Integrand size = 42, antiderivative size = 540 \[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\frac {63 b^3+1034 a^2 b^2 x^2-1652 a^4 b x^4+112 a^6 x^6+\sqrt {-b+a^2 x^2} \left (250 a b^2 x-1596 a^3 b x^3+112 a^5 x^5\right )}{63 x \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+\frac {1}{4} \sqrt {2-\sqrt {2}} a b^{9/8} \arctan \left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{b}-\frac {2 \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{-\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {2}} a b^{9/8} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{-\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )+\frac {1}{4} \sqrt {2-\sqrt {2}} a b^{9/8} \text {arctanh}\left (\frac {\frac {\sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}+\frac {\sqrt {a x+\sqrt {-b+a^2 x^2}}}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}}{\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )+\frac {1}{4} \sqrt {2+\sqrt {2}} a b^{9/8} \text {arctanh}\left (\frac {\frac {\sqrt [8]{b}}{\sqrt {2+\sqrt {2}}}+\frac {\sqrt {a x+\sqrt {-b+a^2 x^2}}}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}}{\sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right ) \]
1/63*(63*b^3+1034*a^2*b^2*x^2-1652*a^4*b*x^4+112*a^6*x^6+(a^2*x^2-b)^(1/2) *(112*a^5*x^5-1596*a^3*b*x^3+250*a*b^2*x))/x/(a*x+(a^2*x^2-b)^(1/2))^(11/4 )+1/4*(2-2^(1/2))^(1/2)*a*b^(9/8)*arctan((2^(1/2)/(2-2^(1/2))^(1/2)*b^(1/8 )-2*b^(1/8)/(2-2^(1/2))^(1/2))*(a*x+(a^2*x^2-b)^(1/2))^(1/4)/(-b^(1/4)+(a* x+(a^2*x^2-b)^(1/2))^(1/2)))-1/4*(2+2^(1/2))^(1/2)*a*b^(9/8)*arctan((2+2^( 1/2))^(1/2)*b^(1/8)*(a*x+(a^2*x^2-b)^(1/2))^(1/4)/(-b^(1/4)+(a*x+(a^2*x^2- b)^(1/2))^(1/2)))+1/4*(2-2^(1/2))^(1/2)*a*b^(9/8)*arctanh((b^(1/8)/(2-2^(1 /2))^(1/2)+(a*x+(a^2*x^2-b)^(1/2))^(1/2)/(2-2^(1/2))^(1/2)/b^(1/8))/(a*x+( a^2*x^2-b)^(1/2))^(1/4))+1/4*(2+2^(1/2))^(1/2)*a*b^(9/8)*arctanh((b^(1/8)/ (2+2^(1/2))^(1/2)+(a*x+(a^2*x^2-b)^(1/2))^(1/2)/(2+2^(1/2))^(1/2)/b^(1/8)) /(a*x+(a^2*x^2-b)^(1/2))^(1/4))
Time = 1.71 (sec) , antiderivative size = 481, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\frac {1}{252} \left (\frac {4 \left (63 b^3+1034 a^2 b^2 x^2-1652 a^4 b x^4+112 a^6 x^6+2 \sqrt {-b+a^2 x^2} \left (125 a b^2 x-798 a^3 b x^3+56 a^5 x^5\right )\right )}{x \left (a x+\sqrt {-b+a^2 x^2}\right )^{11/4}}+63 \sqrt {2-\sqrt {2}} a b^{9/8} \arctan \left (\frac {\sqrt {2-\sqrt {2}} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [4]{b}-\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )+63 \sqrt {2+\sqrt {2}} a b^{9/8} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{\sqrt [4]{b}-\sqrt {a x+\sqrt {-b+a^2 x^2}}}\right )+63 \sqrt {2+\sqrt {2}} a b^{9/8} \text {arctanh}\left (\frac {\sqrt {1-\frac {1}{\sqrt {2}}} \left (\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{\sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )+63 \sqrt {2-\sqrt {2}} a b^{9/8} \text {arctanh}\left (\frac {\sqrt {1+\frac {1}{\sqrt {2}}} \left (\sqrt [4]{b}+\sqrt {a x+\sqrt {-b+a^2 x^2}}\right )}{\sqrt [8]{b} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}\right )\right ) \]
((4*(63*b^3 + 1034*a^2*b^2*x^2 - 1652*a^4*b*x^4 + 112*a^6*x^6 + 2*Sqrt[-b + a^2*x^2]*(125*a*b^2*x - 798*a^3*b*x^3 + 56*a^5*x^5)))/(x*(a*x + Sqrt[-b + a^2*x^2])^(11/4)) + 63*Sqrt[2 - Sqrt[2]]*a*b^(9/8)*ArcTan[(Sqrt[2 - Sqrt [2]]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(b^(1/4) - Sqrt[a*x + Sqrt[ -b + a^2*x^2]])] + 63*Sqrt[2 + Sqrt[2]]*a*b^(9/8)*ArcTan[(Sqrt[2 + Sqrt[2] ]*b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(b^(1/4) - Sqrt[a*x + Sqrt[-b + a^2*x^2]])] + 63*Sqrt[2 + Sqrt[2]]*a*b^(9/8)*ArcTanh[(Sqrt[1 - 1/Sqrt[2] ]*(b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]))/(b^(1/8)*(a*x + Sqrt[-b + a^ 2*x^2])^(1/4))] + 63*Sqrt[2 - Sqrt[2]]*a*b^(9/8)*ArcTanh[(Sqrt[1 + 1/Sqrt[ 2]]*(b^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]))/(b^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4))])/252
Time = 0.67 (sec) , antiderivative size = 522, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2545, 368, 968, 27, 1040, 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.
| \(\displaystyle \int \frac {\left (a^2 x^2-b\right )^{3/2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{x^2} \, dx\) |
| \(\Big \downarrow \) 2545 |
| \(\displaystyle \frac {1}{4} a \int \frac {\left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^4}{\left (a x+\sqrt {a^2 x^2-b}\right )^{11/4} \left (\left (a x+\sqrt {a^2 x^2-b}\right )^2+b\right )^2}d\left (a x+\sqrt {a^2 x^2-b}\right )\) |
| \(\Big \downarrow \) 368 |
| \(\displaystyle a \int \frac {\left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^4}{\left (a x+\sqrt {a^2 x^2-b}\right )^2 \left (\left (a x+\sqrt {a^2 x^2-b}\right )^2+b\right )^2}d\sqrt [4]{a x+\sqrt {a^2 x^2-b}}\) |
| \(\Big \downarrow \) 968 |
| \(\displaystyle a \left (\frac {\left (b-\left (\sqrt {a^2 x^2-b}+a x\right )^2\right )^3}{4 \left (\sqrt {a^2 x^2-b}+a x\right )^{7/4} \left (\left (\sqrt {a^2 x^2-b}+a x\right )^2+b\right )}-\frac {\int -\frac {2 b \left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^2 \left (13 \left (a x+\sqrt {a^2 x^2-b}\right )^2+11 b\right )}{\left (a x+\sqrt {a^2 x^2-b}\right )^2 \left (\left (a x+\sqrt {a^2 x^2-b}\right )^2+b\right )}d\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}{8 b}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle a \left (\frac {1}{4} \int \frac {\left (b-\left (a x+\sqrt {a^2 x^2-b}\right )^2\right )^2 \left (13 \left (a x+\sqrt {a^2 x^2-b}\right )^2+11 b\right )}{\left (a x+\sqrt {a^2 x^2-b}\right )^2 \left (\left (a x+\sqrt {a^2 x^2-b}\right )^2+b\right )}d\sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\frac {\left (b-\left (\sqrt {a^2 x^2-b}+a x\right )^2\right )^3}{4 \left (\sqrt {a^2 x^2-b}+a x\right )^{7/4} \left (\left (\sqrt {a^2 x^2-b}+a x\right )^2+b\right )}\right )\) |
| \(\Big \downarrow \) 1040 |
| \(\displaystyle a \left (\frac {1}{4} \int \left (\frac {8 b^2}{\left (a x+\sqrt {a^2 x^2-b}\right )^2+b}+\frac {11 b^2}{\left (a x+\sqrt {a^2 x^2-b}\right )^2}-28 b+13 \left (a x+\sqrt {a^2 x^2-b}\right )^2\right )d\sqrt [4]{a x+\sqrt {a^2 x^2-b}}+\frac {\left (b-\left (\sqrt {a^2 x^2-b}+a x\right )^2\right )^3}{4 \left (\sqrt {a^2 x^2-b}+a x\right )^{7/4} \left (\left (\sqrt {a^2 x^2-b}+a x\right )^2+b\right )}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle a \left (\frac {1}{4} \left (-2 (-b)^{9/8} \arctan \left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )+\sqrt {2} (-b)^{9/8} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )-\sqrt {2} (-b)^{9/8} \arctan \left (\frac {\sqrt {2} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}+1\right )-2 (-b)^{9/8} \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {a^2 x^2-b}+a x}}{\sqrt [8]{-b}}\right )-\frac {11 b^2}{7 \left (\sqrt {a^2 x^2-b}+a x\right )^{7/4}}+\frac {13}{9} \left (\sqrt {a^2 x^2-b}+a x\right )^{9/4}-28 b \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\frac {(-b)^{9/8} \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}-\sqrt {2} \sqrt [8]{-b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{-b}\right )}{\sqrt {2}}-\frac {(-b)^{9/8} \log \left (\sqrt {\sqrt {a^2 x^2-b}+a x}+\sqrt {2} \sqrt [8]{-b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}+\sqrt [4]{-b}\right )}{\sqrt {2}}\right )+\frac {\left (b-\left (\sqrt {a^2 x^2-b}+a x\right )^2\right )^3}{4 \left (\sqrt {a^2 x^2-b}+a x\right )^{7/4} \left (\left (\sqrt {a^2 x^2-b}+a x\right )^2+b\right )}\right )\) |
a*((b - (a*x + Sqrt[-b + a^2*x^2])^2)^3/(4*(a*x + Sqrt[-b + a^2*x^2])^(7/4 )*(b + (a*x + Sqrt[-b + a^2*x^2])^2)) + ((-11*b^2)/(7*(a*x + Sqrt[-b + a^2 *x^2])^(7/4)) - 28*b*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + (13*(a*x + Sqrt[-b + a^2*x^2])^(9/4))/9 - 2*(-b)^(9/8)*ArcTan[(a*x + Sqrt[-b + a^2*x^2])^(1/ 4)/(-b)^(1/8)] + Sqrt[2]*(-b)^(9/8)*ArcTan[1 - (Sqrt[2]*(a*x + Sqrt[-b + a ^2*x^2])^(1/4))/(-b)^(1/8)] - Sqrt[2]*(-b)^(9/8)*ArcTan[1 + (Sqrt[2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4))/(-b)^(1/8)] - 2*(-b)^(9/8)*ArcTanh[(a*x + Sqr t[-b + a^2*x^2])^(1/4)/(-b)^(1/8)] + ((-b)^(9/8)*Log[(-b)^(1/4) - Sqrt[2]* (-b)^(1/8)*(a*x + Sqrt[-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2 ]]])/Sqrt[2] - ((-b)^(9/8)*Log[(-b)^(1/4) + Sqrt[2]*(-b)^(1/8)*(a*x + Sqrt [-b + a^2*x^2])^(1/4) + Sqrt[a*x + Sqrt[-b + a^2*x^2]]])/Sqrt[2])/4)
3.31.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) *((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1)) Int [(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c *b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ (g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c , d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
Int[(x_)^(p_.)*((g_) + (i_.)*(x_)^2)^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Simp[(1/(2^(2*m + p + 1)*e^(p + 1)*f^(2* m)))*(i/c)^m Subst[Int[x^(n - 2*m - p - 2)*((-a)*f^2 + x^2)^p*(a*f^2 + x^ 2)^(2*m + 1), x], x, e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && EqQ[c*g - a*i, 0] && IntegersQ[p, 2*m] && (IntegerQ[m] || GtQ[i/c, 0])
\[\int \frac {\left (a^{2} x^{2}-b \right )^{\frac {3}{2}} \left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}}{x^{2}}d x\]
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 486, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\frac {\left (63 i + 63\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left (2 \, {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b + \left (i + 1\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) - \left (63 i - 63\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left (2 \, {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b - \left (i - 1\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) + \left (63 i - 63\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left (2 \, {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b + \left (i - 1\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) - \left (63 i + 63\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left (2 \, {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b - \left (i + 1\right ) \, \sqrt {2} \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) + 126 \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b + \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) + 126 i \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b + i \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) - 126 i \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b - i \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) - 126 \, \left (-a^{8} b^{9}\right )^{\frac {1}{8}} x \log \left ({\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}} a b - \left (-a^{8} b^{9}\right )^{\frac {1}{8}}\right ) - 8 \, {\left (4 \, a^{3} x^{3} + 439 \, a b x - {\left (32 \, a^{2} x^{2} + 63 \, b\right )} \sqrt {a^{2} x^{2} - b}\right )} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}}{504 \, x} \]
1/504*((63*I + 63)*sqrt(2)*(-a^8*b^9)^(1/8)*x*log(2*(a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b + (I + 1)*sqrt(2)*(-a^8*b^9)^(1/8)) - (63*I - 63)*sqrt(2)*(- a^8*b^9)^(1/8)*x*log(2*(a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b - (I - 1)*sqrt( 2)*(-a^8*b^9)^(1/8)) + (63*I - 63)*sqrt(2)*(-a^8*b^9)^(1/8)*x*log(2*(a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b + (I - 1)*sqrt(2)*(-a^8*b^9)^(1/8)) - (63*I + 63)*sqrt(2)*(-a^8*b^9)^(1/8)*x*log(2*(a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b - (I + 1)*sqrt(2)*(-a^8*b^9)^(1/8)) + 126*(-a^8*b^9)^(1/8)*x*log((a*x + s qrt(a^2*x^2 - b))^(1/4)*a*b + (-a^8*b^9)^(1/8)) + 126*I*(-a^8*b^9)^(1/8)*x *log((a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b + I*(-a^8*b^9)^(1/8)) - 126*I*(-a ^8*b^9)^(1/8)*x*log((a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b - I*(-a^8*b^9)^(1/ 8)) - 126*(-a^8*b^9)^(1/8)*x*log((a*x + sqrt(a^2*x^2 - b))^(1/4)*a*b - (-a ^8*b^9)^(1/8)) - 8*(4*a^3*x^3 + 439*a*b*x - (32*a^2*x^2 + 63*b)*sqrt(a^2*x ^2 - b))*(a*x + sqrt(a^2*x^2 - b))^(1/4))/x
\[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\int \frac {\sqrt [4]{a x + \sqrt {a^{2} x^{2} - b}} \left (a^{2} x^{2} - b\right )^{\frac {3}{2}}}{x^{2}}\, dx \]
\[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\int { \frac {{\left (a^{2} x^{2} - b\right )}^{\frac {3}{2}} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\left (-b+a^2 x^2\right )^{3/2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}}{x^2} \, dx=\int \frac {{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/4}\,{\left (a^2\,x^2-b\right )}^{3/2}}{x^2} \,d x \]