Integrand size = 31, antiderivative size = 84 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=-\frac {b \sqrt {-c+d x} \sqrt {c+d x}}{x}+\frac {a (-c+d x)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}+2 b d \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right ) \] Output:
-b*(d*x-c)^(1/2)*(d*x+c)^(1/2)/x+1/3*a*(d*x-c)^(3/2)*(d*x+c)^(3/2)/c^2/x^3 +2*b*d*arctanh((d*x-c)^(1/2)/(d*x+c)^(1/2))
Time = 0.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=-\frac {\sqrt {-c+d x} \sqrt {c+d x} \left (3 b c^2 x^2+a \left (c^2-d^2 x^2\right )\right )}{3 c^2 x^3}+2 b d \text {arctanh}\left (\frac {\sqrt {-c+d x}}{\sqrt {c+d x}}\right ) \] Input:
Integrate[(Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2))/x^4,x]
Output:
-1/3*(Sqrt[-c + d*x]*Sqrt[c + d*x]*(3*b*c^2*x^2 + a*(c^2 - d^2*x^2)))/(c^2 *x^3) + 2*b*d*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]]
Time = 0.38 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {956, 108, 27, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \sqrt {d x-c} \sqrt {c+d x}}{x^4} \, dx\) |
\(\Big \downarrow \) 956 |
\(\displaystyle b \int \frac {\sqrt {d x-c} \sqrt {c+d x}}{x^2}dx+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}\) |
\(\Big \downarrow \) 108 |
\(\displaystyle b \left (\int \frac {d^2}{\sqrt {d x-c} \sqrt {c+d x}}dx-\frac {\sqrt {d x-c} \sqrt {c+d x}}{x}\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b \left (d^2 \int \frac {1}{\sqrt {d x-c} \sqrt {c+d x}}dx-\frac {\sqrt {d x-c} \sqrt {c+d x}}{x}\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}\) |
\(\Big \downarrow \) 45 |
\(\displaystyle b \left (2 d^2 \int \frac {1}{d-\frac {d (d x-c)}{c+d x}}d\frac {\sqrt {d x-c}}{\sqrt {c+d x}}-\frac {\sqrt {d x-c} \sqrt {c+d x}}{x}\right )+\frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {a (d x-c)^{3/2} (c+d x)^{3/2}}{3 c^2 x^3}+b \left (2 d \text {arctanh}\left (\frac {\sqrt {d x-c}}{\sqrt {c+d x}}\right )-\frac {\sqrt {d x-c} \sqrt {c+d x}}{x}\right )\) |
Input:
Int[(Sqrt[-c + d*x]*Sqrt[c + d*x]*(a + b*x^2))/x^4,x]
Output:
(a*(-c + d*x)^(3/2)*(c + d*x)^(3/2))/(3*c^2*x^3) + b*(-((Sqrt[-c + d*x]*Sq rt[c + d*x])/x) + 2*d*ArcTanh[Sqrt[-c + d*x]/Sqrt[c + d*x]])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) , x] - Simp[1/(b*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c , d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 *n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
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 = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.48
method | result | size |
risch | \(\frac {\sqrt {d x +c}\, \left (-d x +c \right ) \left (-a \,d^{2} x^{2}+3 b \,c^{2} x^{2}+a \,c^{2}\right )}{3 x^{3} c^{2} \sqrt {d x -c}}+\frac {b \,d^{2} \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-c^{2}}\right ) \sqrt {\left (d x -c \right ) \left (d x +c \right )}}{\sqrt {d^{2}}\, \sqrt {d x -c}\, \sqrt {d x +c}}\) | \(124\) |
default | \(-\frac {\sqrt {d x -c}\, \sqrt {d x +c}\, \left (-3 \ln \left (\left (\sqrt {d^{2} x^{2}-c^{2}}\, \operatorname {csgn}\left (d \right )+d x \right ) \operatorname {csgn}\left (d \right )\right ) b \,c^{2} d \,x^{3}-\operatorname {csgn}\left (d \right ) a \,d^{2} x^{2} \sqrt {d^{2} x^{2}-c^{2}}+3 \,\operatorname {csgn}\left (d \right ) b \,c^{2} x^{2} \sqrt {d^{2} x^{2}-c^{2}}+\operatorname {csgn}\left (d \right ) a \,c^{2} \sqrt {d^{2} x^{2}-c^{2}}\right ) \operatorname {csgn}\left (d \right )}{3 \sqrt {d^{2} x^{2}-c^{2}}\, c^{2} x^{3}}\) | \(153\) |
Input:
int((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^4,x,method=_RETURNVERBOSE)
Output:
1/3*(d*x+c)^(1/2)*(-d*x+c)*(-a*d^2*x^2+3*b*c^2*x^2+a*c^2)/x^3/c^2/(d*x-c)^ (1/2)+b*d^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 = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=-\frac {3 \, b c^{2} d x^{3} \log \left (-d x + \sqrt {d x + c} \sqrt {d x - c}\right ) + {\left (3 \, b c^{2} d - a d^{3}\right )} x^{3} + {\left (a c^{2} + {\left (3 \, b c^{2} - a d^{2}\right )} x^{2}\right )} \sqrt {d x + c} \sqrt {d x - c}}{3 \, c^{2} x^{3}} \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^4,x, algorithm="fricas")
Output:
-1/3*(3*b*c^2*d*x^3*log(-d*x + sqrt(d*x + c)*sqrt(d*x - c)) + (3*b*c^2*d - a*d^3)*x^3 + (a*c^2 + (3*b*c^2 - a*d^2)*x^2)*sqrt(d*x + c)*sqrt(d*x - c)) /(c^2*x^3)
Timed out. \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=\text {Timed out} \] Input:
integrate((d*x-c)**(1/2)*(d*x+c)**(1/2)*(b*x**2+a)/x**4,x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=b d \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - c^{2}} d\right ) - \frac {\sqrt {d^{2} x^{2} - c^{2}} b}{x} + \frac {{\left (d^{2} x^{2} - c^{2}\right )}^{\frac {3}{2}} a}{3 \, c^{2} x^{3}} \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^4,x, algorithm="maxima")
Output:
b*d*log(2*d^2*x + 2*sqrt(d^2*x^2 - c^2)*d) - sqrt(d^2*x^2 - c^2)*b/x + 1/3 *(d^2*x^2 - c^2)^(3/2)*a/(c^2*x^3)
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (70) = 140\).
Time = 0.17 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.96 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=-\frac {1}{6} \, d^{3} {\left (\frac {3 \, b \log \left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4}\right )}{d^{2}} + \frac {16 \, {\left (3 \, b c^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} - 3 \, a d^{2} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{8} + 24 \, b c^{4} {\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 48 \, b c^{6} - 16 \, a c^{4} d^{2}\right )}}{{\left ({\left (\sqrt {d x + c} - \sqrt {d x - c}\right )}^{4} + 4 \, c^{2}\right )}^{3} d^{2}}\right )} \] Input:
integrate((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^4,x, algorithm="giac")
Output:
-1/6*d^3*(3*b*log((sqrt(d*x + c) - sqrt(d*x - c))^4)/d^2 + 16*(3*b*c^2*(sq rt(d*x + c) - sqrt(d*x - c))^8 - 3*a*d^2*(sqrt(d*x + c) - sqrt(d*x - c))^8 + 24*b*c^4*(sqrt(d*x + c) - sqrt(d*x - c))^4 + 48*b*c^6 - 16*a*c^4*d^2)/( ((sqrt(d*x + c) - sqrt(d*x - c))^4 + 4*c^2)^3*d^2))
Time = 6.77 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.81 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=\frac {b\,d+\frac {5\,b\,d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^2}}{\frac {4\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{\sqrt {-c}-\sqrt {d\,x-c}}+\frac {4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}{{\left (\sqrt {-c}-\sqrt {d\,x-c}\right )}^3}}-4\,b\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}-\sqrt {c}}{\sqrt {-c}-\sqrt {d\,x-c}}\right )-\frac {\left (\frac {a\,\sqrt {c+d\,x}}{3}-\frac {a\,d^2\,x^2\,\sqrt {c+d\,x}}{3\,c^2}\right )\,\sqrt {d\,x-c}}{x^3}+\frac {b\,d\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}{4\,\left (\sqrt {-c}-\sqrt {d\,x-c}\right )} \] Input:
int(((a + b*x^2)*(c + d*x)^(1/2)*(d*x - c)^(1/2))/x^4,x)
Output:
(b*d + (5*b*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*b*d*a tanh(((c + d*x)^(1/2) - c^(1/2))/((-c)^(1/2) - (d*x - c)^(1/2))) - (((a*(c + d*x)^(1/2))/3 - (a*d^2*x^2*(c + d*x)^(1/2))/(3*c^2))*(d*x - c)^(1/2))/x ^3 + (b*d*((c + d*x)^(1/2) - c^(1/2)))/(4*((-c)^(1/2) - (d*x - c)^(1/2)))
Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {-c+d x} \sqrt {c+d x} \left (a+b x^2\right )}{x^4} \, dx=\frac {-\sqrt {d x +c}\, \sqrt {d x -c}\, a \,c^{2}+\sqrt {d x +c}\, \sqrt {d x -c}\, a \,d^{2} x^{2}-3 \sqrt {d x +c}\, \sqrt {d x -c}\, b \,c^{2} x^{2}+6 \,\mathrm {log}\left (\frac {\sqrt {d x -c}+\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right ) b \,c^{2} d \,x^{3}+a \,d^{3} x^{3}+b \,c^{2} d \,x^{3}}{3 c^{2} x^{3}} \] Input:
int((d*x-c)^(1/2)*(d*x+c)^(1/2)*(b*x^2+a)/x^4,x)
Output:
( - sqrt(c + d*x)*sqrt( - c + d*x)*a*c**2 + sqrt(c + d*x)*sqrt( - c + d*x) *a*d**2*x**2 - 3*sqrt(c + d*x)*sqrt( - c + d*x)*b*c**2*x**2 + 6*log((sqrt( - c + d*x) + sqrt(c + d*x))/(sqrt(c)*sqrt(2)))*b*c**2*d*x**3 + a*d**3*x** 3 + b*c**2*d*x**3)/(3*c**2*x**3)