Integrand size = 24, antiderivative size = 75 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=-\frac {b (b c-2 a d) \sqrt {c+d x^2}}{d^2}+\frac {b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \] Output:
-b*(-2*a*d+b*c)*(d*x^2+c)^(1/2)/d^2+1/3*b^2*(d*x^2+c)^(3/2)/d^2-a^2*arctan h((d*x^2+c)^(1/2)/c^(1/2))/c^(1/2)
Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {b \sqrt {c+d x^2} \left (-2 b c+6 a d+b d x^2\right )}{3 d^2}-\frac {a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}} \] Input:
Integrate[(a + b*x^2)^2/(x*Sqrt[c + d*x^2]),x]
Output:
(b*Sqrt[c + d*x^2]*(-2*b*c + 6*a*d + b*d*x^2))/(3*d^2) - (a^2*ArcTanh[Sqrt [c + d*x^2]/Sqrt[c]])/Sqrt[c]
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {354, 99, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2}{x^2 \sqrt {d x^2+c}}dx^2\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {1}{2} \int \left (\frac {a^2}{x^2 \sqrt {d x^2+c}}+\frac {b^2 \sqrt {d x^2+c}}{d}-\frac {b (b c-2 a d)}{d \sqrt {d x^2+c}}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {2 a^2 \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {2 b \sqrt {c+d x^2} (b c-2 a d)}{d^2}+\frac {2 b^2 \left (c+d x^2\right )^{3/2}}{3 d^2}\right )\) |
Input:
Int[(a + b*x^2)^2/(x*Sqrt[c + d*x^2]),x]
Output:
((-2*b*(b*c - 2*a*d)*Sqrt[c + d*x^2])/d^2 + (2*b^2*(c + d*x^2)^(3/2))/(3*d ^2) - (2*a^2*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/Sqrt[c])/2
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.46 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84
method | result | size |
pseudoelliptic | \(-\frac {a^{2} d^{2} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{\sqrt {c}}\right )-2 \left (-\frac {c^{\frac {3}{2}} b}{3}+d \sqrt {c}\, \left (\frac {b \,x^{2}}{6}+a \right )\right ) \sqrt {x^{2} d +c}\, b}{\sqrt {c}\, d^{2}}\) | \(63\) |
default | \(-\frac {a^{2} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {x^{2} d +c}}{x}\right )}{\sqrt {c}}+b^{2} \left (\frac {x^{2} \sqrt {x^{2} d +c}}{3 d}-\frac {2 c \sqrt {x^{2} d +c}}{3 d^{2}}\right )+\frac {2 a b \sqrt {x^{2} d +c}}{d}\) | \(86\) |
Input:
int((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-(a^2*d^2*arctanh((d*x^2+c)^(1/2)/c^(1/2))-2*(-1/3*c^(3/2)*b+d*c^(1/2)*(1/ 6*b*x^2+a))*(d*x^2+c)^(1/2)*b)/c^(1/2)/d^2
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.13 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\left [\frac {3 \, a^{2} \sqrt {c} d^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt {d x^{2} + c}}{6 \, c d^{2}}, \frac {3 \, a^{2} \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {d x^{2} + c} \sqrt {-c}}{c}\right ) + {\left (b^{2} c d x^{2} - 2 \, b^{2} c^{2} + 6 \, a b c d\right )} \sqrt {d x^{2} + c}}{3 \, c d^{2}}\right ] \] Input:
integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="fricas")
Output:
[1/6*(3*a^2*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2 ) + 2*(b^2*c*d*x^2 - 2*b^2*c^2 + 6*a*b*c*d)*sqrt(d*x^2 + c))/(c*d^2), 1/3* (3*a^2*sqrt(-c)*d^2*arctan(sqrt(d*x^2 + c)*sqrt(-c)/c) + (b^2*c*d*x^2 - 2* b^2*c^2 + 6*a*b*c*d)*sqrt(d*x^2 + c))/(c*d^2)]
Time = 5.72 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.45 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {\begin {cases} \frac {2 a^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{2}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} + \frac {2 b^{2} \left (c + d x^{2}\right )^{\frac {3}{2}}}{3 d^{2}} + \frac {2 \sqrt {c + d x^{2}} \cdot \left (2 a b d - b^{2} c\right )}{d^{2}} & \text {for}\: d \neq 0 \\\frac {a^{2} \log {\left (x^{2} \right )} + 2 a b x^{2} + \frac {b^{2} x^{4}}{2}}{\sqrt {c}} & \text {otherwise} \end {cases}}{2} \] Input:
integrate((b*x**2+a)**2/x/(d*x**2+c)**(1/2),x)
Output:
Piecewise((2*a**2*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + 2*b**2*(c + d *x**2)**(3/2)/(3*d**2) + 2*sqrt(c + d*x**2)*(2*a*b*d - b**2*c)/d**2, Ne(d, 0)), ((a**2*log(x**2) + 2*a*b*x**2 + b**2*x**4/2)/sqrt(c), True))/2
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{2}}{3 \, d} - \frac {a^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{\sqrt {c}} - \frac {2 \, \sqrt {d x^{2} + c} b^{2} c}{3 \, d^{2}} + \frac {2 \, \sqrt {d x^{2} + c} a b}{d} \] Input:
integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="maxima")
Output:
1/3*sqrt(d*x^2 + c)*b^2*x^2/d - a^2*arcsinh(c/(sqrt(c*d)*abs(x)))/sqrt(c) - 2/3*sqrt(d*x^2 + c)*b^2*c/d^2 + 2*sqrt(d*x^2 + c)*a*b/d
Time = 0.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {a^{2} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{4} - 3 \, \sqrt {d x^{2} + c} b^{2} c d^{4} + 6 \, \sqrt {d x^{2} + c} a b d^{5}}{3 \, d^{6}} \] Input:
integrate((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x, algorithm="giac")
Output:
a^2*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/3*((d*x^2 + c)^(3/2)*b^2 *d^4 - 3*sqrt(d*x^2 + c)*b^2*c*d^4 + 6*sqrt(d*x^2 + c)*a*b*d^5)/d^6
Time = 0.96 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {b^2\,{\left (d\,x^2+c\right )}^{3/2}}{3\,d^2}-\frac {a^2\,\mathrm {atanh}\left (\frac {\sqrt {d\,x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}}-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d^2}-\frac {b^2\,c}{d^2}\right )\,\sqrt {d\,x^2+c} \] Input:
int((a + b*x^2)^2/(x*(c + d*x^2)^(1/2)),x)
Output:
(b^2*(c + d*x^2)^(3/2))/(3*d^2) - (a^2*atanh((c + d*x^2)^(1/2)/c^(1/2)))/c ^(1/2) - ((2*b^2*c - 2*a*b*d)/d^2 - (b^2*c)/d^2)*(c + d*x^2)^(1/2)
Time = 0.20 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.60 \[ \int \frac {\left (a+b x^2\right )^2}{x \sqrt {c+d x^2}} \, dx=\frac {6 \sqrt {d \,x^{2}+c}\, a b c d -2 \sqrt {d \,x^{2}+c}\, b^{2} c^{2}+\sqrt {d \,x^{2}+c}\, b^{2} c d \,x^{2}+3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}-\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{2}-3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a^{2} d^{2}}{3 c \,d^{2}} \] Input:
int((b*x^2+a)^2/x/(d*x^2+c)^(1/2),x)
Output:
(6*sqrt(c + d*x**2)*a*b*c*d - 2*sqrt(c + d*x**2)*b**2*c**2 + sqrt(c + d*x* *2)*b**2*c*d*x**2 + 3*sqrt(c)*log((sqrt(c + d*x**2) - sqrt(c) + sqrt(d)*x) /sqrt(c))*a**2*d**2 - 3*sqrt(c)*log((sqrt(c + d*x**2) + sqrt(c) + sqrt(d)* x)/sqrt(c))*a**2*d**2)/(3*c*d**2)