx+a \right ) }{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(-2*b*x+7*a)*x^(1/2)*(b*x+a)^(1/2)/b^3+(15/8/b^(7/2)*a^2*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))-2/b^4*
a^2/(x+1/b*a)*(b*(x+1/b*a)^2-a*(x+1/b*a))^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89192, size = 429, normalized size = 4.47 \begin{align*} \left [\frac{15 \,{\left (a^{2} b x + a^{3}\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{8 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{15 \,{\left (a^{2} b x + a^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{4 \,{\left (b^{5} x + a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(a^2*b*x + a^3)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*b^3*x^2 - 5*a*b^2*x -
15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^4), -1/4*(15*(a^2*b*x + a^3)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt
(-b)/(b*sqrt(x))) - (2*b^3*x^2 - 5*a*b^2*x - 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^4)]

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Sympy [A]  time = 15.9217, size = 105, normalized size = 1.09 \begin{align*} - \frac{15 a^{\frac{3}{2}} \sqrt{x}}{4 b^{3} \sqrt{1 + \frac{b x}{a}}} - \frac{5 \sqrt{a} x^{\frac{3}{2}}}{4 b^{2} \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 b^{\frac{7}{2}}} + \frac{x^{\frac{5}{2}}}{2 \sqrt{a} b \sqrt{1 + \frac{b x}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b*x/a)) + 15*a**2*asinh(sq
rt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a))

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Giac [A]  time = 60.6907, size = 177, normalized size = 1.84 \begin{align*} \frac{{\left (2 \, \sqrt{{\left (b x + a\right )} b - a b} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )}}{b^{3}} - \frac{9 \, a}{b^{3}}\right )} - \frac{32 \, a^{3}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac{3}{2}}} - \frac{15 \, a^{2} \log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{b^{\frac{5}{2}}}\right )}{\left | b \right |}}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)/b^3 - 9*a/b^3) - 32*a^3/(((sqrt(b*x + a)*sqrt(b) - s
qrt((b*x + a)*b - a*b))^2 + a*b)*b^(3/2)) - 15*a^2*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^
(5/2))*abs(b)/b^2