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

Optimal. Leaf size=68 \[ \frac{2 a^2 \sqrt{x}}{b^3}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}}-\frac{2 a x^{3/2}}{3 b^2}+\frac{2 x^{5/2}}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0319405, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {50, 63, 205} \[ \frac{2 a^2 \sqrt{x}}{b^3}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}}-\frac{2 a x^{3/2}}{3 b^2}+\frac{2 x^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x),x]

[Out]

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

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0296076, size = 61, normalized size = 0.9 \[ \frac{2 \sqrt{x} \left (15 a^2-5 a b x+3 b^2 x^2\right )}{15 b^3}-\frac{2 a^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x),x]

[Out]

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

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Maple [A]  time = 0.007, size = 54, normalized size = 0.8 \begin{align*}{\frac{2}{5\,b}{x}^{{\frac{5}{2}}}}-{\frac{2\,a}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}+2\,{\frac{{a}^{2}\sqrt{x}}{{b}^{3}}}-2\,{\frac{{a}^{3}}{{b}^{3}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*x^(5/2)/b-2/3*a*x^(3/2)/b^2+2*a^2*x^(1/2)/b^3-2*a^3/b^3/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52884, size = 308, normalized size = 4.53 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}}{15 \, b^{3}}, -\frac{2 \,{\left (15 \, a^{2} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (3 \, b^{2} x^{2} - 5 \, a b x + 15 \, a^{2}\right )} \sqrt{x}\right )}}{15 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 16.3615, size = 121, normalized size = 1.78 \begin{align*} \begin{cases} \frac{i a^{\frac{5}{2}} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{b^{4} \sqrt{\frac{1}{b}}} - \frac{i a^{\frac{5}{2}} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{b^{4} \sqrt{\frac{1}{b}}} + \frac{2 a^{2} \sqrt{x}}{b^{3}} - \frac{2 a x^{\frac{3}{2}}}{3 b^{2}} + \frac{2 x^{\frac{5}{2}}}{5 b} & \text{for}\: b \neq 0 \\\frac{2 x^{\frac{7}{2}}}{7 a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*a**(5/2)*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(b**4*sqrt(1/b)) - I*a**(5/2)*log(I*sqrt(a)*sqrt(1/b
) + sqrt(x))/(b**4*sqrt(1/b)) + 2*a**2*sqrt(x)/b**3 - 2*a*x**(3/2)/(3*b**2) + 2*x**(5/2)/(5*b), Ne(b, 0)), (2*
x**(7/2)/(7*a), True))

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Giac [A]  time = 1.19747, size = 80, normalized size = 1.18 \begin{align*} -\frac{2 \, a^{3} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} + \frac{2 \,{\left (3 \, b^{4} x^{\frac{5}{2}} - 5 \, a b^{3} x^{\frac{3}{2}} + 15 \, a^{2} b^{2} \sqrt{x}\right )}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/15*(3*b^4*x^(5/2) - 5*a*b^3*x^(3/2) + 15*a^2*b^2*sqrt(x
))/b^5