Integrand size = 26, antiderivative size = 391 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=-\frac {3 b^7 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{4 \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{4/3}}-\frac {7 a b^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{\left (a+\frac {b}{\sqrt [3]{x}}\right ) x}-\frac {63 a^2 b^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{2/3}}-\frac {105 a^3 b^4 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{\left (a+\frac {b}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}+\frac {63 a^5 b^2 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \sqrt [3]{x}}{a+\frac {b}{\sqrt [3]{x}}}+\frac {21 a^6 b \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right )}+\frac {a^7 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x}{a+\frac {b}{\sqrt [3]{x}}}+\frac {105 a^4 b^3 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \log \left (\sqrt [3]{x}\right )}{a+\frac {b}{\sqrt [3]{x}}} \]
-3/4*b^7*(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x^(4/3)-7*a*b ^6*(a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x-63/2*a^2*b^5*(a^2 +b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x^(2/3)-105*a^3*b^4*(a^2+b ^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))/x^(1/3)+63*a^5*b^2*x^(1/3)*( a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))+21/2*a^6*b*x^(2/3)*(a^2 +b^2/x^(2/3)+2*a*b/x^(1/3))^(1/2)/(a+b/x^(1/3))+a^7*x*(a^2+b^2/x^(2/3)+2*a *b/x^(1/3))^(1/2)/(a+b/x^(1/3))+35*a^4*b^3*ln(x)*(a^2+b^2/x^(2/3)+2*a*b/x^ (1/3))^(1/2)/(a+b/x^(1/3))
Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.32 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\frac {\sqrt {\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}} \left (-3 b^7-28 a b^6 \sqrt [3]{x}-126 a^2 b^5 x^{2/3}-420 a^3 b^4 x+252 a^5 b^2 x^{5/3}+42 a^6 b x^2+4 a^7 x^{7/3}+140 a^4 b^3 x^{4/3} \log (x)\right )}{4 \left (b+a \sqrt [3]{x}\right ) x} \]
(Sqrt[(b + a*x^(1/3))^2/x^(2/3)]*(-3*b^7 - 28*a*b^6*x^(1/3) - 126*a^2*b^5* x^(2/3) - 420*a^3*b^4*x + 252*a^5*b^2*x^(5/3) + 42*a^6*b*x^2 + 4*a^7*x^(7/ 3) + 140*a^4*b^3*x^(4/3)*Log[x]))/(4*(b + a*x^(1/3))*x)
Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.38, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1384, 774, 27, 795, 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.
| \(\displaystyle \int \left (a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}\right )^{7/2} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {\sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (\frac {b^2}{\sqrt [3]{x}}+a b\right )^7dx}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
| \(\Big \downarrow \) 774 |
| \(\displaystyle \frac {3 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int b^7 \left (a+\frac {b}{\sqrt [3]{x}}\right )^7 x^{2/3}d\sqrt [3]{x}}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 b^7 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (a+\frac {b}{\sqrt [3]{x}}\right )^7 x^{2/3}d\sqrt [3]{x}}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
| \(\Big \downarrow \) 795 |
| \(\displaystyle \frac {3 b^7 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \frac {\left (\sqrt [3]{x} a+b\right )^7}{x^{5/3}}d\sqrt [3]{x}}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {3 b^7 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \int \left (x^{2/3} a^7+7 b \sqrt [3]{x} a^6+21 b^2 a^5+\frac {35 b^3 a^4}{\sqrt [3]{x}}+\frac {35 b^4 a^3}{x^{2/3}}+\frac {21 b^5 a^2}{x}+\frac {7 b^6 a}{x^{4/3}}+\frac {b^7}{x^{5/3}}\right )d\sqrt [3]{x}}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 b^7 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \left (\frac {a^7 x}{3}+\frac {7}{2} a^6 b x^{2/3}+21 a^5 b^2 \sqrt [3]{x}+35 a^4 b^3 \log \left (\sqrt [3]{x}\right )-\frac {35 a^3 b^4}{\sqrt [3]{x}}-\frac {21 a^2 b^5}{2 x^{2/3}}-\frac {7 a b^6}{3 x}-\frac {b^7}{4 x^{4/3}}\right )}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\) |
(3*b^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*(-1/4*b^7/x^(4/3) - (7*a* b^6)/(3*x) - (21*a^2*b^5)/(2*x^(2/3)) - (35*a^3*b^4)/x^(1/3) + 21*a^5*b^2* x^(1/3) + (7*a^6*b*x^(2/3))/2 + (a^7*x)/3 + 35*a^4*b^3*Log[x^(1/3)]))/(a*b ^7 + b^8/x^(1/3))
3.5.82.3.1 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[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[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 = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.29
| method | result | size |
| derivativedivides | \(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}} x \left (4 a^{7} x^{\frac {7}{3}}+42 x^{2} a^{6} b +140 a^{4} b^{3} \ln \left (x \right ) x^{\frac {4}{3}}+252 b^{2} a^{5} x^{\frac {5}{3}}-420 b^{4} a^{3} x -126 a^{2} b^{5} x^{\frac {2}{3}}-28 a \,b^{6} x^{\frac {1}{3}}-3 b^{7}\right )}{4 \left (b +a \,x^{\frac {1}{3}}\right )^{7}}\) | \(113\) |
| default | \(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}} \left (42 a^{6} b \,x^{3}+252 b^{2} a^{5} x^{\frac {8}{3}}+140 a^{4} b^{3} \ln \left (x \right ) x^{\frac {7}{3}}+4 a^{7} x^{\frac {10}{3}}-28 a \,b^{6} x^{\frac {4}{3}}-420 b^{4} a^{3} x^{2}-126 a^{2} b^{5} x^{\frac {5}{3}}-3 b^{7} x \right )}{4 \left (b +a \,x^{\frac {1}{3}}\right )^{7}}\) | \(115\) |
1/4*((x^(2/3)*a^2+2*a*b*x^(1/3)+b^2)/x^(2/3))^(7/2)*x*(4*a^7*x^(7/3)+42*x^ 2*a^6*b+140*a^4*b^3*ln(x)*x^(4/3)+252*b^2*a^5*x^(5/3)-420*b^4*a^3*x-126*a^ 2*b^5*x^(2/3)-28*a*b^6*x^(1/3)-3*b^7)/(b+a*x^(1/3))^7
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\text {Timed out} \]
\[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.20 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=35 \, a^{4} b^{3} \log \left (x\right ) + \frac {4 \, a^{7} x^{\frac {7}{3}} + 42 \, a^{6} b x^{2} + 252 \, a^{5} b^{2} x^{\frac {5}{3}} - 420 \, a^{3} b^{4} x - 126 \, a^{2} b^{5} x^{\frac {2}{3}} - 28 \, a b^{6} x^{\frac {1}{3}} - 3 \, b^{7}}{4 \, x^{\frac {4}{3}}} \]
35*a^4*b^3*log(x) + 1/4*(4*a^7*x^(7/3) + 42*a^6*b*x^2 + 252*a^5*b^2*x^(5/3 ) - 420*a^3*b^4*x - 126*a^2*b^5*x^(2/3) - 28*a*b^6*x^(1/3) - 3*b^7)/x^(4/3 )
Time = 0.32 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.44 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=a^{7} x \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 35 \, a^{4} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + \frac {21}{2} \, a^{6} b x^{\frac {2}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 63 \, a^{5} b^{2} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) - \frac {420 \, a^{3} b^{4} x \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 126 \, a^{2} b^{5} x^{\frac {2}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 28 \, a b^{6} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 3 \, b^{7} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )}{4 \, x^{\frac {4}{3}}} \]
a^7*x*sgn(a*x + b*x^(2/3))*sgn(x) + 35*a^4*b^3*log(abs(x))*sgn(a*x + b*x^( 2/3))*sgn(x) + 21/2*a^6*b*x^(2/3)*sgn(a*x + b*x^(2/3))*sgn(x) + 63*a^5*b^2 *x^(1/3)*sgn(a*x + b*x^(2/3))*sgn(x) - 1/4*(420*a^3*b^4*x*sgn(a*x + b*x^(2 /3))*sgn(x) + 126*a^2*b^5*x^(2/3)*sgn(a*x + b*x^(2/3))*sgn(x) + 28*a*b^6*x ^(1/3)*sgn(a*x + b*x^(2/3))*sgn(x) + 3*b^7*sgn(a*x + b*x^(2/3))*sgn(x))/x^ (4/3)
Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\int {\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2\,a\,b}{x^{1/3}}\right )}^{7/2} \,d x \]