\(\int \frac {(a+b x)^{9/2}}{x} \, dx\) [317]

Optimal result

Integrand size = 13, antiderivative size = 97 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=2 a^4 \sqrt {a+b x}+\frac {2}{3} a^3 (a+b x)^{3/2}+\frac {2}{5} a^2 (a+b x)^{5/2}+\frac {2}{7} a (a+b x)^{7/2}+\frac {2}{9} (a+b x)^{9/2}-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {52, 65, 214} \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )+2 a^4 \sqrt {a+b x}+\frac {2}{3} a^3 (a+b x)^{3/2}+\frac {2}{5} a^2 (a+b x)^{5/2}+\frac {2}{7} a (a+b x)^{7/2}+\frac {2}{9} (a+b x)^{9/2} \]

[In]

Int[(a + b*x)^(9/2)/x,x]

[Out]

2*a^4*Sqrt[a + b*x] + (2*a^3*(a + b*x)^(3/2))/3 + (2*a^2*(a + b*x)^(5/2))/5 + (2*a*(a + b*x)^(7/2))/7 + (2*(a
+ b*x)^(9/2))/9 - 2*a^(9/2)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2}{315} \sqrt {a+b x} \left (563 a^4+506 a^3 b x+408 a^2 b^2 x^2+185 a b^3 x^3+35 b^4 x^4\right )-2 a^{9/2} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right ) \]

[In]

Integrate[(a + b*x)^(9/2)/x,x]

[Out]

(2*Sqrt[a + b*x]*(563*a^4 + 506*a^3*b*x + 408*a^2*b^2*x^2 + 185*a*b^3*x^3 + 35*b^4*x^4))/315 - 2*a^(9/2)*ArcTa
nh[Sqrt[a + b*x]/Sqrt[a]]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (35 b^{4} x^{4}+185 a \,b^{3} x^{3}+408 a^{2} b^{2} x^{2}+506 a^{3} b x +563 a^{4}\right )}{315}\)	\(69\)
derivativedivides	\(\frac {2 a^{3} \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 a^{2} \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x +a \right )^{\frac {9}{2}}}{9}-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{4} \sqrt {b x +a}\)	\(74\)
default	\(\frac {2 a^{3} \left (b x +a \right )^{\frac {3}{2}}}{3}+\frac {2 a^{2} \left (b x +a \right )^{\frac {5}{2}}}{5}+\frac {2 a \left (b x +a \right )^{\frac {7}{2}}}{7}+\frac {2 \left (b x +a \right )^{\frac {9}{2}}}{9}-2 a^{\frac {9}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+2 a^{4} \sqrt {b x +a}\)	\(74\)

[In]

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

[Out]

-2*a^(9/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2/315*(b*x+a)^(1/2)*(35*b^4*x^4+185*a*b^3*x^3+408*a^2*b^2*x^2+506*a^
3*b*x+563*a^4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\left [a^{\frac {9}{2}} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \frac {2}{315} \, {\left (35 \, b^{4} x^{4} + 185 \, a b^{3} x^{3} + 408 \, a^{2} b^{2} x^{2} + 506 \, a^{3} b x + 563 \, a^{4}\right )} \sqrt {b x + a}, 2 \, \sqrt {-a} a^{4} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) + \frac {2}{315} \, {\left (35 \, b^{4} x^{4} + 185 \, a b^{3} x^{3} + 408 \, a^{2} b^{2} x^{2} + 506 \, a^{3} b x + 563 \, a^{4}\right )} \sqrt {b x + a}\right ] \]

[In]

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

[Out]

[a^(9/2)*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2/315*(35*b^4*x^4 + 185*a*b^3*x^3 + 408*a^2*b^2*x^2 +
506*a^3*b*x + 563*a^4)*sqrt(b*x + a), 2*sqrt(-a)*a^4*arctan(sqrt(b*x + a)*sqrt(-a)/a) + 2/315*(35*b^4*x^4 + 18
5*a*b^3*x^3 + 408*a^2*b^2*x^2 + 506*a^3*b*x + 563*a^4)*sqrt(b*x + a)]

Sympy [A] (verification not implemented)

Time = 15.93 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {1126 a^{\frac {9}{2}} \sqrt {1 + \frac {b x}{a}}}{315} + a^{\frac {9}{2}} \log {\left (\frac {b x}{a} \right )} - 2 a^{\frac {9}{2}} \log {\left (\sqrt {1 + \frac {b x}{a}} + 1 \right )} + \frac {1012 a^{\frac {7}{2}} b x \sqrt {1 + \frac {b x}{a}}}{315} + \frac {272 a^{\frac {5}{2}} b^{2} x^{2} \sqrt {1 + \frac {b x}{a}}}{105} + \frac {74 a^{\frac {3}{2}} b^{3} x^{3} \sqrt {1 + \frac {b x}{a}}}{63} + \frac {2 \sqrt {a} b^{4} x^{4} \sqrt {1 + \frac {b x}{a}}}{9} \]

[In]

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

[Out]

1126*a**(9/2)*sqrt(1 + b*x/a)/315 + a**(9/2)*log(b*x/a) - 2*a**(9/2)*log(sqrt(1 + b*x/a) + 1) + 1012*a**(7/2)*
b*x*sqrt(1 + b*x/a)/315 + 272*a**(5/2)*b**2*x**2*sqrt(1 + b*x/a)/105 + 74*a**(3/2)*b**3*x**3*sqrt(1 + b*x/a)/6
3 + 2*sqrt(a)*b**4*x**4*sqrt(1 + b*x/a)/9

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.91 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=a^{\frac {9}{2}} \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right ) + \frac {2}{9} \, {\left (b x + a\right )}^{\frac {9}{2}} + \frac {2}{7} \, {\left (b x + a\right )}^{\frac {7}{2}} a + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 2 \, \sqrt {b x + a} a^{4} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2 \, a^{5} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2}{9} \, {\left (b x + a\right )}^{\frac {9}{2}} + \frac {2}{7} \, {\left (b x + a\right )}^{\frac {7}{2}} a + \frac {2}{5} \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} + \frac {2}{3} \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 2 \, \sqrt {b x + a} a^{4} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b x)^{9/2}}{x} \, dx=\frac {2\,a\,{\left (a+b\,x\right )}^{7/2}}{7}+\frac {2\,{\left (a+b\,x\right )}^{9/2}}{9}+2\,a^4\,\sqrt {a+b\,x}+\frac {2\,a^3\,{\left (a+b\,x\right )}^{3/2}}{3}+\frac {2\,a^2\,{\left (a+b\,x\right )}^{5/2}}{5}+a^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {a+b\,x}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,2{}\mathrm {i} \]

[In]

int((a + b*x)^(9/2)/x,x)

[Out]

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