Integrand size = 40, antiderivative size = 62 \[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=-\frac {2 b}{5 a^2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}}+\frac {2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{3/4}}{3 a^2} \] Output:
-2/5*b/a^2/(a*x+(a^2*x^2-b)^(1/2))^(5/4)+2/3*(a*x+(a^2*x^2-b)^(1/2))^(3/4) /a^2
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\frac {4 \left (-4 b+5 a x \left (a x+\sqrt {-b+a^2 x^2}\right )\right )}{15 a^2 \left (a x+\sqrt {-b+a^2 x^2}\right )^{5/4}} \] Input:
Integrate[x/(Sqrt[-b + a^2*x^2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4)),x]
Output:
(4*(-4*b + 5*a*x*(a*x + Sqrt[-b + a^2*x^2])))/(15*a^2*(a*x + Sqrt[-b + a^2 *x^2])^(5/4))
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2545, 244, 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.
\(\displaystyle \int \frac {x}{\sqrt {a^2 x^2-b} \sqrt [4]{\sqrt {a^2 x^2-b}+a x}} \, dx\) |
\(\Big \downarrow \) 2545 |
\(\displaystyle \frac {\int \frac {\left (a x+\sqrt {a^2 x^2-b}\right )^2+b}{\left (a x+\sqrt {a^2 x^2-b}\right )^{9/4}}d\left (a x+\sqrt {a^2 x^2-b}\right )}{2 a^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\frac {b}{\left (a x+\sqrt {a^2 x^2-b}\right )^{9/4}}+\frac {1}{\sqrt [4]{a x+\sqrt {a^2 x^2-b}}}\right )d\left (a x+\sqrt {a^2 x^2-b}\right )}{2 a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {4}{3} \left (\sqrt {a^2 x^2-b}+a x\right )^{3/4}-\frac {4 b}{5 \left (\sqrt {a^2 x^2-b}+a x\right )^{5/4}}}{2 a^2}\) |
Input:
Int[x/(Sqrt[-b + a^2*x^2]*(a*x + Sqrt[-b + a^2*x^2])^(1/4)),x]
Output:
((-4*b)/(5*(a*x + Sqrt[-b + a^2*x^2])^(5/4)) + (4*(a*x + Sqrt[-b + a^2*x^2 ])^(3/4))/3)/(2*a^2)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 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])
\[\int \frac {x}{\sqrt {a^{2} x^{2}-b}\, \left (a x +\sqrt {a^{2} x^{2}-b}\right )^{\frac {1}{4}}}d x\]
Input:
int(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x)
Output:
int(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x)
Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.90 \[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=-\frac {4 \, {\left (3 \, a^{2} x^{2} - 3 \, \sqrt {a^{2} x^{2} - b} a x - 4 \, b\right )} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {3}{4}}}{15 \, a^{2} b} \] Input:
integrate(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x, algorithm=" fricas")
Output:
-4/15*(3*a^2*x^2 - 3*sqrt(a^2*x^2 - b)*a*x - 4*b)*(a*x + sqrt(a^2*x^2 - b) )^(3/4)/(a^2*b)
\[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\int \frac {x}{\sqrt [4]{a x + \sqrt {a^{2} x^{2} - b}} \sqrt {a^{2} x^{2} - b}}\, dx \] Input:
integrate(x/(a**2*x**2-b)**(1/2)/(a*x+(a**2*x**2-b)**(1/2))**(1/4),x)
Output:
Integral(x/((a*x + sqrt(a**2*x**2 - b))**(1/4)*sqrt(a**2*x**2 - b)), x)
\[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\int { \frac {x}{\sqrt {a^{2} x^{2} - b} {\left (a x + \sqrt {a^{2} x^{2} - b}\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x, algorithm=" maxima")
Output:
integrate(x/(sqrt(a^2*x^2 - b)*(a*x + sqrt(a^2*x^2 - b))^(1/4)), x)
Timed out. \[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\text {Timed out} \] Input:
integrate(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x, algorithm=" giac")
Output:
Timed out
Timed out. \[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\int \frac {x}{{\left (a\,x+\sqrt {a^2\,x^2-b}\right )}^{1/4}\,\sqrt {a^2\,x^2-b}} \,d x \] Input:
int(x/((a*x + (a^2*x^2 - b)^(1/2))^(1/4)*(a^2*x^2 - b)^(1/2)),x)
Output:
int(x/((a*x + (a^2*x^2 - b)^(1/2))^(1/4)*(a^2*x^2 - b)^(1/2)), x)
\[ \int \frac {x}{\sqrt {-b+a^2 x^2} \sqrt [4]{a x+\sqrt {-b+a^2 x^2}}} \, dx=\frac {4 \sqrt {a^{2} x^{2}-b}\, \left (\sqrt {a^{2} x^{2}-b}+a x \right )^{\frac {1}{4}} b -2 \sqrt {\sqrt {a^{2} x^{2}-b}+a x}\, \left (\int \frac {\left (\sqrt {a^{2} x^{2}-b}+a x \right )^{\frac {3}{4}} x^{3}}{a^{2} x^{2}-b}d x \right ) a^{4}+2 \sqrt {\sqrt {a^{2} x^{2}-b}+a x}\, \left (\int \frac {\left (\sqrt {a^{2} x^{2}-b}+a x \right )^{\frac {3}{4}} x}{a^{2} x^{2}-b}d x \right ) a^{2} b +2 \sqrt {\sqrt {a^{2} x^{2}-b}+a x}\, \left (\int \frac {\sqrt {a^{2} x^{2}-b}\, \left (\sqrt {a^{2} x^{2}-b}+a x \right )^{\frac {3}{4}} x^{2}}{a^{2} x^{2}-b}d x \right ) a^{3}-\sqrt {\sqrt {a^{2} x^{2}-b}+a x}\, \left (\int \frac {\sqrt {a^{2} x^{2}-b}\, \left (\sqrt {a^{2} x^{2}-b}+a x \right )^{\frac {3}{4}}}{a^{2} x^{2}-b}d x \right ) a b}{5 \sqrt {\sqrt {a^{2} x^{2}-b}+a x}\, a^{2} b} \] Input:
int(x/(a^2*x^2-b)^(1/2)/(a*x+(a^2*x^2-b)^(1/2))^(1/4),x)
Output:
(4*sqrt(a**2*x**2 - b)*(sqrt(a**2*x**2 - b) + a*x)**(1/4)*b - 2*sqrt(sqrt( a**2*x**2 - b) + a*x)*int(((sqrt(a**2*x**2 - b) + a*x)**(3/4)*x**3)/(a**2* x**2 - b),x)*a**4 + 2*sqrt(sqrt(a**2*x**2 - b) + a*x)*int(((sqrt(a**2*x**2 - b) + a*x)**(3/4)*x)/(a**2*x**2 - b),x)*a**2*b + 2*sqrt(sqrt(a**2*x**2 - b) + a*x)*int((sqrt(a**2*x**2 - b)*(sqrt(a**2*x**2 - b) + a*x)**(3/4)*x** 2)/(a**2*x**2 - b),x)*a**3 - sqrt(sqrt(a**2*x**2 - b) + a*x)*int((sqrt(a** 2*x**2 - b)*(sqrt(a**2*x**2 - b) + a*x)**(3/4))/(a**2*x**2 - b),x)*a*b)/(5 *sqrt(sqrt(a**2*x**2 - b) + a*x)*a**2*b)