Integrand size = 26, antiderivative size = 143 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\frac {(b c+3 a d) \sqrt {a+b x^2} \sqrt {c+d x^2}}{8 a c^2 x^2}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{4 a c x^4}+\frac {(b c-a d) (b c+3 a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{3/2} c^{5/2}} \] Output:
1/8*(3*a*d+b*c)*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/a/c^2/x^2-1/4*(b*x^2+a)^(3 /2)*(d*x^2+c)^(1/2)/a/c/x^4+1/8*(-a*d+b*c)*(3*a*d+b*c)*arctanh(c^(1/2)*(b* x^2+a)^(1/2)/a^(1/2)/(d*x^2+c)^(1/2))/a^(3/2)/c^(5/2)
Time = 1.11 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\frac {\sqrt {a+b x^2} \sqrt {c+d x^2} \left (-2 a c-b c x^2+3 a d x^2\right )}{8 a c^2 x^4}+\frac {\left (b^2 c^2+2 a b c d-3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{8 a^{3/2} c^{5/2}} \] Input:
Integrate[Sqrt[a + b*x^2]/(x^5*Sqrt[c + d*x^2]),x]
Output:
(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c - b*c*x^2 + 3*a*d*x^2))/(8*a*c^2* x^4) + ((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2] )/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(3/2)*c^(5/2))
Time = 0.24 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {354, 107, 105, 104, 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.
\(\displaystyle \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {b x^2+a}}{x^6 \sqrt {d x^2+c}}dx^2\) |
\(\Big \downarrow \) 107 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 a d+b c) \int \frac {\sqrt {b x^2+a}}{x^4 \sqrt {d x^2+c}}dx^2}{4 a c}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 a c x^4}\right )\) |
\(\Big \downarrow \) 105 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 a d+b c) \left (\frac {(b c-a d) \int \frac {1}{x^2 \sqrt {b x^2+a} \sqrt {d x^2+c}}dx^2}{2 c}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{c x^2}\right )}{4 a c}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 a c x^4}\right )\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 a d+b c) \left (\frac {(b c-a d) \int \frac {1}{c x^4-a}d\frac {\sqrt {b x^2+a}}{\sqrt {d x^2+c}}}{c}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{c x^2}\right )}{4 a c}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 a c x^4}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{2} \left (-\frac {(3 a d+b c) \left (-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{\sqrt {a} c^{3/2}}-\frac {\sqrt {a+b x^2} \sqrt {c+d x^2}}{c x^2}\right )}{4 a c}-\frac {\left (a+b x^2\right )^{3/2} \sqrt {c+d x^2}}{2 a c x^4}\right )\) |
Input:
Int[Sqrt[a + b*x^2]/(x^5*Sqrt[c + d*x^2]),x]
Output:
(-1/2*((a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(a*c*x^4) - ((b*c + 3*a*d)*(-((S qrt[a + b*x^2]*Sqrt[c + d*x^2])/(c*x^2)) - ((b*c - a*d)*ArcTanh[(Sqrt[c]*S qrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(Sqrt[a]*c^(3/2))))/(4*a*c))/2
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
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 + 1)/((m + 1)*(b*e - a*f))), x] - Simp[n*((d*e - c*f)/((m + 1)*(b*e - a*f))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p + 2, 0] && GtQ[n, 0] && (SumSimplerQ[m, 1] || !SumSimplerQ[p, 1]) && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x ] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 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]
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 = 0.58 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.20
method | result | size |
risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, \left (-3 a d \,x^{2}+x^{2} b c +2 a c \right )}{8 c^{2} x^{4} a}-\frac {\left (3 a^{2} d^{2}-2 a b c d -b^{2} c^{2}\right ) \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}}{16 a \,c^{2} \sqrt {a c}\, \sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\) | \(171\) |
default | \(-\frac {\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, \left (3 \ln \left (\frac {a d \,x^{2}+x^{2} b c +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}+2 a c}{x^{2}}\right ) a^{2} d^{2} x^{4}-2 \ln \left (\frac {a d \,x^{2}+x^{2} b c +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}+2 a c}{x^{2}}\right ) a b c d \,x^{4}-\ln \left (\frac {a d \,x^{2}+x^{2} b c +2 \sqrt {a c}\, \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}+2 a c}{x^{2}}\right ) b^{2} c^{2} x^{4}-6 d \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, a \,x^{2} \sqrt {a c}+2 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, b c \,x^{2} \sqrt {a c}+4 \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, a c \sqrt {a c}\right )}{16 a \,c^{2} \sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, x^{4} \sqrt {a c}}\) | \(306\) |
elliptic | \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \left (-\frac {b \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{8 a c \,x^{2}}+\frac {b \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d}{8 c \sqrt {a c}}+\frac {\ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) b^{2}}{16 a \sqrt {a c}}-\frac {\sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{4 c \,x^{4}}+\frac {3 \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}\, d}{8 c^{2} x^{2}}-\frac {3 a \ln \left (\frac {2 a c +\left (a d +b c \right ) x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+\left (a d +b c \right ) x^{2}+a c}}{x^{2}}\right ) d^{2}}{16 c^{2} \sqrt {a c}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}}\) | \(328\) |
Input:
int((b*x^2+a)^(1/2)/x^5/(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/8*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)*(-3*a*d*x^2+b*c*x^2+2*a*c)/c^2/x^4/a- 1/16/a/c^2*(3*a^2*d^2-2*a*b*c*d-b^2*c^2)/(a*c)^(1/2)*ln((2*a*c+(a*d+b*c)*x ^2+2*(a*c)^(1/2)*(b*d*x^4+(a*d+b*c)*x^2+a*c)^(1/2))/x^2)*((b*x^2+a)*(d*x^2 +c))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)
Time = 0.20 (sec) , antiderivative size = 358, normalized size of antiderivative = 2.50 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\left [-\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {a c} x^{4} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {a c}}{x^{4}}\right ) + 4 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{32 \, a^{2} c^{3} x^{4}}, -\frac {{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c} \sqrt {-a c}}{2 \, {\left (a b c d x^{4} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) + 2 \, {\left (2 \, a^{2} c^{2} + {\left (a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {d x^{2} + c}}{16 \, a^{2} c^{3} x^{4}}\right ] \] Input:
integrate((b*x^2+a)^(1/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="fricas")
Output:
[-1/32*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(a*c)*x^4*log(((b^2*c^2 + 6* a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4) + 4*(2* a^2*c^2 + (a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a^2 *c^3*x^4), -1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*sqrt(-a*c)*x^4*arctan( 1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-a*c)/( a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2)) + 2*(2*a^2*c^2 + (a*b*c^ 2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a^2*c^3*x^4)]
\[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\int \frac {\sqrt {a + b x^{2}}}{x^{5} \sqrt {c + d x^{2}}}\, dx \] Input:
integrate((b*x**2+a)**(1/2)/x**5/(d*x**2+c)**(1/2),x)
Output:
Integral(sqrt(a + b*x**2)/(x**5*sqrt(c + d*x**2)), x)
Exception generated. \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*x^2+a)^(1/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 1107 vs. \(2 (117) = 234\).
Time = 0.42 (sec) , antiderivative size = 1107, normalized size of antiderivative = 7.74 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx =\text {Too large to display} \] Input:
integrate((b*x^2+a)^(1/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="giac")
Output:
1/8*b*((sqrt(b*d)*b^3*c^2 + 2*sqrt(b*d)*a*b^2*c*d - 3*sqrt(b*d)*a^2*b*d^2) *arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b *x^2 + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*a*b*c^2) - 2*(sqrt(b*d)*b^9*c^5 - 7*sqrt(b*d)*a*b^8*c^4*d + 18*sqrt(b*d)*a^2*b^7*c^3* d^2 - 22*sqrt(b*d)*a^3*b^6*c^2*d^3 + 13*sqrt(b*d)*a^4*b^5*c*d^4 - 3*sqrt(b *d)*a^5*b^4*d^5 - 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b *x^2 + a)*b*d - a*b*d))^2*b^7*c^4 + 16*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d ) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a*b^6*c^3*d - 14*sqrt(b*d)*(s qrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^2*b^ 5*c^2*d^2 - 8*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^3*b^4*c*d^3 + 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^2*a^4*b^3*d^4 + 3*sqrt(b*d)*(sqr t(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*b^5*c^3 - 7*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a*b^4*c^2*d - 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2* c + (b*x^2 + a)*b*d - a*b*d))^4*a^2*b^3*c*d^2 - 9*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^4*a^3*b^2*d^3 - sqrt (b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^ 6*b^3*c^2 - 2*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) - sqrt(b^2*c + (b*x^2 + a)*b*d - a*b*d))^6*a*b^2*c*d + 3*sqrt(b*d)*(sqrt(b*x^2 + a)*sqrt(b*d) ...
Time = 22.21 (sec) , antiderivative size = 955, normalized size of antiderivative = 6.68 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx =\text {Too large to display} \] Input:
int((a + b*x^2)^(1/2)/(x^5*(c + d*x^2)^(1/2)),x)
Output:
(log(((a + b*x^2)^(1/2) - a^(1/2))/((c + d*x^2)^(1/2) - c^(1/2)))*(a^(1/2) *b^2*c^(5/2) - 3*a^(5/2)*c^(1/2)*d^2 + 2*a^(3/2)*b*c^(3/2)*d))/(16*a^2*c^3 ) - (((a + b*x^2)^(1/2) - a^(1/2))*((b*d)/(8*a*c) - (3*d*(a*d + b*c))/(32* a*c^2)))/((c + d*x^2)^(1/2) - c^(1/2)) - ((((a + b*x^2)^(1/2) - a^(1/2))^5 *((a^2*d^2)/8 + (5*b^2*c^2)/32 - (11*a*b*c*d)/32))/(a*c^3*((c + d*x^2)^(1/ 2) - c^(1/2))^5) - b^4/(64*a^(1/2)*c^(3/2)*d^2) + (((a + b*x^2)^(1/2) - a^ (1/2))^2*((11*a^2*b^2*d^2)/64 - (5*b^4*c^2)/64 + (a*b^3*c*d)/16))/(a^(3/2) *c^(5/2)*d^2*((c + d*x^2)^(1/2) - c^(1/2))^2) + (((a + b*x^2)^(1/2) - a^(1 /2))^3*((b^4*c^3)/32 + (a^3*b*d^3)/32 - (9*a^2*b^2*c*d^2)/16 + (3*a*b^3*c^ 2*d)/16))/(a^2*c^3*d^2*((c + d*x^2)^(1/2) - c^(1/2))^3) + (((a + b*x^2)^(1 /2) - a^(1/2))*((b^4*c)/16 - (a*b^3*d)/16))/(a*c^2*d^2*((c + d*x^2)^(1/2) - c^(1/2))) + (((a + b*x^2)^(1/2) - a^(1/2))^4*((b^4*c^4)/64 - (7*a^4*d^4) /64 + (21*a^2*b^2*c^2*d^2)/64 - (a*b^3*c^3*d)/4 + (a^3*b*c*d^3)/4))/(a^(5/ 2)*c^(7/2)*d^2*((c + d*x^2)^(1/2) - c^(1/2))^4))/(((a + b*x^2)^(1/2) - a^( 1/2))^6/((c + d*x^2)^(1/2) - c^(1/2))^6 + (b^2*((a + b*x^2)^(1/2) - a^(1/2 ))^2)/(d^2*((c + d*x^2)^(1/2) - c^(1/2))^2) - (((a + b*x^2)^(1/2) - a^(1/2 ))^3*(2*b^2*c + 2*a*b*d))/(a^(1/2)*c^(1/2)*d^2*((c + d*x^2)^(1/2) - c^(1/2 ))^3) - (((a + b*x^2)^(1/2) - a^(1/2))^5*(2*a*d + 2*b*c))/(a^(1/2)*c^(1/2) *d*((c + d*x^2)^(1/2) - c^(1/2))^5) + (((a + b*x^2)^(1/2) - a^(1/2))^4*(a^ 2*d^2 + b^2*c^2 + 4*a*b*c*d))/(a*c*d^2*((c + d*x^2)^(1/2) - c^(1/2))^4)...
Time = 0.24 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {a+b x^2}}{x^5 \sqrt {c+d x^2}} \, dx=\frac {-2 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} c^{2}+3 \sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a^{2} c d \,x^{2}-\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}\, a b \,c^{2} x^{2}-3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a^{2} d^{2} x^{4}+2 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) a b c d \,x^{4}+\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {a}\, \sqrt {b \,x^{2}+a}\, c +\sqrt {c}\, \sqrt {d \,x^{2}+c}\, a \right ) b^{2} c^{2} x^{4}+3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a^{2} d^{2} x^{4}-2 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) a b c d \,x^{4}-\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (x \right ) b^{2} c^{2} x^{4}}{8 a^{2} c^{3} x^{4}} \] Input:
int((b*x^2+a)^(1/2)/x^5/(d*x^2+c)^(1/2),x)
Output:
( - 2*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*c**2 + 3*sqrt(c + d*x**2)*sqr t(a + b*x**2)*a**2*c*d*x**2 - sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b*c**2*x **2 - 3*sqrt(c)*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*a**2*d**2*x**4 + 2*sqrt(c)*sqrt(a)*log(sqrt(a)*sqrt(a + b*x**2) *c + sqrt(c)*sqrt(c + d*x**2)*a)*a*b*c*d*x**4 + sqrt(c)*sqrt(a)*log(sqrt(a )*sqrt(a + b*x**2)*c + sqrt(c)*sqrt(c + d*x**2)*a)*b**2*c**2*x**4 + 3*sqrt (c)*sqrt(a)*log(x)*a**2*d**2*x**4 - 2*sqrt(c)*sqrt(a)*log(x)*a*b*c*d*x**4 - sqrt(c)*sqrt(a)*log(x)*b**2*c**2*x**4)/(8*a**2*c**3*x**4)