3.5.79 \(\int \frac {1}{(a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x})^{3/2}} \, dx\) [479]

Optimal. Leaf size=176 \[ -\frac {12 a^2}{b^4 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}+\frac {2 a^3}{b^4 \left (a+b \sqrt [4]{x}\right ) \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}+\frac {4 \left (a+b \sqrt [4]{x}\right ) \sqrt [4]{x}}{b^3 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}}-\frac {12 a \left (a+b \sqrt [4]{x}\right ) \log \left (a+b \sqrt [4]{x}\right )}{b^4 \sqrt {a^2+2 a b \sqrt [4]{x}+b^2 \sqrt {x}}} \]

[Out]

-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)

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Rubi [A]
time = 0.06, antiderivative size = 176, 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 {12 a^2}{b^4 \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}}}+\frac {4 \sqrt [4]{x} \left (a+b \sqrt [4]{x}\right )}{b^3 \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}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*a^2)/(b^4*Sqrt[a^2 + 2*a*b*x^(1/4) + b^2*Sqrt[x]]) + (2*a^3)/(b^4*(a + b*x^(1/4))*Sqrt[a^2 + 2*a*b*x^(1/4
) + b^2*Sqrt[x]]) + (4*(a + b*x^(1/4))*x^(1/4))/(b^3*Sqrt[a^2 + 2*a*b*x^(1/4) + b^2*Sqrt[x]]) - (12*a*(a + b*x
^(1/4))*Log[a + b*x^(1/4)])/(b^4*Sqrt[a^2 + 2*a*b*x^(1/4) + b^2*Sqrt[x]])

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

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Mathematica [A]
time = 0.07, size = 93, normalized size = 0.53 \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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])

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Maple [A]
time = 0.08, size = 114, normalized size = 0.65

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	\(\frac {2 \sqrt {a^{2}+2 a b \,x^{\frac {1}{4}}+b^{2} \sqrt {x}}\, \left (2 b^{3} x^{\frac {3}{4}}-6 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a \,b^{2} \sqrt {x}+4 a \,b^{2} \sqrt {x}-12 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a^{2} b \,x^{\frac {1}{4}}-4 a^{2} b \,x^{\frac {1}{4}}-6 \ln \left (a +b \,x^{\frac {1}{4}}\right ) a^{3}-5 a^{3}\right )}{\left (a +b \,x^{\frac {1}{4}}\right )^{3} b^{4}}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2*(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(1/2)*(2*b^3*x^(3/4)-6*ln(a+b*x^(1/4))*a*b^2*x^(1/2)+4*a*b^2*x^(1/2)-12*ln(a
+b*x^(1/4))*a^2*b*x^(1/4)-4*a^2*b*x^(1/4)-6*ln(a+b*x^(1/4))*a^3-5*a^3)/(a+b*x^(1/4))^3/b^4

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Maxima [A]
time = 0.29, size = 114, normalized size = 0.65 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(3/2),x, algorithm="maxima")

[Out]

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)

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Fricas [A]
time = 2.12, size = 147, normalized size = 0.84 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(3/2),x, algorithm="fricas")

[Out]

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)

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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 [4]{x} + b^{2} \sqrt {x}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a**2+2*a*b*x**(1/4)+b**2*x**(1/2))**(3/2),x)

[Out]

Integral((a**2 + 2*a*b*x**(1/4) + b**2*sqrt(x))**(-3/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(3/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a^2+b^2\,\sqrt {x}+2\,a\,b\,x^{1/4}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^(1/2) + 2*a*b*x^(1/4))^(3/2),x)

[Out]

int(1/(a^2 + b^2*x^(1/2) + 2*a*b*x^(1/4))^(3/2), x)

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