Integrand size = 31, antiderivative size = 104 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=\frac {1}{2} \left (b-\frac {2 a d^2}{c^2}\right ) x \sqrt {-c+d x} \sqrt {c+d x}+\frac {a (-c+d x)^{3/2} (c+d x)^{3/2}}{c^2 x}-\frac {\left (b c^2-2 a d^2\right ) \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right )}{d} \] Output:
1/2*(b-2*a*d^2/c^2)*x*(d*x-c)^(1/2)*(d*x+c)^(1/2)+a*(d*x-c)^(3/2)*(d*x+c)^ (3/2)/c^2/x-(-2*a*d^2+b*c^2)*arctanh((d*x-c)^(1/2)/(d*x+c)^(1/2))/d
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=\frac {\sqrt {-c+d x} \sqrt {c+d x} \left (-2 a+b x^2\right )}{2 x}+\left (-\frac {b c^2}{d}+2 a d\right ) \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right ) \] Input:
Integrate[(Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2))/x^2,x]
Output:
(Sqrt[-c + d*x]*Sqrt[c + d*x]*(-2*a + b*x^2))/(2*x) + (-((b*c^2)/d) + 2*a* d)*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]]
Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {956, 40, 45, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \sqrt {d x-c} \sqrt {c+d x}}{x^2} \, dx\) |
\(\Big \downarrow \) 956 |
\(\displaystyle \left (b-\frac {2 a d^2}{c^2}\right ) \int \sqrt {d x-c} \sqrt {c+d x}dx+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x}\) |
\(\Big \downarrow \) 40 |
\(\displaystyle \left (b-\frac {2 a d^2}{c^2}\right ) \left (\frac {1}{2} x \sqrt {d x-c} \sqrt {c+d x}-\frac {1}{2} c^2 \int \frac {1}{\sqrt {d x-c} \sqrt {c+d x}}dx\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle \left (b-\frac {2 a d^2}{c^2}\right ) \left (\frac {1}{2} x \sqrt {d x-c} \sqrt {c+d x}-c^2 \int \frac {1}{d-\frac {d (d x-c)}{c+d x}}d\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \left (b-\frac {2 a d^2}{c^2}\right ) \left (\frac {1}{2} x \sqrt {d x-c} \sqrt {c+d x}-\frac {c^2 \text {arctanh}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )}{d}\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{c^2 x}\) |
Input:
Int[(Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2))/x^2,x]
Output:
(a*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(c^2*x) + (b - (2*a*d^2)/c^2)*((x*Sqr t[-c + d*x]*Sqrt[c + d*x])/2 - (c^2*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]]) /d)
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* (a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1)) Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ b*c + a*d, 0] && IGtQ[m + 1/2, 0]
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
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]
Int[((e_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.) *(x_)^(non2_.))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^( m + 1)*(a1 + b1*x^(n/2))^(p + 1)*((a2 + b2*x^(n/2))^(p + 1)/(a1*a2*e*(m + 1 ))), x] + Simp[(a1*a2*d*(m + 1) - b1*b2*c*(m + n*(p + 1) + 1))/(a1*a2*e^n*( m + 1)) Int[(e*x)^(m + n)*(a1 + b1*x^(n/2))^p*(a2 + b2*x^(n/2))^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, d, e, p}, x] && EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || ( LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (-b \,x^{2}+2 a \right )}{2 x \sqrt {d x -c}}-\frac {\left (-a \,d^{2}+\frac {b \,c^{2}}{2}\right ) \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\) | \(117\) |
default | \(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-\operatorname {csgn}\left (d \right ) b d \,x^{2} \sqrt {d^{2} x^{2}-c^{2}}-2 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) a \,d^{2} x +\ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b \,c^{2} x +2 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-c^{2}}\, a \right ) \operatorname {csgn}\left (d \right )}{2 \sqrt {d^{2} x^{2}-c^{2}}\, x d}\) | \(153\) |
Input:
int((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^2,x,method=_RETURNVERBOSE)
Output:
1/2*(d*x+c)^(1/2)*(-d*x+c)*(-b*x^2+2*a)/x/(d*x-c)^(1/2)-(-a*d^2+1/2*b*c^2) *ln(d^2*x/(d^2)^(1/2)+(d^2*x^2-c^2)^(1/2))/(d^2)^(1/2)*((d*x-c)*(d*x+c))^( 1/2)/(d*x-c)^(1/2)/(d*x+c)^(1/2)
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=-\frac {2 \, a d^{2} x - {\left (b c^{2} - 2 \, a d^{2}\right )} x \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right ) - {\left (b d x^{2} - 2 \, a d\right )} \sqrt {d x + c} \sqrt {d x - c}}{2 \, d x} \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^2,x, algorithm="fricas")
Output:
-1/2*(2*a*d^2*x - (b*c^2 - 2*a*d^2)*x*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)) - (b*d*x^2 - 2*a*d)*sqrt(d*x + c)*sqrt(d*x - c))/(d*x)
\[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=\int \frac {\left (a + b x^{2}\right ) \sqrt {- c + d x} \sqrt {c + d x}}{x^{2}}\, dx \] Input:
integrate((d*x-c)**(1/2)*(d*x+c)**(1/2)*(b*x**2+a)/x**2,x)
Output:
Integral((a + b*x**2)*sqrt(-c + d*x)*sqrt(c + d*x)/x**2, x)
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=-\frac {b c^{2} \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right )}{2 \, d} + a d \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right ) + \frac {1}{2} \, \sqrt {d^{2} x^{2} - c^{2}} b x - \frac {\sqrt {d^{2} x^{2} - c^{2}} a}{x} \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^2,x, algorithm="maxima")
Output:
-1/2*b*c^2*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*d)/d + a*d*log(2*d^2*x + 2* sqrt(d^2*x^2 - c^2)*d) + 1/2*sqrt(d^2*x^2 - c^2)*b*x - sqrt(d^2*x^2 - c^2) *a/x
Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.10 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=-\frac {1}{4} \, {\left (\frac {32 \, a c^{2}}{{\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}} - 2 \, \sqrt {d x + c} \sqrt {d x - c} {\left (\frac {{\left (d x + c\right )} b}{d^{2}} - \frac {b c}{d^{2}}\right )} - \frac {{\left (b c^{2} - 2 \, a d^{2}\right )} \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4}\right )}{d^{2}}\right )} d \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^2,x, algorithm="giac")
Output:
-1/4*(32*a*c^2/((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2) - 2*sqrt(d*x + c)*sqrt(d*x - c)*((d*x + c)*b/d^2 - b*c/d^2) - (b*c^2 - 2*a*d^2)*log((sqrt (d*x + c) - sqrt(d*x - c))^4)/d^2)*d
Time = 5.43 (sec) , antiderivative size = 243, normalized size of antiderivative = 2.34 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=\frac {a\,d+\frac {5\,a\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}}{\frac {4\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-c}-\sqrt {d\,x-c}}+\frac {4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}}-4\,a\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )+\frac {b\,x\,\sqrt {c+d\,x}\,\sqrt {d\,x-c}}{2}-\frac {b\,c^2\,\ln \left (d\,x+\sqrt {c+d\,x}\,\sqrt {d\,x-c}\right )}{2\,d}+\frac {a\,d\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{4\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )} \] Input:
int(((a + b*x^2)*(c + d*x)^(1/2)*(d*x - c)^(1/2))/x^2,x)
Output:
(a*d + (5*a*d*((c + d*x)^(1/2) - c^(1/2))^2)/((-c)^(1/2) - (d*x - c)^(1/2) )^2)/((4*((c + d*x)^(1/2) - c^(1/2)))/((-c)^(1/2) - (d*x - c)^(1/2)) + (4* ((c + d*x)^(1/2) - c^(1/2))^3)/((-c)^(1/2) - (d*x - c)^(1/2))^3) - 4*a*d*a tanh(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c)^(1/2))) + (b*x*(c + d*x)^(1/2)*(d*x - c)^(1/2))/2 - (b*c^2*log(d*x + (c + d*x)^(1/2)*(d*x - c)^(1/2)))/(2*d) + (a*d*((c + d*x)^(1/2) - c^(1/2)))/(4*((-c)^(1/2) - (d* x - c)^(1/2)))
Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^2} \, dx=\frac {-8 \sqrt {d x +c}\, \sqrt {d x -c}\, a d +4 \sqrt {d x +c}\, \sqrt {d x -c}\, b d \,x^{2}+16 \,\mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) a \,d^{2} x -8 \,\mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{2} x -8 a \,d^{2} x +b \,c^{2} x}{8 d x} \] Input:
int((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^2,x)
Output:
( - 8*sqrt(c + d*x)*sqrt( - c + d*x)*a*d + 4*sqrt(c + d*x)*sqrt( - c + d*x )*b*d*x**2 + 16*log((sqrt( - c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))* a*d**2*x - 8*log((sqrt( - c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*b*c **2*x - 8*a*d**2*x + b*c**2*x)/(8*d*x)