3.108 \(\int \frac{1}{x^3 \sqrt{b \sqrt{x}+a x}} \, dx\)

Optimal. Leaf size=142 \[ -\frac{64 a^2 \sqrt{a x+b \sqrt{x}}}{105 b^3 x^{3/2}}-\frac{512 a^4 \sqrt{a x+b \sqrt{x}}}{315 b^5 \sqrt{x}}+\frac{256 a^3 \sqrt{a x+b \sqrt{x}}}{315 b^4 x}+\frac{32 a \sqrt{a x+b \sqrt{x}}}{63 b^2 x^2}-\frac{4 \sqrt{a x+b \sqrt{x}}}{9 b x^{5/2}} \]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(9*b*x^(5/2)) + (32*a*Sqrt[b*Sqrt[x] + a*x])/(63*b^2*x^2) - (64*a^2*Sqrt[b*Sqrt[x]
+ a*x])/(105*b^3*x^(3/2)) + (256*a^3*Sqrt[b*Sqrt[x] + a*x])/(315*b^4*x) - (512*a^4*Sqrt[b*Sqrt[x] + a*x])/(315
*b^5*Sqrt[x])

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Rubi [A]  time = 0.203379, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2016, 2014} \[ -\frac{64 a^2 \sqrt{a x+b \sqrt{x}}}{105 b^3 x^{3/2}}-\frac{512 a^4 \sqrt{a x+b \sqrt{x}}}{315 b^5 \sqrt{x}}+\frac{256 a^3 \sqrt{a x+b \sqrt{x}}}{315 b^4 x}+\frac{32 a \sqrt{a x+b \sqrt{x}}}{63 b^2 x^2}-\frac{4 \sqrt{a x+b \sqrt{x}}}{9 b x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[b*Sqrt[x] + a*x]),x]

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x])/(9*b*x^(5/2)) + (32*a*Sqrt[b*Sqrt[x] + a*x])/(63*b^2*x^2) - (64*a^2*Sqrt[b*Sqrt[x]
+ a*x])/(105*b^3*x^(3/2)) + (256*a^3*Sqrt[b*Sqrt[x] + a*x])/(315*b^4*x) - (512*a^4*Sqrt[b*Sqrt[x] + a*x])/(315
*b^5*Sqrt[x])

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{b \sqrt{x}+a x}} \, dx &=-\frac{4 \sqrt{b \sqrt{x}+a x}}{9 b x^{5/2}}-\frac{(8 a) \int \frac{1}{x^{5/2} \sqrt{b \sqrt{x}+a x}} \, dx}{9 b}\\ &=-\frac{4 \sqrt{b \sqrt{x}+a x}}{9 b x^{5/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{63 b^2 x^2}+\frac{\left (16 a^2\right ) \int \frac{1}{x^2 \sqrt{b \sqrt{x}+a x}} \, dx}{21 b^2}\\ &=-\frac{4 \sqrt{b \sqrt{x}+a x}}{9 b x^{5/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{63 b^2 x^2}-\frac{64 a^2 \sqrt{b \sqrt{x}+a x}}{105 b^3 x^{3/2}}-\frac{\left (64 a^3\right ) \int \frac{1}{x^{3/2} \sqrt{b \sqrt{x}+a x}} \, dx}{105 b^3}\\ &=-\frac{4 \sqrt{b \sqrt{x}+a x}}{9 b x^{5/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{63 b^2 x^2}-\frac{64 a^2 \sqrt{b \sqrt{x}+a x}}{105 b^3 x^{3/2}}+\frac{256 a^3 \sqrt{b \sqrt{x}+a x}}{315 b^4 x}+\frac{\left (128 a^4\right ) \int \frac{1}{x \sqrt{b \sqrt{x}+a x}} \, dx}{315 b^4}\\ &=-\frac{4 \sqrt{b \sqrt{x}+a x}}{9 b x^{5/2}}+\frac{32 a \sqrt{b \sqrt{x}+a x}}{63 b^2 x^2}-\frac{64 a^2 \sqrt{b \sqrt{x}+a x}}{105 b^3 x^{3/2}}+\frac{256 a^3 \sqrt{b \sqrt{x}+a x}}{315 b^4 x}-\frac{512 a^4 \sqrt{b \sqrt{x}+a x}}{315 b^5 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0528787, size = 72, normalized size = 0.51 \[ -\frac{4 \sqrt{a x+b \sqrt{x}} \left (48 a^2 b^2 x-64 a^3 b x^{3/2}+128 a^4 x^2-40 a b^3 \sqrt{x}+35 b^4\right )}{315 b^5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(35*b^4 - 40*a*b^3*Sqrt[x] + 48*a^2*b^2*x - 64*a^3*b*x^(3/2) + 128*a^4*x^2))/(315*b^
5*x^(5/2))

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Maple [C]  time = 0.013, size = 262, normalized size = 1.9 \begin{align*} -{\frac{1}{315\,{b}^{6}}\sqrt{b\sqrt{x}+ax} \left ( 1260\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{9/2}{x}^{9/2}-630\,\sqrt{b\sqrt{x}+ax}{a}^{11/2}{x}^{11/2}-315\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{11/2}{a}^{5}b-630\,{a}^{11/2}{x}^{11/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }+315\,\ln \left ( 1/2\,{\frac{2\,\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }\sqrt{a}+2\,a\sqrt{x}+b}{\sqrt{a}}} \right ){x}^{11/2}{a}^{5}b+492\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{7/2}{b}^{2}+140\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}{x}^{5/2}{b}^{4}-748\,{a}^{7/2} \left ( b\sqrt{x}+ax \right ) ^{3/2}b{x}^{4}-300\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{3/2}{x}^{3}{b}^{3} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}}{x}^{-{\frac{11}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/315*(b*x^(1/2)+a*x)^(1/2)*(1260*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)*x^(9/2)-630*(b*x^(1/2)+a*x)^(1/2)*a^(11/2)*x^
(11/2)-315*ln(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a^(1/2))*x^(11/2)*a^5*b-630*a^(11/2)*x^(11/2
)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)+315*ln(1/2*(2*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*a^(1/2)+2*a*x^(1/2)+b)/a^(1/2))*x^
(11/2)*a^5*b+492*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*x^(7/2)*b^2+140*(b*x^(1/2)+a*x)^(3/2)*a^(1/2)*x^(5/2)*b^4-748*a
^(7/2)*(b*x^(1/2)+a*x)^(3/2)*b*x^4-300*(b*x^(1/2)+a*x)^(3/2)*a^(3/2)*x^3*b^3)/(x^(1/2)*(b+a*x^(1/2)))^(1/2)/b^
6/a^(1/2)/x^(11/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a x + b \sqrt{x}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a*x + b*sqrt(x))*x^3), x)

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Fricas [A]  time = 2.33588, size = 155, normalized size = 1.09 \begin{align*} \frac{4 \,{\left (64 \, a^{3} b x^{2} + 40 \, a b^{3} x -{\left (128 \, a^{4} x^{2} + 48 \, a^{2} b^{2} x + 35 \, b^{4}\right )} \sqrt{x}\right )} \sqrt{a x + b \sqrt{x}}}{315 \, b^{5} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(64*a^3*b*x^2 + 40*a*b^3*x - (128*a^4*x^2 + 48*a^2*b^2*x + 35*b^4)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(b^5*x
^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \sqrt{a x + b \sqrt{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16857, size = 197, normalized size = 1.39 \begin{align*} \frac{4 \,{\left (1008 \, a^{2}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )}^{4} + 1680 \, a^{\frac{3}{2}} b{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )}^{3} + 1080 \, a b^{2}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )}^{2} + 315 \, \sqrt{a} b^{3}{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )} + 35 \, b^{4}\right )}}{315 \,{\left (\sqrt{a} \sqrt{x} - \sqrt{a x + b \sqrt{x}}\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/315*(1008*a^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^4 + 1680*a^(3/2)*b*(sqrt(a)*sqrt(x) - sqrt(a*x + b*s
qrt(x)))^3 + 1080*a*b^2*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^2 + 315*sqrt(a)*b^3*(sqrt(a)*sqrt(x) - sqrt(
a*x + b*sqrt(x))) + 35*b^4)/(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x)))^9