Integrand size = 20, antiderivative size = 71 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=\frac {2 b^2 \sqrt {c+d x}}{d}-\frac {a^2 \sqrt {c+d x}}{c x}-\frac {a (4 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Output:
2*b^2*(d*x+c)^(1/2)/d-a^2*(d*x+c)^(1/2)/c/x-a*(-a*d+4*b*c)*arctanh((d*x+c) ^(1/2)/c^(1/2))/c^(3/2)
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=\frac {\left (-a^2 d+2 b^2 c x\right ) \sqrt {c+d x}}{c d x}+\frac {a (-4 b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{c^{3/2}} \] Input:
Integrate[(a + b*x)^2/(x^2*Sqrt[c + d*x]),x]
Output:
((-(a^2*d) + 2*b^2*c*x)*Sqrt[c + d*x])/(c*d*x) + (a*(-4*b*c + a*d)*ArcTanh [Sqrt[c + d*x]/Sqrt[c]])/c^(3/2)
Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {100, 27, 90, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\int \frac {2 c x b^2+a (4 b c-a d)}{2 x \sqrt {c+d x}}dx}{c}-\frac {a^2 \sqrt {c+d x}}{c x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {2 c x b^2+a (4 b c-a d)}{x \sqrt {c+d x}}dx}{2 c}-\frac {a^2 \sqrt {c+d x}}{c x}\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {a (4 b c-a d) \int \frac {1}{x \sqrt {c+d x}}dx+\frac {4 b^2 c \sqrt {c+d x}}{d}}{2 c}-\frac {a^2 \sqrt {c+d x}}{c x}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {2 a (4 b c-a d) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{d}+\frac {4 b^2 c \sqrt {c+d x}}{d}}{2 c}-\frac {a^2 \sqrt {c+d x}}{c x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {4 b^2 c \sqrt {c+d x}}{d}-\frac {2 a (4 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}}{2 c}-\frac {a^2 \sqrt {c+d x}}{c x}\) |
Input:
Int[(a + b*x)^2/(x^2*Sqrt[c + d*x]),x]
Output:
-((a^2*Sqrt[c + d*x])/(c*x)) + ((4*b^2*c*Sqrt[c + d*x])/d - (2*a*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqrt[c])/(2*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
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]
Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(\frac {a c d x \left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )+2 \left (b^{2} c x -\frac {d \,a^{2}}{2}\right ) \sqrt {x d +c}\, c^{\frac {3}{2}}}{d \,c^{\frac {5}{2}} x}\) | \(62\) |
derivativedivides | \(\frac {2 b^{2} \sqrt {x d +c}+2 a d \left (-\frac {a \sqrt {x d +c}}{2 c x}+\frac {\left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d}\) | \(63\) |
default | \(\frac {2 b^{2} \sqrt {x d +c}+2 a d \left (-\frac {a \sqrt {x d +c}}{2 c x}+\frac {\left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{2 c^{\frac {3}{2}}}\right )}{d}\) | \(63\) |
risch | \(-\frac {a^{2} \sqrt {x d +c}}{c x}+\frac {2 b^{2} c \sqrt {x d +c}+\frac {a d \left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x d +c}}{\sqrt {c}}\right )}{\sqrt {c}}}{c d}\) | \(67\) |
Input:
int((b*x+a)^2/x^2/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
(a*c*d*x*(a*d-4*b*c)*arctanh((d*x+c)^(1/2)/c^(1/2))+2*(b^2*c*x-1/2*d*a^2)* (d*x+c)^(1/2)*c^(3/2))/d/c^(5/2)/x
Time = 0.08 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.18 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=\left [-\frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {c} x \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c} + 2 \, c}{x}\right ) - 2 \, {\left (2 \, b^{2} c^{2} x - a^{2} c d\right )} \sqrt {d x + c}}{2 \, c^{2} d x}, \frac {{\left (4 \, a b c d - a^{2} d^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x + c}}\right ) + {\left (2 \, b^{2} c^{2} x - a^{2} c d\right )} \sqrt {d x + c}}{c^{2} d x}\right ] \] Input:
integrate((b*x+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="fricas")
Output:
[-1/2*((4*a*b*c*d - a^2*d^2)*sqrt(c)*x*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*b^2*c^2*x - a^2*c*d)*sqrt(d*x + c))/(c^2*d*x), ((4*a*b*c* d - a^2*d^2)*sqrt(-c)*x*arctan(sqrt(-c)/sqrt(d*x + c)) + (2*b^2*c^2*x - a^ 2*c*d)*sqrt(d*x + c))/(c^2*d*x)]
Time = 17.46 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.58 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=- \frac {a^{2} \sqrt {d} \sqrt {\frac {c}{d x} + 1}}{c \sqrt {x}} + \frac {a^{2} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} \sqrt {x}} \right )}}{c^{\frac {3}{2}}} + 2 a b \left (\begin {cases} \frac {2 \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c}} \right )}}{\sqrt {- c}} & \text {for}\: d \neq 0 \\\frac {\log {\left (x \right )}}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \frac {2 \sqrt {c + d x}}{d} & \text {for}\: d \neq 0 \\\frac {x}{\sqrt {c}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((b*x+a)**2/x**2/(d*x+c)**(1/2),x)
Output:
-a**2*sqrt(d)*sqrt(c/(d*x) + 1)/(c*sqrt(x)) + a**2*d*asinh(sqrt(c)/(sqrt(d )*sqrt(x)))/c**(3/2) + 2*a*b*Piecewise((2*atan(sqrt(c + d*x)/sqrt(-c))/sqr t(-c), Ne(d, 0)), (log(x)/sqrt(c), True)) + b**2*Piecewise((2*sqrt(c + d*x )/d, Ne(d, 0)), (x/sqrt(c), True))
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, \sqrt {d x + c} a^{2}}{{\left (d x + c\right )} c - c^{2}} - \frac {4 \, \sqrt {d x + c} b^{2}}{d^{2}} - \frac {{\left (4 \, b c - a d\right )} a \log \left (\frac {\sqrt {d x + c} - \sqrt {c}}{\sqrt {d x + c} + \sqrt {c}}\right )}{c^{\frac {3}{2}} d}\right )} d \] Input:
integrate((b*x+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="maxima")
Output:
-1/2*(2*sqrt(d*x + c)*a^2/((d*x + c)*c - c^2) - 4*sqrt(d*x + c)*b^2/d^2 - (4*b*c - a*d)*a*log((sqrt(d*x + c) - sqrt(c))/(sqrt(d*x + c) + sqrt(c)))/( c^(3/2)*d))*d
Time = 0.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=d {\left (\frac {2 \, \sqrt {d x + c} b^{2}}{d^{2}} - \frac {\sqrt {d x + c} a^{2}}{c d x} + \frac {{\left (4 \, a b c - a^{2} d\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{\sqrt {-c} c d}\right )} \] Input:
integrate((b*x+a)^2/x^2/(d*x+c)^(1/2),x, algorithm="giac")
Output:
d*(2*sqrt(d*x + c)*b^2/d^2 - sqrt(d*x + c)*a^2/(c*d*x) + (4*a*b*c - a^2*d) *arctan(sqrt(d*x + c)/sqrt(-c))/(sqrt(-c)*c*d))
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=\frac {2\,b^2\,\sqrt {c+d\,x}}{d}+\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c}}\right )\,\left (a\,d-4\,b\,c\right )}{c^{3/2}}-\frac {a^2\,\sqrt {c+d\,x}}{c\,x} \] Input:
int((a + b*x)^2/(x^2*(c + d*x)^(1/2)),x)
Output:
(2*b^2*(c + d*x)^(1/2))/d + (a*atanh((c + d*x)^(1/2)/c^(1/2))*(a*d - 4*b*c ))/c^(3/2) - (a^2*(c + d*x)^(1/2))/(c*x)
Time = 0.18 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.73 \[ \int \frac {(a+b x)^2}{x^2 \sqrt {c+d x}} \, dx=\frac {-2 \sqrt {d x +c}\, a^{2} c d +4 \sqrt {d x +c}\, b^{2} c^{2} x -\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a^{2} d^{2} x +4 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}-\sqrt {c}\right ) a b c d x +\sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a^{2} d^{2} x -4 \sqrt {c}\, \mathrm {log}\left (\sqrt {d x +c}+\sqrt {c}\right ) a b c d x}{2 c^{2} d x} \] Input:
int((b*x+a)^2/x^2/(d*x+c)^(1/2),x)
Output:
( - 2*sqrt(c + d*x)*a**2*c*d + 4*sqrt(c + d*x)*b**2*c**2*x - sqrt(c)*log(s qrt(c + d*x) - sqrt(c))*a**2*d**2*x + 4*sqrt(c)*log(sqrt(c + d*x) - sqrt(c ))*a*b*c*d*x + sqrt(c)*log(sqrt(c + d*x) + sqrt(c))*a**2*d**2*x - 4*sqrt(c )*log(sqrt(c + d*x) + sqrt(c))*a*b*c*d*x)/(2*c**2*d*x)