Integrand size = 26, antiderivative size = 176 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=-\frac {12 a^2}{b^4 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}+\frac {2 a^3}{b^4 \left (a+b \sqrt [4]{x}\right ) \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}+\frac {4 \left (a+b \sqrt [4]{x}\right ) \sqrt [4]{x}}{b^3 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}-\frac {12 a \left (a+b \sqrt [4]{x}\right ) \log \left (a+b \sqrt [4]{x}\right )}{b^4 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}} \]
-12*a^2/b^4/(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(1/2)+2*a^3/b^4/(a+b*x^(1/4))/ (a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(1/2)+4*(a+b*x^(1/4))*x^(1/4)/b^3/(a^2+2*a *b*x^(1/4)+b^2*x^(1/2))^(1/2)-12*a*(a+b*x^(1/4))*ln(a+b*x^(1/4))/b^4/(a^2+ 2*a*b*x^(1/4)+b^2*x^(1/2))^(1/2)
Time = 0.09 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\frac {2 \left (-5 a^3-4 a^2 b \sqrt [4]{x}+4 a b^2 \sqrt {x}+2 b^3 x^{3/4}-6 a \left (a+b \sqrt [4]{x}\right )^2 \log \left (a+b \sqrt [4]{x}\right )\right )}{b^4 \left (a+b \sqrt [4]{x}\right ) \sqrt {\left (a+b \sqrt [4]{x}\right )^2}} \]
(2*(-5*a^3 - 4*a^2*b*x^(1/4) + 4*a*b^2*Sqrt[x] + 2*b^3*x^(3/4) - 6*a*(a + b*x^(1/4))^2*Log[a + b*x^(1/4)]))/(b^4*(a + b*x^(1/4))*Sqrt[(a + b*x^(1/4) )^2])
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.64, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 774, 27, 49, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\left (a b^3+b^4 \sqrt [4]{x}\right ) \int \frac {1}{\left (\sqrt [4]{x} b^2+a b\right )^3}dx}{\sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}\) |
\(\Big \downarrow \) 774 |
\(\displaystyle \frac {4 \left (a b^3+b^4 \sqrt [4]{x}\right ) \int \frac {x^{3/4}}{b^3 \left (a+b \sqrt [4]{x}\right )^3}d\sqrt [4]{x}}{\sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \left (a b^3+b^4 \sqrt [4]{x}\right ) \int \frac {x^{3/4}}{\left (a+b \sqrt [4]{x}\right )^3}d\sqrt [4]{x}}{b^3 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {4 \left (a b^3+b^4 \sqrt [4]{x}\right ) \int \left (-\frac {a^3}{b^3 \left (a+b \sqrt [4]{x}\right )^3}+\frac {3 a^2}{b^3 \left (a+b \sqrt [4]{x}\right )^2}-\frac {3 a}{b^3 \left (a+b \sqrt [4]{x}\right )}+\frac {1}{b^3}\right )d\sqrt [4]{x}}{b^3 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \left (a b^3+b^4 \sqrt [4]{x}\right ) \left (\frac {a^3}{2 b^4 \left (a+b \sqrt [4]{x}\right )^2}-\frac {3 a^2}{b^4 \left (a+b \sqrt [4]{x}\right )}-\frac {3 a \log \left (a+b \sqrt [4]{x}\right )}{b^4}+\frac {\sqrt [4]{x}}{b^3}\right )}{b^3 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}\) |
(4*(a*b^3 + b^4*x^(1/4))*(a^3/(2*b^4*(a + b*x^(1/4))^2) - (3*a^2)/(b^4*(a + b*x^(1/4))) + x^(1/4)/b^3 - (3*a*Log[a + b*x^(1/4)])/b^4))/(b^3*Sqrt[a^2 + 2*a*b*x^(1/4) + b^2*Sqrt[x]])
3.5.79.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.59
method | result | size |
derivativedivides | \(-\frac {2 \left (6 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a \,b^{2} \sqrt {x}-2 b^{3} x^{\frac {3}{4}}+12 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a^{2} b \,x^{\frac {1}{4}}-4 a \,b^{2} \sqrt {x}+6 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a^{3}+4 a^{2} b \,x^{\frac {1}{4}}+5 a^{3}\right ) \left (a +b \,x^{\frac {1}{4}}\right )}{b^{4} {\left (\left (a +b \,x^{\frac {1}{4}}\right )^{2}\right )}^{\frac {3}{2}}}\) | \(103\) |
default | \(\text {Expression too large to display}\) | \(1473\) |
-2*(6*ln(a+b*x^(1/4))*a*b^2*x^(1/2)-2*b^3*x^(3/4)+12*ln(a+b*x^(1/4))*a^2*b *x^(1/4)-4*a*b^2*x^(1/2)+6*ln(a+b*x^(1/4))*a^3+4*a^2*b*x^(1/4)+5*a^3)*(a+b *x^(1/4))/b^4/((a+b*x^(1/4))^2)^(3/2)
Time = 1.99 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\frac {2 \, {\left (9 \, a^{5} b^{4} x - 5 \, a^{9} - 6 \, {\left (a b^{8} x^{2} - 2 \, a^{5} b^{4} x + a^{9}\right )} \log \left (b x^{\frac {1}{4}} + a\right ) - 2 \, {\left (3 \, a^{2} b^{7} x - a^{6} b^{3}\right )} x^{\frac {3}{4}} + {\left (7 \, a^{3} b^{6} x - 3 \, a^{7} b^{2}\right )} \sqrt {x} + 2 \, {\left (b^{9} x^{2} - 6 \, a^{4} b^{5} x + 3 \, a^{8} b\right )} x^{\frac {1}{4}}\right )}}{b^{12} x^{2} - 2 \, a^{4} b^{8} x + a^{8} b^{4}} \]
2*(9*a^5*b^4*x - 5*a^9 - 6*(a*b^8*x^2 - 2*a^5*b^4*x + a^9)*log(b*x^(1/4) + a) - 2*(3*a^2*b^7*x - a^6*b^3)*x^(3/4) + (7*a^3*b^6*x - 3*a^7*b^2)*sqrt(x ) + 2*(b^9*x^2 - 6*a^4*b^5*x + 3*a^8*b)*x^(1/4))/(b^12*x^2 - 2*a^4*b^8*x + a^8*b^4)
\[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b \sqrt [4]{x} + b^{2} \sqrt {x}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\frac {4 \, \sqrt {x}}{\sqrt {b^{2} \sqrt {x} + 2 \, a b x^{\frac {1}{4}} + a^{2}} b^{2}} - \frac {12 \, a \log \left (x^{\frac {1}{4}} + \frac {a}{b}\right )}{b^{4}} + \frac {8 \, a^{2}}{\sqrt {b^{2} \sqrt {x} + 2 \, a b x^{\frac {1}{4}} + a^{2}} b^{4}} - \frac {24 \, a^{2} x^{\frac {1}{4}}}{b^{5} {\left (x^{\frac {1}{4}} + \frac {a}{b}\right )}^{2}} - \frac {22 \, a^{3}}{b^{6} {\left (x^{\frac {1}{4}} + \frac {a}{b}\right )}^{2}} \]
4*sqrt(x)/(sqrt(b^2*sqrt(x) + 2*a*b*x^(1/4) + a^2)*b^2) - 12*a*log(x^(1/4) + a/b)/b^4 + 8*a^2/(sqrt(b^2*sqrt(x) + 2*a*b*x^(1/4) + a^2)*b^4) - 24*a^2 *x^(1/4)/(b^5*(x^(1/4) + a/b)^2) - 22*a^3/(b^6*(x^(1/4) + a/b)^2)
Timed out. \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}\right )^{3/2}} \, dx=\int \frac {1}{{\left (a^2+b^2\,\sqrt {x}+2\,a\,b\,x^{1/4}\right )}^{3/2}} \,d x \]