3.129 \(\int \frac{1}{x^{5/2} (b \sqrt{x}+a x)^{3/2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{128 a^2 \sqrt{a x+b \sqrt{x}}}{21 b^4 x^{3/2}}-\frac{1024 a^4 \sqrt{a x+b \sqrt{x}}}{63 b^6 \sqrt{x}}+\frac{512 a^3 \sqrt{a x+b \sqrt{x}}}{63 b^5 x}+\frac{320 a \sqrt{a x+b \sqrt{x}}}{63 b^3 x^2}-\frac{40 \sqrt{a x+b \sqrt{x}}}{9 b^2 x^{5/2}}+\frac{4}{b x^2 \sqrt{a x+b \sqrt{x}}} \]

[Out]

4/(b*x^2*Sqrt[b*Sqrt[x] + a*x]) - (40*Sqrt[b*Sqrt[x] + a*x])/(9*b^2*x^(5/2)) + (320*a*Sqrt[b*Sqrt[x] + a*x])/(
63*b^3*x^2) - (128*a^2*Sqrt[b*Sqrt[x] + a*x])/(21*b^4*x^(3/2)) + (512*a^3*Sqrt[b*Sqrt[x] + a*x])/(63*b^5*x) -
(1024*a^4*Sqrt[b*Sqrt[x] + a*x])/(63*b^6*Sqrt[x])

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Rubi [A]  time = 0.257146, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2015, 2016, 2014} \[ -\frac{128 a^2 \sqrt{a x+b \sqrt{x}}}{21 b^4 x^{3/2}}-\frac{1024 a^4 \sqrt{a x+b \sqrt{x}}}{63 b^6 \sqrt{x}}+\frac{512 a^3 \sqrt{a x+b \sqrt{x}}}{63 b^5 x}+\frac{320 a \sqrt{a x+b \sqrt{x}}}{63 b^3 x^2}-\frac{40 \sqrt{a x+b \sqrt{x}}}{9 b^2 x^{5/2}}+\frac{4}{b x^2 \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(b*Sqrt[x] + a*x)^(3/2)),x]

[Out]

4/(b*x^2*Sqrt[b*Sqrt[x] + a*x]) - (40*Sqrt[b*Sqrt[x] + a*x])/(9*b^2*x^(5/2)) + (320*a*Sqrt[b*Sqrt[x] + a*x])/(
63*b^3*x^2) - (128*a^2*Sqrt[b*Sqrt[x] + a*x])/(21*b^4*x^(3/2)) + (512*a^3*Sqrt[b*Sqrt[x] + a*x])/(63*b^5*x) -
(1024*a^4*Sqrt[b*Sqrt[x] + a*x])/(63*b^6*Sqrt[x])

Rule 2015

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] + Dist[(c^j*(m + n*p + n - j + 1))/(a*(n - j)*(p + 1)),
Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, j, m, n}, x] &&  !IntegerQ[p] && NeQ[n, j
] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && LtQ[p, -1] && (IntegerQ[j] || GtQ[c, 0])

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^{5/2} \left (b \sqrt{x}+a x\right )^{3/2}} \, dx &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}+\frac{10 \int \frac{1}{x^3 \sqrt{b \sqrt{x}+a x}} \, dx}{b}\\ &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}-\frac{40 \sqrt{b \sqrt{x}+a x}}{9 b^2 x^{5/2}}-\frac{(80 a) \int \frac{1}{x^{5/2} \sqrt{b \sqrt{x}+a x}} \, dx}{9 b^2}\\ &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}-\frac{40 \sqrt{b \sqrt{x}+a x}}{9 b^2 x^{5/2}}+\frac{320 a \sqrt{b \sqrt{x}+a x}}{63 b^3 x^2}+\frac{\left (160 a^2\right ) \int \frac{1}{x^2 \sqrt{b \sqrt{x}+a x}} \, dx}{21 b^3}\\ &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}-\frac{40 \sqrt{b \sqrt{x}+a x}}{9 b^2 x^{5/2}}+\frac{320 a \sqrt{b \sqrt{x}+a x}}{63 b^3 x^2}-\frac{128 a^2 \sqrt{b \sqrt{x}+a x}}{21 b^4 x^{3/2}}-\frac{\left (128 a^3\right ) \int \frac{1}{x^{3/2} \sqrt{b \sqrt{x}+a x}} \, dx}{21 b^4}\\ &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}-\frac{40 \sqrt{b \sqrt{x}+a x}}{9 b^2 x^{5/2}}+\frac{320 a \sqrt{b \sqrt{x}+a x}}{63 b^3 x^2}-\frac{128 a^2 \sqrt{b \sqrt{x}+a x}}{21 b^4 x^{3/2}}+\frac{512 a^3 \sqrt{b \sqrt{x}+a x}}{63 b^5 x}+\frac{\left (256 a^4\right ) \int \frac{1}{x \sqrt{b \sqrt{x}+a x}} \, dx}{63 b^5}\\ &=\frac{4}{b x^2 \sqrt{b \sqrt{x}+a x}}-\frac{40 \sqrt{b \sqrt{x}+a x}}{9 b^2 x^{5/2}}+\frac{320 a \sqrt{b \sqrt{x}+a x}}{63 b^3 x^2}-\frac{128 a^2 \sqrt{b \sqrt{x}+a x}}{21 b^4 x^{3/2}}+\frac{512 a^3 \sqrt{b \sqrt{x}+a x}}{63 b^5 x}-\frac{1024 a^4 \sqrt{b \sqrt{x}+a x}}{63 b^6 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0522184, size = 83, normalized size = 0.5 \[ -\frac{4 \left (-32 a^3 b^2 x^{3/2}+16 a^2 b^3 x+128 a^4 b x^2+256 a^5 x^{5/2}-10 a b^4 \sqrt{x}+7 b^5\right )}{63 b^6 x^2 \sqrt{a x+b \sqrt{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(b*Sqrt[x] + a*x)^(3/2)),x]

[Out]

(-4*(7*b^5 - 10*a*b^4*Sqrt[x] + 16*a^2*b^3*x - 32*a^3*b^2*x^(3/2) + 128*a^4*b*x^2 + 256*a^5*x^(5/2)))/(63*b^6*
x^2*Sqrt[b*Sqrt[x] + a*x])

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Maple [C]  time = 0.01, size = 592, normalized size = 3.6 \begin{align*} -{\frac{4}{63\,{b}^{7}}\sqrt{b\sqrt{x}+ax} \left ( 63\,\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}^{3}-10\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{3/2}{x}^{3}{b}^{5}+63\,\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}^{13/2}{a}^{7}b-126\,\sqrt{b\sqrt{x}+ax}{a}^{15/2}{x}^{13/2}-63\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{13/2}{a}^{7}b-126\,{a}^{15/2}{x}^{13/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }-126\,\sqrt{b\sqrt{x}+ax}{a}^{11/2}{x}^{11/2}{b}^{2}+128\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{9/2}{x}^{9/2}{b}^{2}+508\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{11/2}{x}^{5}b-252\,\sqrt{b\sqrt{x}+ax}{a}^{13/2}{x}^{6}b-252\,{a}^{13/2}{x}^{6}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }b-32\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{7/2}{x}^{4}{b}^{3}+7\, \left ( b\sqrt{x}+ax \right ) ^{3/2}\sqrt{a}{x}^{5/2}{b}^{6}+315\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{13/2}{x}^{11/2}-126\,{a}^{11/2}{x}^{11/2}\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }{b}^{2}-126\,\ln \left ( 1/2\,{\frac{2\,a\sqrt{x}+2\,\sqrt{b\sqrt{x}+ax}\sqrt{a}+b}{\sqrt{a}}} \right ){x}^{6}{a}^{6}{b}^{2}-63\,\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}^{3}-63\,{a}^{13/2}{x}^{11/2} \left ( \sqrt{x} \left ( b+a\sqrt{x} \right ) \right ) ^{3/2}+126\,\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}^{6}{a}^{6}{b}^{2}+16\, \left ( b\sqrt{x}+ax \right ) ^{3/2}{a}^{5/2}{x}^{7/2}{b}^{4} \right ){\frac{1}{\sqrt{\sqrt{x} \left ( b+a\sqrt{x} \right ) }}}{\frac{1}{\sqrt{a}}}{x}^{-{\frac{11}{2}}} \left ( b+a\sqrt{x} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(b*x^(1/2)+a*x)^(3/2),x)

[Out]

-4/63*(b*x^(1/2)+a*x)^(1/2)*(63*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^3-10*(b*x^(1/2)+a*x)^(3/2)*a^(3/2)*x^3*b^5+63*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^(13/2)*a^7*b-126*(b*x^(1/2)+a*x)^(1/2)*a^(15/2)*x^(13/2)-63*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^(13/2)*a^7*b-126*a^(15/2)*x^(13/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)-126*(b*x^(
1/2)+a*x)^(1/2)*a^(11/2)*x^(11/2)*b^2+128*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)*x^(9/2)*b^2+508*(b*x^(1/2)+a*x)^(3/2)*
a^(11/2)*x^5*b-252*(b*x^(1/2)+a*x)^(1/2)*a^(13/2)*x^6*b-252*a^(13/2)*x^6*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b-32*(b
*x^(1/2)+a*x)^(3/2)*a^(7/2)*x^4*b^3+7*(b*x^(1/2)+a*x)^(3/2)*a^(1/2)*x^(5/2)*b^6+315*(b*x^(1/2)+a*x)^(3/2)*a^(1
3/2)*x^(11/2)-126*a^(11/2)*x^(11/2)*(x^(1/2)*(b+a*x^(1/2)))^(1/2)*b^2-126*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^6*a^6*b^2-63*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^3-63*a^(13/2)*x^(11/2)*(x^(1/2)*(b+a*x^(1/2)))^(3/2)+126*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^6*a^6*b^2+16*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*x^(7/2)*b^4)/(x^(1/2)*(b+a*x^(1/2)
))^(1/2)/b^7/a^(1/2)/x^(11/2)/(b+a*x^(1/2))^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a x + b \sqrt{x}\right )}^{\frac{3}{2}} x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((a*x + b*sqrt(x))^(3/2)*x^(5/2)), x)

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Fricas [A]  time = 2.39902, size = 220, normalized size = 1.33 \begin{align*} \frac{4 \,{\left (128 \, a^{5} b x^{3} - 48 \, a^{3} b^{3} x^{2} - 17 \, a b^{5} x -{\left (256 \, a^{6} x^{3} - 160 \, a^{4} b^{2} x^{2} - 26 \, a^{2} b^{4} x - 7 \, b^{6}\right )} \sqrt{x}\right )} \sqrt{a x + b \sqrt{x}}}{63 \,{\left (a^{2} b^{6} x^{4} - b^{8} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="fricas")

[Out]

4/63*(128*a^5*b*x^3 - 48*a^3*b^3*x^2 - 17*a*b^5*x - (256*a^6*x^3 - 160*a^4*b^2*x^2 - 26*a^2*b^4*x - 7*b^6)*sqr
t(x))*sqrt(a*x + b*sqrt(x))/(a^2*b^6*x^4 - b^8*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{\frac{5}{2}} \left (a x + b \sqrt{x}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(b*x**(1/2)+a*x)**(3/2),x)

[Out]

Integral(1/(x**(5/2)*(a*x + b*sqrt(x))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a x + b \sqrt{x}\right )}^{\frac{3}{2}} x^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*x + b*sqrt(x))^(3/2)*x^(5/2)), x)