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

Optimal. Leaf size=91 \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{a+b x}}+\frac{5 \sqrt{x} \sqrt{a+b x}}{b^3}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b (a+b x)^{3/2}} \]

[Out]

(-2*x^(5/2))/(3*b*(a + b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[a + b*x]) + (5*Sqrt[x]*Sqrt[a + b*x])/b^3 - (5*a
*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/b^(7/2)

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Rubi [A]  time = 0.0281517, antiderivative size = 91, 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} \[ -\frac{10 x^{3/2}}{3 b^2 \sqrt{a+b x}}+\frac{5 \sqrt{x} \sqrt{a+b x}}{b^3}-\frac{5 a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{b^{7/2}}-\frac{2 x^{5/2}}{3 b (a+b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x)^(5/2),x]

[Out]

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

Mathematica [C]  time = 0.0104971, size = 50, normalized size = 0.55 \[ \frac{2 x^{7/2} \sqrt{\frac{b x}{a}+1} \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )}{7 a^2 \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x)^(5/2),x]

[Out]

(2*x^(7/2)*Sqrt[1 + (b*x)/a]*Hypergeometric2F1[5/2, 7/2, 9/2, -((b*x)/a)])/(7*a^2*Sqrt[a + b*x])

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Maple [B]  time = 0.031, size = 147, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{3}}\sqrt{x}\sqrt{bx+a}}+{ \left ( -{\frac{5\,a}{2}\ln \left ({ \left ({\frac{a}{2}}+bx \right ){\frac{1}{\sqrt{b}}}}+\sqrt{b{x}^{2}+ax} \right ){b}^{-{\frac{7}{2}}}}-{\frac{2\,{a}^{2}}{3\,{b}^{5}}\sqrt{b \left ( x+{\frac{a}{b}} \right ) ^{2}-a \left ( x+{\frac{a}{b}} \right ) } \left ( x+{\frac{a}{b}} \right ) ^{-2}}+{\frac{14\,a}{3\,{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)^(5/2),x)

[Out]

x^(1/2)*(b*x+a)^(1/2)/b^3+(-5/2/b^(7/2)*a*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))-2/3/b^5*a^2/(x+1/b*a)^2*(b
*(x+1/b*a)^2-a*(x+1/b*a))^(1/2)+14/3/b^4*a/(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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.11829, size = 512, normalized size = 5.63 \begin{align*} \left [\frac{15 \,{\left (a b^{2} x^{2} + 2 \, 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 (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{6 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{15 \,{\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt{b x + a} \sqrt{x}}{3 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(3*b^3*x^2
+ 20*a*b^2*x + 15*a^2*b)*sqrt(b*x + a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(15*(a*b^2*x^2 + 2*a^2*b
*x + a^3)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (3*b^3*x^2 + 20*a*b^2*x + 15*a^2*b)*sqrt(b*x +
a)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]

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Sympy [B]  time = 15.8031, size = 396, normalized size = 4.35 \begin{align*} - \frac{15 a^{\frac{81}{2}} b^{22} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{3 a^{\frac{79}{2}} b^{\frac{51}{2}} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{77}{2}} b^{\frac{53}{2}} x^{\frac{53}{2}} \sqrt{1 + \frac{b x}{a}}} - \frac{15 a^{\frac{79}{2}} b^{23} x^{\frac{53}{2}} \sqrt{1 + \frac{b x}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{3 a^{\frac{79}{2}} b^{\frac{51}{2}} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{77}{2}} b^{\frac{53}{2}} x^{\frac{53}{2}} \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{40} b^{\frac{45}{2}} x^{26}}{3 a^{\frac{79}{2}} b^{\frac{51}{2}} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{77}{2}} b^{\frac{53}{2}} x^{\frac{53}{2}} \sqrt{1 + \frac{b x}{a}}} + \frac{20 a^{39} b^{\frac{47}{2}} x^{27}}{3 a^{\frac{79}{2}} b^{\frac{51}{2}} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{77}{2}} b^{\frac{53}{2}} x^{\frac{53}{2}} \sqrt{1 + \frac{b x}{a}}} + \frac{3 a^{38} b^{\frac{49}{2}} x^{28}}{3 a^{\frac{79}{2}} b^{\frac{51}{2}} x^{\frac{51}{2}} \sqrt{1 + \frac{b x}{a}} + 3 a^{\frac{77}{2}} b^{\frac{53}{2}} x^{\frac{53}{2}} \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)**(5/2),x)

[Out]

-15*a**(81/2)*b**22*x**(51/2)*sqrt(1 + b*x/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/2)*x**(51/2)*
sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) - 15*a**(79/2)*b**23*x**(53/2)*sqrt(1 + b*x
/a)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x*
*(53/2)*sqrt(1 + b*x/a)) + 15*a**40*b**(45/2)*x**26/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 + b*x/a) + 3*a**(7
7/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) + 20*a**39*b**(47/2)*x**27/(3*a**(79/2)*b**(51/2)*x**(51/2)*sqrt(1 +
b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a)) + 3*a**38*b**(49/2)*x**28/(3*a**(79/2)*b**(51/2)*x*
*(51/2)*sqrt(1 + b*x/a) + 3*a**(77/2)*b**(53/2)*x**(53/2)*sqrt(1 + b*x/a))

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Giac [B]  time = 59.5621, size = 266, normalized size = 2.92 \begin{align*} \frac{{\left (\frac{15 \, a \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}}} + \frac{6 \, \sqrt{{\left (b x + a\right )} b - a b} \sqrt{b x + a}}{b^{3}} + \frac{8 \,{\left (9 \, a^{2}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{4} \sqrt{b} + 12 \, a^{3}{\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} b^{\frac{3}{2}} + 7 \, a^{4} 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} b^{2}}\right )}{\left | b \right |}}{6 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(15*a*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^(5/2) + 6*sqrt((b*x + a)*b - a*b)*sqrt(b*
x + a)/b^3 + 8*(9*a^2*(sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^4*sqrt(b) + 12*a^3*(sqrt(b*x + a)*sqrt
(b) - sqrt((b*x + a)*b - a*b))^2*b^(3/2) + 7*a^4*b^(5/2))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^
2 + a*b)^3*b^2))*abs(b)/b^2