\(\int \frac {(a+b x)^{5/2}}{x^3} \, dx\) [306]

Optimal result

Integrand size = 13, antiderivative size = 78 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=\frac {15}{4} b^2 \sqrt {a+b x}-\frac {5 b (a+b x)^{3/2}}{4 x}-\frac {(a+b x)^{5/2}}{2 x^2}-\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 78, 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, 214} \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=-\frac {15}{4} \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+\frac {15}{4} b^2 \sqrt {a+b x}-\frac {(a+b x)^{5/2}}{2 x^2}-\frac {5 b (a+b x)^{3/2}}{4 x} \]

[In]

Int[(a + b*x)^(5/2)/x^3,x]

[Out]

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

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 214

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.81 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {a+b x} \left (-2 a^2-9 a b x+8 b^2 x^2\right )}{x^2}-15 \sqrt {a} b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right ) \]

[In]

Integrate[(a + b*x)^(5/2)/x^3,x]

[Out]

((Sqrt[a + b*x]*(-2*a^2 - 9*a*b*x + 8*b^2*x^2))/x^2 - 15*Sqrt[a]*b^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/4

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {a \sqrt {b x +a}\, \left (9 b x +2 a \right )}{4 x^{2}}+\frac {b^{2} \left (16 \sqrt {b x +a}-30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {a}\right )}{8}\)	\(55\)
derivativedivides	\(2 b^{2} \left (\sqrt {b x +a}-a \left (\frac {\frac {9 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(62\)
default	\(2 b^{2} \left (\sqrt {b x +a}-a \left (\frac {\frac {9 \left (b x +a \right )^{\frac {3}{2}}}{8}-\frac {7 a \sqrt {b x +a}}{8}}{b^{2} x^{2}}+\frac {15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )\)	\(62\)
pseudoelliptic	\(\frac {-2 \sqrt {b x +a}\, a^{\frac {5}{2}}-9 a^{\frac {3}{2}} b x \sqrt {b x +a}+8 b^{2} x^{2} \sqrt {b x +a}\, \sqrt {a}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{2} x^{2}}{4 x^{2} \sqrt {a}}\)	\(75\)

[In]

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

[Out]

-1/4*a*(b*x+a)^(1/2)*(9*b*x+2*a)/x^2+1/8*b^2*(16*(b*x+a)^(1/2)-30*arctanh((b*x+a)^(1/2)/a^(1/2))*a^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=\left [\frac {15 \, \sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{8 \, x^{2}}, \frac {15 \, \sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (8 \, b^{2} x^{2} - 9 \, a b x - 2 \, a^{2}\right )} \sqrt {b x + a}}{4 \, x^{2}}\right ] \]

[In]

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

[Out]

[1/8*(15*sqrt(a)*b^2*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2, 1/4*(15*sqrt(-a)*b^2*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*b^2*x^2 - 9*a*b*x - 2*a^2)*sqrt(b
*x + a))/x^2]

Sympy [A] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=- \frac {15 \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4} - \frac {a^{3}}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {11 a^{2} \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} + 1}} - \frac {a b^{\frac {3}{2}}}{4 \sqrt {x} \sqrt {\frac {a}{b x} + 1}} + \frac {2 b^{\frac {5}{2}} \sqrt {x}}{\sqrt {\frac {a}{b x} + 1}} \]

[In]

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

[Out]

-15*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/4 - a**3/(2*sqrt(b)*x**(5/2)*sqrt(a/(b*x) + 1)) - 11*a**2*sq
rt(b)/(4*x**(3/2)*sqrt(a/(b*x) + 1)) - a*b**(3/2)/(4*sqrt(x)*sqrt(a/(b*x) + 1)) + 2*b**(5/2)*sqrt(x)/sqrt(a/(b
*x) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.29 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=\frac {15}{8} \, \sqrt {a} b^{2} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + 2 \, \sqrt {b x + a} b^{2} - \frac {9 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{2} - 7 \, \sqrt {b x + a} a^{2} b^{2}}{4 \, {\left ({\left (b x + a\right )}^{2} - 2 \, {\left (b x + a\right )} a + a^{2}\right )}} \]

[In]

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

[Out]

15/8*sqrt(a)*b^2*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2*sqrt(b*x + a)*b^2 - 1/4*(9*(b*x
+ a)^(3/2)*a*b^2 - 7*sqrt(b*x + a)*a^2*b^2)/((b*x + a)^2 - 2*(b*x + a)*a + a^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.03 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=\frac {\frac {15 \, a b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, \sqrt {b x + a} b^{3} - \frac {9 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{3} - 7 \, \sqrt {b x + a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \]

[In]

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

[Out]

1/4*(15*a*b^3*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(b*x + a)*b^3 - (9*(b*x + a)^(3/2)*a*b^3 - 7*sqr
t(b*x + a)*a^2*b^3)/(b^2*x^2))/b

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^{5/2}}{x^3} \, dx=2\,b^2\,\sqrt {a+b\,x}+\frac {7\,a^2\,\sqrt {a+b\,x}}{4\,x^2}-\frac {9\,a\,{\left (a+b\,x\right )}^{3/2}}{4\,x^2}+\frac {\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{4} \]

[In]

int((a + b*x)^(5/2)/x^3,x)

[Out]

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