\(\int \frac {x^{5/2}}{(a+b x)^2} \, dx\) [456]

Optimal result

Integrand size = 13, antiderivative size = 70 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=-\frac {5 a \sqrt {x}}{b^3}+\frac {5 x^{3/2}}{3 b^2}-\frac {x^{5/2}}{b (a+b x)}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 211} \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}}-\frac {5 a \sqrt {x}}{b^3}-\frac {x^{5/2}}{b (a+b x)}+\frac {5 x^{3/2}}{3 b^2} \]

[In]

Int[x^(5/2)/(a + b*x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.97 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\frac {\sqrt {x} \left (-15 a^2-10 a b x+2 b^2 x^2\right )}{3 b^3 (a+b x)}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

[In]

Integrate[x^(5/2)/(a + b*x)^2,x]

[Out]

(Sqrt[x]*(-15*a^2 - 10*a*b*x + 2*b^2*x^2))/(3*b^3*(a + b*x)) + (5*a^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/b
^(7/2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {2 \left (-b x +6 a \right ) \sqrt {x}}{3 b^{3}}+\frac {a^{2} \left (-\frac {\sqrt {x}}{b x +a}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{b^{3}}\)	\(56\)
derivativedivides	\(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}\right )}{b^{3}}+\frac {2 a^{2} \left (-\frac {\sqrt {x}}{2 \left (b x +a \right )}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\)	\(59\)
default	\(-\frac {2 \left (-\frac {b \,x^{\frac {3}{2}}}{3}+2 a \sqrt {x}\right )}{b^{3}}+\frac {2 a^{2} \left (-\frac {\sqrt {x}}{2 \left (b x +a \right )}+\frac {5 \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}\)	\(59\)

[In]

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

[Out]

-2/3*(-b*x+6*a)*x^(1/2)/b^3+a^2/b^3*(-x^(1/2)/(b*x+a)+5/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.30 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\left [\frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x + 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - a}{b x + a}\right ) + 2 \, {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{6 \, {\left (b^{4} x + a b^{3}\right )}}, \frac {15 \, {\left (a b x + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {x} \sqrt {\frac {a}{b}}}{a}\right ) + {\left (2 \, b^{2} x^{2} - 10 \, a b x - 15 \, a^{2}\right )} \sqrt {x}}{3 \, {\left (b^{4} x + a b^{3}\right )}}\right ] \]

[In]

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

[Out]

[1/6*(15*(a*b*x + a^2)*sqrt(-a/b)*log((b*x + 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) + 2*(2*b^2*x^2 - 10*a*b*x
- 15*a^2)*sqrt(x))/(b^4*x + a*b^3), 1/3*(15*(a*b*x + a^2)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (2*b^2*x^2
- 10*a*b*x - 15*a^2)*sqrt(x))/(b^4*x + a*b^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (63) = 126\).

Time = 12.89 (sec) , antiderivative size = 389, normalized size of antiderivative = 5.56 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a^{2}} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b^{2}} & \text {for}\: a = 0 \\\frac {15 a^{3} \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} - \frac {15 a^{3} \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} - \frac {30 a^{2} b \sqrt {x} \sqrt {- \frac {a}{b}}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} + \frac {15 a^{2} b x \log {\left (\sqrt {x} - \sqrt {- \frac {a}{b}} \right )}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} - \frac {15 a^{2} b x \log {\left (\sqrt {x} + \sqrt {- \frac {a}{b}} \right )}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} - \frac {20 a b^{2} x^{\frac {3}{2}} \sqrt {- \frac {a}{b}}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} + \frac {4 b^{3} x^{\frac {5}{2}} \sqrt {- \frac {a}{b}}}{6 a b^{4} \sqrt {- \frac {a}{b}} + 6 b^{5} x \sqrt {- \frac {a}{b}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a**2), Eq(b, 0)), (2*x**(3/2)/(3*b**2), Eq(a, 0)
), (15*a**3*log(sqrt(x) - sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 15*a**3*log(sqrt(x) + sqrt
(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 30*a**2*b*sqrt(x)*sqrt(-a/b)/(6*a*b**4*sqrt(-a/b) + 6*b*
*5*x*sqrt(-a/b)) + 15*a**2*b*x*log(sqrt(x) - sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 15*a**2
*b*x*log(sqrt(x) + sqrt(-a/b))/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) - 20*a*b**2*x**(3/2)*sqrt(-a/b)/(6*
a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/b)) + 4*b**3*x**(5/2)*sqrt(-a/b)/(6*a*b**4*sqrt(-a/b) + 6*b**5*x*sqrt(-a/
b)), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=-\frac {a^{2} \sqrt {x}}{b^{4} x + a b^{3}} + \frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {2 \, {\left (b x^{\frac {3}{2}} - 6 \, a \sqrt {x}\right )}}{3 \, b^{3}} \]

[In]

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

[Out]

-a^2*sqrt(x)/(b^4*x + a*b^3) + 5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) + 2/3*(b*x^(3/2) - 6*a*sqrt(x
))/b^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.93 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\frac {5 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {a^{2} \sqrt {x}}{{\left (b x + a\right )} b^{3}} + \frac {2 \, {\left (b^{4} x^{\frac {3}{2}} - 6 \, a b^{3} \sqrt {x}\right )}}{3 \, b^{6}} \]

[In]

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

[Out]

5*a^2*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - a^2*sqrt(x)/((b*x + a)*b^3) + 2/3*(b^4*x^(3/2) - 6*a*b^3*s
qrt(x))/b^6

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.83 \[ \int \frac {x^{5/2}}{(a+b x)^2} \, dx=\frac {2\,x^{3/2}}{3\,b^2}-\frac {4\,a\,\sqrt {x}}{b^3}-\frac {a^2\,\sqrt {x}}{x\,b^4+a\,b^3}+\frac {5\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{b^{7/2}} \]

[In]

int(x^(5/2)/(a + b*x)^2,x)

[Out]

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