\(\int \frac {(a+b x)^{9/2}}{x^2} \, dx\) [318]

Optimal result

Integrand size = 13, antiderivative size = 98 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=9 a^3 b \sqrt {a+b x}+3 a^2 b (a+b x)^{3/2}+\frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}-9 a^{7/2} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {43, 52, 65, 214} \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=-9 a^{7/2} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+9 a^3 b \sqrt {a+b x}+3 a^2 b (a+b x)^{3/2}-\frac {(a+b x)^{9/2}}{x}+\frac {9}{7} b (a+b x)^{7/2}+\frac {9}{5} a b (a+b x)^{5/2} \]

[In]

Int[(a + b*x)^(9/2)/x^2,x]

[Out]

9*a^3*b*Sqrt[a + b*x] + 3*a^2*b*(a + b*x)^(3/2) + (9*a*b*(a + b*x)^(5/2))/5 + (9*b*(a + b*x)^(7/2))/7 - (a + b
*x)^(9/2)/x - 9*a^(7/2)*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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)^{9/2}}{x}+\frac {1}{2} (9 b) \int \frac {(a+b x)^{7/2}}{x} \, dx \\ & = \frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}+\frac {1}{2} (9 a b) \int \frac {(a+b x)^{5/2}}{x} \, dx \\ & = \frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}+\frac {1}{2} \left (9 a^2 b\right ) \int \frac {(a+b x)^{3/2}}{x} \, dx \\ & = 3 a^2 b (a+b x)^{3/2}+\frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}+\frac {1}{2} \left (9 a^3 b\right ) \int \frac {\sqrt {a+b x}}{x} \, dx \\ & = 9 a^3 b \sqrt {a+b x}+3 a^2 b (a+b x)^{3/2}+\frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}+\frac {1}{2} \left (9 a^4 b\right ) \int \frac {1}{x \sqrt {a+b x}} \, dx \\ & = 9 a^3 b \sqrt {a+b x}+3 a^2 b (a+b x)^{3/2}+\frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}+\left (9 a^4\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x}\right ) \\ & = 9 a^3 b \sqrt {a+b x}+3 a^2 b (a+b x)^{3/2}+\frac {9}{5} a b (a+b x)^{5/2}+\frac {9}{7} b (a+b x)^{7/2}-\frac {(a+b x)^{9/2}}{x}-9 a^{7/2} b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=\frac {\sqrt {a+b x} \left (-35 a^4+388 a^3 b x+156 a^2 b^2 x^2+58 a b^3 x^3+10 b^4 x^4\right )}{35 x}-9 a^{7/2} b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

Integrate[(a + b*x)^(9/2)/x^2,x]

[Out]

(Sqrt[a + b*x]*(-35*a^4 + 388*a^3*b*x + 156*a^2*b^2*x^2 + 58*a*b^3*x^3 + 10*b^4*x^4))/(35*x) - 9*a^(7/2)*b*Arc
Tanh[Sqrt[a + b*x]/Sqrt[a]]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.83

method	result	size
risch	\(-\frac {a^{4} \sqrt {b x +a}}{x}+\frac {b \left (\frac {4 \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {8 a \left (b x +a \right )^{\frac {5}{2}}}{5}+4 a^{2} \left (b x +a \right )^{\frac {3}{2}}+16 a^{3} \sqrt {b x +a}-18 a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )\right )}{2}\)	\(81\)
pseudoelliptic	\(-\frac {9 \left (\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{4} b x -\frac {2 \left (\sqrt {a}\, b^{4} x^{4}+\frac {29 a^{\frac {3}{2}} b^{3} x^{3}}{5}+\frac {78 a^{\frac {5}{2}} b^{2} x^{2}}{5}+\frac {194 a^{\frac {7}{2}} b x}{5}-\frac {7 a^{\frac {9}{2}}}{2}\right ) \sqrt {b x +a}}{63}\right )}{\sqrt {a}\, x}\)	\(82\)
derivativedivides	\(2 b \left (\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}+a^{2} \left (b x +a \right )^{\frac {3}{2}}+4 a^{3} \sqrt {b x +a}-a^{4} \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {9 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(85\)
default	\(2 b \left (\frac {\left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 a \left (b x +a \right )^{\frac {5}{2}}}{5}+a^{2} \left (b x +a \right )^{\frac {3}{2}}+4 a^{3} \sqrt {b x +a}-a^{4} \left (\frac {\sqrt {b x +a}}{2 b x}+\frac {9 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )\right )\)	\(85\)

[In]

int((b*x+a)^(9/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-a^4*(b*x+a)^(1/2)/x+1/2*b*(4/7*(b*x+a)^(7/2)+8/5*a*(b*x+a)^(5/2)+4*a^2*(b*x+a)^(3/2)+16*a^3*(b*x+a)^(1/2)-18*
a^(7/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.76 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=\left [\frac {315 \, a^{\frac {7}{2}} b x \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (10 \, b^{4} x^{4} + 58 \, a b^{3} x^{3} + 156 \, a^{2} b^{2} x^{2} + 388 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{70 \, x}, \frac {315 \, \sqrt {-a} a^{3} b x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + {\left (10 \, b^{4} x^{4} + 58 \, a b^{3} x^{3} + 156 \, a^{2} b^{2} x^{2} + 388 \, a^{3} b x - 35 \, a^{4}\right )} \sqrt {b x + a}}{35 \, x}\right ] \]

[In]

integrate((b*x+a)^(9/2)/x^2,x, algorithm="fricas")

[Out]

[1/70*(315*a^(7/2)*b*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(10*b^4*x^4 + 58*a*b^3*x^3 + 156*a^2*b
^2*x^2 + 388*a^3*b*x - 35*a^4)*sqrt(b*x + a))/x, 1/35*(315*sqrt(-a)*a^3*b*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) +
(10*b^4*x^4 + 58*a*b^3*x^3 + 156*a^2*b^2*x^2 + 388*a^3*b*x - 35*a^4)*sqrt(b*x + a))/x]

Sympy [A] (verification not implemented)

Time = 13.43 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=- \frac {a^{\frac {9}{2}} \sqrt {1 + \frac {b x}{a}}}{x} + \frac {388 a^{\frac {7}{2}} b \sqrt {1 + \frac {b x}{a}}}{35} + \frac {9 a^{\frac {7}{2}} b \log {\left (\frac {b x}{a} \right )}}{2} - 9 a^{\frac {7}{2}} b \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {156 a^{\frac {5}{2}} b^{2} x \sqrt {1 + \frac {b x}{a}}}{35} + \frac {58 a^{\frac {3}{2}} b^{3} x^{2} \sqrt {1 + \frac {b x}{a}}}{35} + \frac {2 \sqrt {a} b^{4} x^{3} \sqrt {1 + \frac {b x}{a}}}{7} \]

[In]

integrate((b*x+a)**(9/2)/x**2,x)

[Out]

-a**(9/2)*sqrt(1 + b*x/a)/x + 388*a**(7/2)*b*sqrt(1 + b*x/a)/35 + 9*a**(7/2)*b*log(b*x/a)/2 - 9*a**(7/2)*b*log
(sqrt(1 + b*x/a) + 1) + 156*a**(5/2)*b**2*x*sqrt(1 + b*x/a)/35 + 58*a**(3/2)*b**3*x**2*sqrt(1 + b*x/a)/35 + 2*
sqrt(a)*b**4*x**3*sqrt(1 + b*x/a)/7

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=\frac {9}{2} \, a^{\frac {7}{2}} b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{7} \, {\left (b x + a\right )}^{\frac {7}{2}} b + \frac {4}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} a b + 2 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b + 8 \, \sqrt {b x + a} a^{3} b - \frac {\sqrt {b x + a} a^{4}}{x} \]

[In]

integrate((b*x+a)^(9/2)/x^2,x, algorithm="maxima")

[Out]

9/2*a^(7/2)*b*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a))) + 2/7*(b*x + a)^(7/2)*b + 4/5*(b*x + a)
^(5/2)*a*b + 2*(b*x + a)^(3/2)*a^2*b + 8*sqrt(b*x + a)*a^3*b - sqrt(b*x + a)*a^4/x

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=\frac {\frac {315 \, a^{4} b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 10 \, {\left (b x + a\right )}^{\frac {7}{2}} b^{2} + 28 \, {\left (b x + a\right )}^{\frac {5}{2}} a b^{2} + 70 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} b^{2} + 280 \, \sqrt {b x + a} a^{3} b^{2} - \frac {35 \, \sqrt {b x + a} a^{4} b}{x}}{35 \, b} \]

[In]

integrate((b*x+a)^(9/2)/x^2,x, algorithm="giac")

[Out]

1/35*(315*a^4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) + 10*(b*x + a)^(7/2)*b^2 + 28*(b*x + a)^(5/2)*a*b^2
+ 70*(b*x + a)^(3/2)*a^2*b^2 + 280*sqrt(b*x + a)*a^3*b^2 - 35*sqrt(b*x + a)*a^4*b/x)/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^{9/2}}{x^2} \, dx=\frac {2\,b\,{\left (a+b\,x\right )}^{7/2}}{7}-\frac {a^4\,\sqrt {a+b\,x}}{x}+\frac {4\,a\,b\,{\left (a+b\,x\right )}^{5/2}}{5}+8\,a^3\,b\,\sqrt {a+b\,x}+2\,a^2\,b\,{\left (a+b\,x\right )}^{3/2}+a^{7/2}\,b\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,9{}\mathrm {i} \]

[In]

int((a + b*x)^(9/2)/x^2,x)

[Out]

(2*b*(a + b*x)^(7/2))/7 - (a^4*(a + b*x)^(1/2))/x + a^(7/2)*b*atan(((a + b*x)^(1/2)*1i)/a^(1/2))*9i + (4*a*b*(
a + b*x)^(5/2))/5 + 8*a^3*b*(a + b*x)^(1/2) + 2*a^2*b*(a + b*x)^(3/2)