3.5.69 \(\int \frac {1}{(a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^{7/2}} \, dx\) [469]

Optimal. Leaf size=137 \[ -\frac {a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{5 b^3 \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {3}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \]

[Out]

-1/2*a^2/b^3/(a+b*x^(1/3))^5/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)+6/5*a/b^3/(a+b*x^(1/3))^4/(a^2+2*a*b*x^(1/3
)+b^2*x^(2/3))^(1/2)-3/4/b^3/(a+b*x^(1/3))^3/(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1355, 660, 45} \begin {gather*} -\frac {a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{5 b^3 \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {3}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a^2/(b^3*(a + b*x^(1/3))^5*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) + (6*a)/(5*b^3*(a + b*x^(1/3))^4*Sqrt
[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)]) - 3/(4*b^3*(a + b*x^(1/3))^3*Sqrt[a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3)])

Rule 45

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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 1355

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^{7/2}} \, dx &=3 \text {Subst}\left (\int \frac {x^2}{\left (a^2+2 a b x+b^2 x^2\right )^{7/2}} \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {\left (3 b^7 \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \frac {x^2}{\left (a b+b^2 x\right )^7} \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=\frac {\left (3 b^7 \left (a+b \sqrt [3]{x}\right )\right ) \text {Subst}\left (\int \left (\frac {a^2}{b^9 (a+b x)^7}-\frac {2 a}{b^9 (a+b x)^6}+\frac {1}{b^9 (a+b x)^5}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ &=-\frac {a^2}{2 b^3 \left (a+b \sqrt [3]{x}\right )^5 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}+\frac {6 a}{5 b^3 \left (a+b \sqrt [3]{x}\right )^4 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}-\frac {3}{4 b^3 \left (a+b \sqrt [3]{x}\right )^3 \sqrt {a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 56, normalized size = 0.41 \begin {gather*} \frac {\left (a+b \sqrt [3]{x}\right ) \left (-a^2-6 a b \sqrt [3]{x}-15 b^2 x^{2/3}\right )}{20 b^3 \left (\left (a+b \sqrt [3]{x}\right )^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^(1/3))*(-a^2 - 6*a*b*x^(1/3) - 15*b^2*x^(2/3)))/(20*b^3*((a + b*x^(1/3))^2)^(7/2))

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Maple [A]
time = 0.05, size = 54, normalized size = 0.39

method	result	size
derivativedivides	\(-\frac {\left (15 b^{2} x^{\frac {2}{3}}+6 a b \,x^{\frac {1}{3}}+a^{2}\right ) \left (a +b \,x^{\frac {1}{3}}\right )}{20 b^{3} \left (\left (a +b \,x^{\frac {1}{3}}\right )^{2}\right )^{\frac {7}{2}}}\)	\(43\)
default	\(-\frac {\sqrt {a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}}\, \left (15 b^{2} x^{\frac {2}{3}}+6 a b \,x^{\frac {1}{3}}+a^{2}\right )}{20 \left (a +b \,x^{\frac {1}{3}}\right )^{7} b^{3}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^(1/2)*(15*b^2*x^(2/3)+6*a*b*x^(1/3)+a^2)/(a+b*x^(1/3))^7/b^3

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Maxima [A]
time = 0.42, size = 53, normalized size = 0.39 \begin {gather*} -\frac {3}{4 \, b^{7} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{4}} + \frac {6 \, a}{5 \, b^{8} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{5}} - \frac {a^{2}}{2 \, b^{9} {\left (x^{\frac {1}{3}} + \frac {a}{b}\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4/(b^7*(x^(1/3) + a/b)^4) + 6/5*a/(b^8*(x^(1/3) + a/b)^5) - 1/2*a^2/(b^9*(x^(1/3) + a/b)^6)

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Fricas [A]
time = 0.40, size = 209, normalized size = 1.53 \begin {gather*} -\frac {280 \, a^{2} b^{12} x^{4} - 1400 \, a^{5} b^{9} x^{3} + 735 \, a^{8} b^{6} x^{2} - 14 \, a^{11} b^{3} x + a^{14} + 3 \, {\left (5 \, b^{14} x^{4} - 210 \, a^{3} b^{11} x^{3} + 483 \, a^{6} b^{8} x^{2} - 112 \, a^{9} b^{5} x\right )} x^{\frac {2}{3}} - 3 \, {\left (28 \, a b^{13} x^{4} - 357 \, a^{4} b^{10} x^{3} + 390 \, a^{7} b^{7} x^{2} - 35 \, a^{10} b^{4} x\right )} x^{\frac {1}{3}}}{20 \, {\left (b^{21} x^{6} + 6 \, a^{3} b^{18} x^{5} + 15 \, a^{6} b^{15} x^{4} + 20 \, a^{9} b^{12} x^{3} + 15 \, a^{12} b^{9} x^{2} + 6 \, a^{15} b^{6} x + a^{18} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(280*a^2*b^12*x^4 - 1400*a^5*b^9*x^3 + 735*a^8*b^6*x^2 - 14*a^11*b^3*x + a^14 + 3*(5*b^14*x^4 - 210*a^3*
b^11*x^3 + 483*a^6*b^8*x^2 - 112*a^9*b^5*x)*x^(2/3) - 3*(28*a*b^13*x^4 - 357*a^4*b^10*x^3 + 390*a^7*b^7*x^2 -
35*a^10*b^4*x)*x^(1/3))/(b^21*x^6 + 6*a^3*b^18*x^5 + 15*a^6*b^15*x^4 + 20*a^9*b^12*x^3 + 15*a^12*b^9*x^2 + 6*a
^15*b^6*x + a^18*b^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac {2}{3}}\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.52, size = 43, normalized size = 0.31 \begin {gather*} -\frac {15 \, b^{2} x^{\frac {2}{3}} + 6 \, a b x^{\frac {1}{3}} + a^{2}}{20 \, {\left (b x^{\frac {1}{3}} + a\right )}^{6} b^{3} \mathrm {sgn}\left (b x^{\frac {1}{3}} + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(15*b^2*x^(2/3) + 6*a*b*x^(1/3) + a^2)/((b*x^(1/3) + a)^6*b^3*sgn(b*x^(1/3) + a))

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Mupad [B]
time = 3.23, size = 53, normalized size = 0.39 \begin {gather*} -\frac {\sqrt {a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}}\,\left (a^2+15\,b^2\,x^{2/3}+6\,a\,b\,x^{1/3}\right )}{20\,b^3\,{\left (a+b\,x^{1/3}\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^(1/2)*(a^2 + 15*b^2*x^(2/3) + 6*a*b*x^(1/3)))/(20*b^3*(a + b*x^(1/3))^7)

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