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

Optimal. Leaf size=57 \[ -\frac{3 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{5/2}}-\frac{x^{3/2}}{b (a+b x)}+\frac{3 \sqrt{x}}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0168536, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {47, 50, 63, 205} \[ -\frac{3 \sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{5/2}}-\frac{x^{3/2}}{b (a+b x)}+\frac{3 \sqrt{x}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b*x)^2,x]

[Out]

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

Mathematica [C]  time = 0.0039348, size = 27, normalized size = 0.47 \[ \frac{2 x^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\frac{b x}{a}\right )}{5 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b*x)^2,x]

[Out]

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

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Maple [A]  time = 0.009, size = 47, normalized size = 0.8 \begin{align*} 2\,{\frac{\sqrt{x}}{{b}^{2}}}+{\frac{a}{{b}^{2} \left ( bx+a \right ) }\sqrt{x}}-3\,{\frac{a}{{b}^{2}\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^(3/2)/(b*x+a)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.49069, size = 300, normalized size = 5.26 \begin{align*} \left [\frac{3 \,{\left (b x + a\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) + 2 \,{\left (2 \, b x + 3 \, a\right )} \sqrt{x}}{2 \,{\left (b^{3} x + a b^{2}\right )}}, -\frac{3 \,{\left (b x + a\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) -{\left (2 \, b x + 3 \, a\right )} \sqrt{x}}{b^{3} x + a b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 19.2066, size = 411, normalized size = 7.21 \begin{align*} \begin{cases} \tilde{\infty } \sqrt{x} & \text{for}\: a = 0 \wedge b = 0 \\\frac{2 x^{\frac{5}{2}}}{5 a^{2}} & \text{for}\: b = 0 \\\frac{2 \sqrt{x}}{b^{2}} & \text{for}\: a = 0 \\\frac{6 i a^{\frac{3}{2}} b \sqrt{x} \sqrt{\frac{1}{b}}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} + \frac{4 i \sqrt{a} b^{2} x^{\frac{3}{2}} \sqrt{\frac{1}{b}}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} - \frac{3 a^{2} \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} + \frac{3 a^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} - \frac{3 a b x \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} + \frac{3 a b x \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + \sqrt{x} \right )}}{2 i a^{\frac{3}{2}} b^{3} \sqrt{\frac{1}{b}} + 2 i \sqrt{a} b^{4} x \sqrt{\frac{1}{b}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(5/2)/(5*a**2), Eq(b, 0)), (2*sqrt(x)/b**2, Eq(a, 0)), (6*
I*a**(3/2)*b*sqrt(x)*sqrt(1/b)/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)) + 4*I*sqrt(a)*b**2
*x**(3/2)*sqrt(1/b)/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)) - 3*a**2*log(-I*sqrt(a)*sqrt(
1/b) + sqrt(x))/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)) + 3*a**2*log(I*sqrt(a)*sqrt(1/b)
+ sqrt(x))/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)) - 3*a*b*x*log(-I*sqrt(a)*sqrt(1/b) + s
qrt(x))/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)) + 3*a*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(
x))/(2*I*a**(3/2)*b**3*sqrt(1/b) + 2*I*sqrt(a)*b**4*x*sqrt(1/b)), True))

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Giac [A]  time = 1.19443, size = 62, normalized size = 1.09 \begin{align*} -\frac{3 \, a \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{2}} + \frac{a \sqrt{x}}{{\left (b x + a\right )} b^{2}} + \frac{2 \, \sqrt{x}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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