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

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

[Out]

(-12*a^2)/(b^4*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]])

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Rubi [A]  time = 0.103783, antiderivative size = 176, normalized size of antiderivative = 1., 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 = {1341, 646, 43} \[ \frac{2 a^3}{b^4 \left (a+b \sqrt [4]{x}\right ) \sqrt{a^2+2 a b \sqrt [4]{x}+b^2 \sqrt{x}}}-\frac{12 a^2}{b^4 \sqrt{a^2+2 a b \sqrt [4]{x}+b^2 \sqrt{x}}}+\frac{4 \sqrt [4]{x} \left (a+b \sqrt [4]{x}\right )}{b^3 \sqrt{a^2+2 a b \sqrt [4]{x}+b^2 \sqrt{x}}}-\frac{12 a \left (a+b \sqrt [4]{x}\right ) \log \left (a+b \sqrt [4]{x}\right )}{b^4 \sqrt{a^2+2 a b \sqrt [4]{x}+b^2 \sqrt{x}}} \]

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 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 646

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 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 \frac{1}{\left (a^2+2 a b \sqrt [4]{x}+b^2 \sqrt{x}\right )^{3/2}} \, dx &=4 \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.0677754, size = 93, normalized size = 0.53 \[ \frac{2 \left (-4 a^2 b \sqrt [4]{x}-5 a^3+4 a b^2 \sqrt{x}-6 a \left (a+b \sqrt [4]{x}\right )^2 \log \left (a+b \sqrt [4]{x}\right )+2 b^3 x^{3/4}\right )}{b^4 \left (a+b \sqrt [4]{x}\right ) \sqrt{\left (a+b \sqrt [4]{x}\right )^2}} \]

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 = 0.016, size = 114, normalized size = 0.7 \begin{align*} 2\,{\frac{\sqrt{{a}^{2}+2\,ab\sqrt [4]{x}+{b}^{2}\sqrt{x}} \left ( 2\,{x}^{3/4}{b}^{3}-6\,\sqrt{x}\ln \left ( a+b\sqrt [4]{x} \right ) a{b}^{2}+4\,\sqrt{x}a{b}^{2}-12\,\sqrt [4]{x}\ln \left ( a+b\sqrt [4]{x} \right ){a}^{2}b-4\,\sqrt [4]{x}{a}^{2}b-6\,\ln \left ( a+b\sqrt [4]{x} \right ){a}^{3}-5\,{a}^{3} \right ) }{ \left ( a+b\sqrt [4]{x} \right ) ^{3}{b}^{4}}} \end{align*}

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)

[Out]

2*(a^2+2*a*b*x^(1/4)+b^2*x^(1/2))^(1/2)*(2*x^(3/4)*b^3-6*x^(1/2)*ln(a+b*x^(1/4))*a*b^2+4*x^(1/2)*a*b^2-12*x^(1
/4)*ln(a+b*x^(1/4))*a^2*b-4*x^(1/4)*a^2*b-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 = 1.0124, size = 200, normalized size = 1.14 \begin{align*} -\frac{12 \, a \log \left (x^{\frac{1}{4}} + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}} b} - \frac{18 \, a^{3} b}{{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x^{\frac{1}{4}} + \frac{a}{b}\right )}^{2}} + \frac{4 \, \sqrt{x}}{\sqrt{b^{2} \sqrt{x} + 2 \, a b x^{\frac{1}{4}} + a^{2}} b^{2}} - \frac{24 \, a^{2} x^{\frac{1}{4}}}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x^{\frac{1}{4}} + \frac{a}{b}\right )}^{2}} + \frac{8 \, a^{2}}{\sqrt{b^{2} \sqrt{x} + 2 \, a b x^{\frac{1}{4}} + a^{2}} b^{4}} - \frac{4 \, a^{3}}{{\left (b^{2}\right )}^{\frac{3}{2}} b^{3}{\left (x^{\frac{1}{4}} + \frac{a}{b}\right )}^{2}} \end{align*}

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]

-12*a*log(x^(1/4) + a/b)/((b^2)^(3/2)*b) - 18*a^3*b/((b^2)^(7/2)*(x^(1/4) + a/b)^2) + 4*sqrt(x)/(sqrt(b^2*sqrt
(x) + 2*a*b*x^(1/4) + a^2)*b^2) - 24*a^2*x^(1/4)/((b^2)^(5/2)*(x^(1/4) + a/b)^2) + 8*a^2/(sqrt(b^2*sqrt(x) + 2
*a*b*x^(1/4) + a^2)*b^4) - 4*a^3/((b^2)^(3/2)*b^3*(x^(1/4) + a/b)^2)

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Fricas [A]  time = 21.1281, size = 315, normalized size = 1.79 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a^{2} + 2 a b \sqrt [4]{x} + b^{2} \sqrt{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

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

[Out]

Timed out