\(\int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx\) [577]

Optimal result

Integrand size = 15, antiderivative size = 96 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=-\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {49, 52, 65, 223, 212} \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}-\frac {2 x^{5/2}}{b \sqrt {a+b x}} \]

[In]

Int[x^(5/2)/(a + b*x)^(3/2),x]

[Out]

(-2*x^(5/2))/(b*Sqrt[a + b*x]) - (15*a*Sqrt[x]*Sqrt[a + b*x])/(4*b^3) + (5*x^(3/2)*Sqrt[a + b*x])/(2*b^2) + (1
5*a^2*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*b^(7/2))

Rule 49

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 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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}+\frac {5 \int \frac {x^{3/2}}{\sqrt {a+b x}} \, dx}{b} \\ & = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}-\frac {(15 a) \int \frac {\sqrt {x}}{\sqrt {a+b x}} \, dx}{4 b^2} \\ & = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x}} \, dx}{8 b^3} \\ & = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3} \\ & = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {\left (15 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^3} \\ & = -\frac {2 x^{5/2}}{b \sqrt {a+b x}}-\frac {15 a \sqrt {x} \sqrt {a+b x}}{4 b^3}+\frac {5 x^{3/2} \sqrt {a+b x}}{2 b^2}+\frac {15 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {\sqrt {x} \left (-15 a^2-5 a b x+2 b^2 x^2\right )}{4 b^3 \sqrt {a+b x}}+\frac {15 a^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )}{2 b^{7/2}} \]

[In]

Integrate[x^(5/2)/(a + b*x)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.24

method	result	size
risch	\(-\frac {\left (-2 b x +7 a \right ) \sqrt {x}\, \sqrt {b x +a}}{4 b^{3}}+\frac {\left (\frac {15 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{8 b^{\frac {7}{2}}}-\frac {2 a^{2} \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{b^{4} \left (x +\frac {a}{b}\right )}\right ) \sqrt {x \left (b x +a \right )}}{\sqrt {x}\, \sqrt {b x +a}}\)	\(119\)

[In]

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

[Out]

-1/4*(-2*b*x+7*a)*x^(1/2)*(b*x+a)^(1/2)/b^3+(15/8*a^2/b^(7/2)*ln((1/2*a+b*x)/b^(1/2)+(b*x^2+a*x)^(1/2))-2*a^2/
b^4/(x+a/b)*(b*(x+a/b)^2-(x+a/b)*a)^(1/2))*(x*(b*x+a))^(1/2)/x^(1/2)/(b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.82 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{8 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {15 \, {\left (a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, b^{3} x^{2} - 5 \, a b^{2} x - 15 \, a^{2} b\right )} \sqrt {b x + a} \sqrt {x}}{4 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 6.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.09 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=- \frac {15 a^{\frac {3}{2}} \sqrt {x}}{4 b^{3} \sqrt {1 + \frac {b x}{a}}} - \frac {5 \sqrt {a} x^{\frac {3}{2}}}{4 b^{2} \sqrt {1 + \frac {b x}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 b^{\frac {7}{2}}} + \frac {x^{\frac {5}{2}}}{2 \sqrt {a} b \sqrt {1 + \frac {b x}{a}}} \]

[In]

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

[Out]

-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b*x/a)) + 15*a**2*asinh(sq
rt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.36 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=-\frac {8 \, a^{2} b^{2} - \frac {25 \, {\left (b x + a\right )} a^{2} b}{x} + \frac {15 \, {\left (b x + a\right )}^{2} a^{2}}{x^{2}}}{4 \, {\left (\frac {\sqrt {b x + a} b^{5}}{\sqrt {x}} - \frac {2 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (b x + a\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} - \frac {15 \, a^{2} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + a}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + a}}{\sqrt {x}}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

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

[Out]

-1/4*(8*a^2*b^2 - 25*(b*x + a)*a^2*b/x + 15*(b*x + a)^2*a^2/x^2)/(sqrt(b*x + a)*b^5/sqrt(x) - 2*(b*x + a)^(3/2
)*b^4/x^(3/2) + (b*x + a)^(5/2)*b^3/x^(5/2)) - 15/8*a^2*log(-(sqrt(b) - sqrt(b*x + a)/sqrt(x))/(sqrt(b) + sqrt
(b*x + a)/sqrt(x)))/b^(7/2)

Giac [A] (verification not implemented)

none

Time = 15.82 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.36 \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\frac {{\left (2 \, \sqrt {{\left (b x + a\right )} b - a b} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )}}{b^{3}} - \frac {9 \, a}{b^{3}}\right )} - \frac {32 \, a^{3}}{{\left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{\frac {3}{2}}} - \frac {15 \, a^{2} \log \left ({\left (\sqrt {b x + a} \sqrt {b} - \sqrt {{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{b^{\frac {5}{2}}}\right )} {\left | b \right |}}{8 \, b^{2}} \]

[In]

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

[Out]

1/8*(2*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)/b^3 - 9*a/b^3) - 32*a^3/(((sqrt(b*x + a)*sqrt(b) - s
qrt((b*x + a)*b - a*b))^2 + a*b)*b^(3/2)) - 15*a^2*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^2)/b^
(5/2))*abs(b)/b^2

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{(a+b x)^{3/2}} \, dx=\int \frac {x^{5/2}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]

[In]

int(x^(5/2)/(a + b*x)^(3/2),x)

[Out]

int(x^(5/2)/(a + b*x)^(3/2), x)