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

Optimal. Leaf size=391 \[ \frac{a^7 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{21 a^6 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{63 a^5 b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}-\frac{105 a^3 b^4 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{\sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{63 a^2 b^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{7 a b^6 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{x \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{3 b^7 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{4 x^{4/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{105 a^4 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

[Out]

(-3*b^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(4*(a + b/x^(1/3))*x^(4/3)) - (7*a*b^6*Sqrt[a^2 + b^2/x^(2/
3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x) - (63*a^2*b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(2*(a + b/
x^(1/3))*x^(2/3)) - (105*a^3*b^4*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x^(1/3)) + (63*a^
5*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1/3))/(a + b/x^(1/3)) + (21*a^6*b*Sqrt[a^2 + b^2/x^(2/3) +
(2*a*b)/x^(1/3)]*x^(2/3))/(2*(a + b/x^(1/3))) + (a^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(1/
3)) + (105*a^4*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*Log[x^(1/3)])/(a + b/x^(1/3))

________________________________________________________________________________________

Rubi [A]  time = 0.187171, antiderivative size = 391, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1341, 1355, 263, 43} \[ \frac{a^7 x \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}+\frac{21 a^6 b x^{2/3} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{63 a^5 b^2 \sqrt [3]{x} \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}}-\frac{105 a^3 b^4 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{\sqrt [3]{x} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{63 a^2 b^5 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{2 x^{2/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{7 a b^6 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{x \left (a+\frac{b}{\sqrt [3]{x}}\right )}-\frac{3 b^7 \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{4 x^{4/3} \left (a+\frac{b}{\sqrt [3]{x}}\right )}+\frac{105 a^4 b^3 \log \left (\sqrt [3]{x}\right ) \sqrt{a^2+\frac{2 a b}{\sqrt [3]{x}}+\frac{b^2}{x^{2/3}}}}{a+\frac{b}{\sqrt [3]{x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(4*(a + b/x^(1/3))*x^(4/3)) - (7*a*b^6*Sqrt[a^2 + b^2/x^(2/
3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x) - (63*a^2*b^5*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/(2*(a + b/
x^(1/3))*x^(2/3)) - (105*a^3*b^4*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)])/((a + b/x^(1/3))*x^(1/3)) + (63*a^
5*b^2*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x^(1/3))/(a + b/x^(1/3)) + (21*a^6*b*Sqrt[a^2 + b^2/x^(2/3) +
(2*a*b)/x^(1/3)]*x^(2/3))/(2*(a + b/x^(1/3))) + (a^7*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*x)/(a + b/x^(1/
3)) + (105*a^4*b^3*Sqrt[a^2 + b^2/x^(2/3) + (2*a*b)/x^(1/3)]*Log[x^(1/3)])/(a + b/x^(1/3))

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 263

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \left (a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}\right )^{7/2} \, dx &=3 \operatorname{Subst}\left (\int \left (a^2+\frac{b^2}{x^2}+\frac{2 a b}{x}\right )^{7/2} x^2 \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (a b+\frac{b^2}{x}\right )^7 x^2 \, dx,x,\sqrt [3]{x}\right )}{b^6 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \frac{\left (b^2+a b x\right )^7}{x^5} \, dx,x,\sqrt [3]{x}\right )}{b^6 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=\frac{\left (3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}\right ) \operatorname{Subst}\left (\int \left (21 a^5 b^9+\frac{b^{14}}{x^5}+\frac{7 a b^{13}}{x^4}+\frac{21 a^2 b^{12}}{x^3}+\frac{35 a^3 b^{11}}{x^2}+\frac{35 a^4 b^{10}}{x}+7 a^6 b^8 x+a^7 b^7 x^2\right ) \, dx,x,\sqrt [3]{x}\right )}{b^6 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right )}\\ &=-\frac{3 b^8 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{4 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) x^{4/3}}-\frac{7 a b^7 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{\left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) x}-\frac{63 a^2 b^6 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{2 \left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) x^{2/3}}-\frac{105 a^3 b^5 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}}}{\left (a b+\frac{b^2}{\sqrt [3]{x}}\right ) \sqrt [3]{x}}+\frac{63 a^5 b^3 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \sqrt [3]{x}}{a b+\frac{b^2}{\sqrt [3]{x}}}+\frac{21 a^6 b^2 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} x^{2/3}}{2 \left (a b+\frac{b^2}{\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^4 \sqrt{a^2+\frac{b^2}{x^{2/3}}+\frac{2 a b}{\sqrt [3]{x}}} \log (x)}{a b+\frac{b^2}{\sqrt [3]{x}}}\\ \end{align*}

Mathematica [A]  time = 0.073284, size = 125, normalized size = 0.32 \[ \frac{\sqrt{\frac{\left (a \sqrt [3]{x}+b\right )^2}{x^{2/3}}} \left (252 a^5 b^2 x^{5/3}-126 a^2 b^5 x^{2/3}+140 a^4 b^3 x^{4/3} \log (x)-420 a^3 b^4 x+42 a^6 b x^2+4 a^7 x^{7/3}-28 a b^6 \sqrt [3]{x}-3 b^7\right )}{4 x \left (a \sqrt [3]{x}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

________________________________________________________________________________________

Maple [A]  time = 0.022, size = 115, normalized size = 0.3 \begin{align*}{\frac{1}{4} \left ({ \left ({a}^{2}{x}^{{\frac{2}{3}}}+2\,ab\sqrt [3]{x}+{b}^{2} \right ){x}^{-{\frac{2}{3}}}} \right ) ^{{\frac{7}{2}}} \left ( 42\,{a}^{6}b{x}^{3}+140\,{a}^{4}{b}^{3}\ln \left ( x \right ){x}^{7/3}+252\,{a}^{5}{b}^{2}{x}^{8/3}+4\,{a}^{7}{x}^{10/3}-28\,a{b}^{6}{x}^{4/3}-420\,{a}^{3}{b}^{4}{x}^{2}-126\,{a}^{2}{b}^{5}{x}^{5/3}-3\,{b}^{7}x \right ) \left ( b+a\sqrt [3]{x} \right ) ^{-7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a^2*x^(2/3)+2*a*b*x^(1/3)+b^2)/x^(2/3))^(7/2)*(42*a^6*b*x^3+140*a^4*b^3*ln(x)*x^(7/3)+252*a^5*b^2*x^(8/3
)+4*a^7*x^(10/3)-28*a*b^6*x^(4/3)-420*a^3*b^4*x^2-126*a^2*b^5*x^(5/3)-3*b^7*x)/(b+a*x^(1/3))^7

________________________________________________________________________________________

Maxima [A]  time = 0.977742, size = 107, normalized size = 0.27 \begin{align*} 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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.30014, size = 234, normalized size = 0.6 \begin{align*} 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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*sg
n(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)