\(\int (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}})^{7/2} \, dx\) [83]

Optimal result

Integrand size = 26, antiderivative size = 387 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=-\frac {3 b^7 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{4 \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{4/3}}-\frac {7 a b^6 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{\left (a+\frac {b}{\sqrt [3]{x}}\right ) x}-\frac {63 a^2 b^5 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right ) x^{2/3}}-\frac {105 a^3 b^4 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}}}{\left (a+\frac {b}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}+\frac {63 a^5 b^2 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \sqrt [3]{x}}{a+\frac {b}{\sqrt [3]{x}}}+\frac {21 a^6 b \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a+\frac {b}{\sqrt [3]{x}}\right )}+\frac {a^7 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} x}{a+\frac {b}{\sqrt [3]{x}}}+\frac {35 a^4 b^3 \sqrt {a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}} \log (x)}{a+\frac {b}{\sqrt [3]{x}}} \] Output:

-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*(a^2+b^2/ 
x^(2/3)+2*a*b/x^(1/3))^(1/2)*x^(1/3)/(a+b/x^(1/3))+21*a^6*b*(a^2+b^2/x^(2/ 
3)+2*a*b/x^(1/3))^(1/2)*x^(2/3)/(2*a+2*b/x^(1/3))+a^7*(a^2+b^2/x^(2/3)+2*a 
*b/x^(1/3))^(1/2)*x/(a+b/x^(1/3))+35*a^4*b^3*(a^2+b^2/x^(2/3)+2*a*b/x^(1/3 
))^(1/2)*ln(x)/(a+b/x^(1/3))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.32 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\frac {\sqrt {\frac {\left (b+a \sqrt [3]{x}\right )^2}{x^{2/3}}} \left (-3 b^7-28 a b^6 \sqrt [3]{x}-126 a^2 b^5 x^{2/3}-420 a^3 b^4 x+252 a^5 b^2 x^{5/3}+42 a^6 b x^2+4 a^7 x^{7/3}+140 a^4 b^3 x^{4/3} \log (x)\right )}{4 \left (b+a \sqrt [3]{x}\right ) x} \] Input:

Integrate[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(7/2),x]
 

Output:

(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)
 

Rubi [A] (verified)

Time = 0.30 (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

\(\displaystyle \frac {3 b^7 \sqrt {a^2+\frac {2 a b}{\sqrt [3]{x}}+\frac {b^2}{x^{2/3}}} \left (\frac {a^7 x}{3}+\frac {7}{2} a^6 b x^{2/3}+21 a^5 b^2 \sqrt [3]{x}+35 a^4 b^3 \log \left (\sqrt [3]{x}\right )-\frac {35 a^3 b^4}{\sqrt [3]{x}}-\frac {21 a^2 b^5}{2 x^{2/3}}-\frac {7 a b^6}{3 x}-\frac {b^7}{4 x^{4/3}}\right )}{a b^7+\frac {b^8}{\sqrt [3]{x}}}\)

Input:

Int[(a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(7/2),x]
 

Output:

(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))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49

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]
 

rule 774

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]
 

rule 795

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]
 

rule 1384

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)])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.29

method	result	size
derivativedivides	\(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}} x \left (4 a^{7} x^{\frac {7}{3}}+42 a^{6} b \,x^{2}+140 a^{4} b^{3} \ln \left (x \right ) x^{\frac {4}{3}}+252 a^{5} b^{2} x^{\frac {5}{3}}-420 a^{3} b^{4} x -126 a^{2} b^{5} x^{\frac {2}{3}}-28 a \,b^{6} x^{\frac {1}{3}}-3 b^{7}\right )}{4 \left (b +a \,x^{\frac {1}{3}}\right )^{7}}\)	\(113\)
default	\(\frac {\left (\frac {x^{\frac {2}{3}} a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}} \left (42 a^{6} b \,x^{3}+252 x^{\frac {8}{3}} a^{5} b^{2}+140 a^{4} b^{3} \ln \left (x \right ) x^{\frac {7}{3}}+4 x^{\frac {10}{3}} a^{7}-28 a \,b^{6} x^{\frac {4}{3}}-420 a^{3} b^{4} x^{2}-126 a^{2} b^{5} x^{\frac {5}{3}}-3 b^{7} x \right )}{4 \left (b +a \,x^{\frac {1}{3}}\right )^{7}}\)	\(115\)

Input:

int((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

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*a^ 
6*b*x^2+140*a^4*b^3*ln(x)*x^(4/3)+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)/(b+a*x^(1/3))^7
 

Fricas [F(-1)]

Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\text {Timed out} \] Input:

integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(7/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

Timed out
 

Sympy [F]

\[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\int \left (a^{2} + \frac {2 a b}{\sqrt [3]{x}} + \frac {b^{2}}{x^{\frac {2}{3}}}\right )^{\frac {7}{2}}\, dx \] Input:

integrate((a**2+b**2/x**(2/3)+2*a*b/x**(1/3))**(7/2),x)
 

Output:

Integral((a**2 + 2*a*b/x**(1/3) + b**2/x**(2/3))**(7/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.20 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=35 \, a^{4} b^{3} \log \left (x\right ) + \frac {4 \, a^{7} x^{\frac {7}{3}} + 42 \, a^{6} b x^{2} + 252 \, a^{5} b^{2} x^{\frac {5}{3}} - 420 \, a^{3} b^{4} x - 126 \, a^{2} b^{5} x^{\frac {2}{3}} - 28 \, a b^{6} x^{\frac {1}{3}} - 3 \, b^{7}}{4 \, x^{\frac {4}{3}}} \] Input:

integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(7/2),x, algorithm="maxima")
 

Output:

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 
)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.45 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=a^{7} x \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 35 \, a^{4} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + \frac {21}{2} \, a^{6} b x^{\frac {2}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 63 \, a^{5} b^{2} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) - \frac {420 \, a^{3} b^{4} x \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 126 \, a^{2} b^{5} x^{\frac {2}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 28 \, a b^{6} x^{\frac {1}{3}} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right ) + 3 \, b^{7} \mathrm {sgn}\left (a x + b x^{\frac {2}{3}}\right ) \mathrm {sgn}\left (x\right )}{4 \, x^{\frac {4}{3}}} \] Input:

integrate((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(7/2),x, algorithm="giac")
 

Output:

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)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\int {\left (a^2+\frac {b^2}{x^{2/3}}+\frac {2\,a\,b}{x^{1/3}}\right )}^{7/2} \,d x \] Input:

int((a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(7/2),x)
 

Output:

int((a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3))^(7/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.21 \[ \int \left (a^2+\frac {b^2}{x^{2/3}}+\frac {2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx=\frac {252 x^{\frac {5}{3}} a^{5} b^{2}-126 x^{\frac {2}{3}} a^{2} b^{5}+420 x^{\frac {4}{3}} \mathrm {log}\left (x^{\frac {1}{3}}\right ) a^{4} b^{3}+4 x^{\frac {7}{3}} a^{7}-28 x^{\frac {1}{3}} a \,b^{6}+42 a^{6} b \,x^{2}-420 a^{3} b^{4} x -3 b^{7}}{4 x^{\frac {4}{3}}} \] Input:

int((a^2+b^2/x^(2/3)+2*a*b/x^(1/3))^(7/2),x)
 

Output:

(252*x**(2/3)*a**5*b**2*x - 126*x**(2/3)*a**2*b**5 + 420*x**(1/3)*log(x**( 
1/3))*a**4*b**3*x + 4*x**(1/3)*a**7*x**2 - 28*x**(1/3)*a*b**6 + 42*a**6*b* 
x**2 - 420*a**3*b**4*x - 3*b**7)/(4*x**(1/3)*x)