3.2.20 \(\int \frac {x^{5/2}}{(a+b x)^3} \, dx\) [120]

Optimal. Leaf size=101 \[ \frac {2 \sqrt {x} \left (\frac {5 a x}{b^2}+\frac {x^2}{b}\right )}{(a+b x)^2}-\frac {15 a^2 \left (\sqrt {x} \left (-\frac {1}{2 b (a+b x)^2}+\frac {1}{4 a b (a+b x)}\right )+\frac {\arctan \left (\frac {b x}{a}\right )}{4 a b \sqrt {a b}}\right )}{b^2} \]

[Out]

2*(x^2/b+5*a*x/b^2)*x^(1/2)/(b*x+a)^2-15*a^2/b^2*((-1/2/b/(b*x+a)^2+1/4/a/b/(b*x+a))*x^(1/2)+1/4/a/b/(a*b)^(1/
2)*arctan(b*x/a))

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Rubi [A]
time = 0.01, antiderivative size = 82, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 211} \begin {gather*} -\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {x^{5/2}}{2 b (a+b x)^2}+\frac {15 \sqrt {x}}{4 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(a + b*x)^3,x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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 65

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 211

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

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=-\frac {x^{5/2}}{2 b (a+b x)^2}+\frac {5 \int \frac {x^{3/2}}{(a+b x)^2} \, dx}{4 b}\\ &=-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}+\frac {15 \int \frac {\sqrt {x}}{a+b x} \, dx}{8 b^2}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {(15 a) \int \frac {1}{\sqrt {x} (a+b x)} \, dx}{8 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {(15 a) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=\frac {15 \sqrt {x}}{4 b^3}-\frac {x^{5/2}}{2 b (a+b x)^2}-\frac {5 x^{3/2}}{4 b^2 (a+b x)}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.10, size = 70, normalized size = 0.69 \begin {gather*} \frac {\sqrt {x} \left (15 a^2+25 a b x+8 b^2 x^2\right )}{4 b^3 (a+b x)^2}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(a + b*x)^3,x]

[Out]

(Sqrt[x]*(15*a^2 + 25*a*b*x + 8*b^2*x^2))/(4*b^3*(a + b*x)^2) - (15*Sqrt[a]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])
/(4*b^(7/2))

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Maple [A]
time = 0.07, size = 56, normalized size = 0.55

method	result	size
derivativedivides	\(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\)	\(56\)
default	\(\frac {2 \sqrt {x}}{b^{3}}-\frac {2 a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {x}}{8}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}\)	\(56\)
risch	\(\frac {2 \sqrt {x}}{b^{3}}-\frac {a \left (\frac {-\frac {9 b \,x^{\frac {3}{2}}}{4}-\frac {7 a \sqrt {x}}{4}}{\left (b x +a \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}}\right )}{b^{3}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^3*x^(1/2)-2*a/b^3*((-9/8*b*x^(3/2)-7/8*a*x^(1/2))/(b*x+a)^2+15/8/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2))
)

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Maxima [A]
time = 0.44, size = 73, normalized size = 0.72 \begin {gather*} \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3) - 15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt
(a*b)*b^3) + 2*sqrt(x)/b^3

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Fricas [A]
time = 0.61, size = 200, normalized size = 1.98 \begin {gather*} \left [\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\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 (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) - {\left (8 \, b^{2} x^{2} + 25 \, a b x + 15 \, a^{2}\right )} \sqrt {x}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(8*b^2*x^2
+ 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(a/b)*a
rctan(b*sqrt(x)*sqrt(a/b)/a) - (8*b^2*x^2 + 25*a*b*x + 15*a^2)*sqrt(x))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (82) = 164\).
time = 16.97, size = 683, normalized size = 6.76 \begin {gather*} \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{3}} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{b^{3}} & \text {for}\: a = 0 \\- \frac {15 a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {15 a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {30 a^{2} b \sqrt {x} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} - \frac {30 a^{2} b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {30 a^{2} b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {50 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} - \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {15 a b^{2} x^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} + \frac {16 b^{3} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}}{8 a^{2} b^{4} \sqrt {- \frac {a}{b}} + 16 a b^{5} x \sqrt {- \frac {a}{b}} + 8 b^{6} x^{2} \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**3), Eq(b, 0)), (2*sqrt(x)/b**3, Eq(a, 0)), (-1
5*a**3*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) +
15*a**3*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) +
30*a**2*b*sqrt(x)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)) - 30*
a**2*b*x*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b))
+ 30*a**2*b*x*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a
/b)) + 50*a*b**2*x**(3/2)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sqrt(-a/b)
) - 15*a*b**2*x**2*log(sqrt(x) - sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2*sq
rt(-a/b)) + 15*a*b**2*x**2*log(sqrt(x) + sqrt(-a/b))/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6
*x**2*sqrt(-a/b)) + 16*b**3*x**(5/2)*sqrt(-a/b)/(8*a**2*b**4*sqrt(-a/b) + 16*a*b**5*x*sqrt(-a/b) + 8*b**6*x**2
*sqrt(-a/b)), True))

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Giac [A]
time = 0.42, size = 59, normalized size = 0.58 \begin {gather*} -\frac {15 \, a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} + \frac {2 \, \sqrt {x}}{b^{3}} + \frac {9 \, a b x^{\frac {3}{2}} + 7 \, a^{2} \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*a*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2*sqrt(x)/b^3 + 1/4*(9*a*b*x^(3/2) + 7*a^2*sqrt(x))/((b*
x + a)^2*b^3)

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Mupad [B]
time = 0.12, size = 69, normalized size = 0.68 \begin {gather*} \frac {\frac {7\,a^2\,\sqrt {x}}{4}+\frac {9\,a\,b\,x^{3/2}}{4}}{a^2\,b^3+2\,a\,b^4\,x+b^5\,x^2}+\frac {2\,\sqrt {x}}{b^3}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(a + b*x)^3,x)

[Out]

((7*a^2*x^(1/2))/4 + (9*a*b*x^(3/2))/4)/(a^2*b^3 + b^5*x^2 + 2*a*b^4*x) + (2*x^(1/2))/b^3 - (15*a^(1/2)*atan((
b^(1/2)*x^(1/2))/a^(1/2)))/(4*b^(7/2))

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