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

Optimal. Leaf size=195 \[ \frac {4096 a^5 \sqrt {a x+b \sqrt {x}}}{231 b^7 \sqrt {x}}-\frac {2048 a^4 \sqrt {a x+b \sqrt {x}}}{231 b^6 x}+\frac {512 a^3 \sqrt {a x+b \sqrt {x}}}{77 b^5 x^{3/2}}-\frac {1280 a^2 \sqrt {a x+b \sqrt {x}}}{231 b^4 x^2}+\frac {160 a \sqrt {a x+b \sqrt {x}}}{33 b^3 x^{5/2}}-\frac {48 \sqrt {a x+b \sqrt {x}}}{11 b^2 x^3}+\frac {4}{b x^{5/2} \sqrt {a x+b \sqrt {x}}} \]

[Out]

4/b/x^(5/2)/(b*x^(1/2)+a*x)^(1/2)-48/11*(b*x^(1/2)+a*x)^(1/2)/b^2/x^3+160/33*a*(b*x^(1/2)+a*x)^(1/2)/b^3/x^(5/
2)-1280/231*a^2*(b*x^(1/2)+a*x)^(1/2)/b^4/x^2+512/77*a^3*(b*x^(1/2)+a*x)^(1/2)/b^5/x^(3/2)-2048/231*a^4*(b*x^(
1/2)+a*x)^(1/2)/b^6/x+4096/231*a^5*(b*x^(1/2)+a*x)^(1/2)/b^7/x^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2015, 2016, 2014} \[ \frac {512 a^3 \sqrt {a x+b \sqrt {x}}}{77 b^5 x^{3/2}}-\frac {1280 a^2 \sqrt {a x+b \sqrt {x}}}{231 b^4 x^2}+\frac {4096 a^5 \sqrt {a x+b \sqrt {x}}}{231 b^7 \sqrt {x}}-\frac {2048 a^4 \sqrt {a x+b \sqrt {x}}}{231 b^6 x}+\frac {160 a \sqrt {a x+b \sqrt {x}}}{33 b^3 x^{5/2}}-\frac {48 \sqrt {a x+b \sqrt {x}}}{11 b^2 x^3}+\frac {4}{b x^{5/2} \sqrt {a x+b \sqrt {x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(b*x^(5/2)*Sqrt[b*Sqrt[x] + a*x]) - (48*Sqrt[b*Sqrt[x] + a*x])/(11*b^2*x^3) + (160*a*Sqrt[b*Sqrt[x] + a*x])/
(33*b^3*x^(5/2)) - (1280*a^2*Sqrt[b*Sqrt[x] + a*x])/(231*b^4*x^2) + (512*a^3*Sqrt[b*Sqrt[x] + a*x])/(77*b^5*x^
(3/2)) - (2048*a^4*Sqrt[b*Sqrt[x] + a*x])/(231*b^6*x) + (4096*a^5*Sqrt[b*Sqrt[x] + a*x])/(231*b^7*Sqrt[x])

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])

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])

Rubi steps

\begin {align*} \int \frac {1}{x^3 \left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}+\frac {12 \int \frac {1}{x^{7/2} \sqrt {b \sqrt {x}+a x}} \, dx}{b}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}-\frac {(120 a) \int \frac {1}{x^3 \sqrt {b \sqrt {x}+a x}} \, dx}{11 b^2}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}+\frac {160 a \sqrt {b \sqrt {x}+a x}}{33 b^3 x^{5/2}}+\frac {\left (320 a^2\right ) \int \frac {1}{x^{5/2} \sqrt {b \sqrt {x}+a x}} \, dx}{33 b^3}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}+\frac {160 a \sqrt {b \sqrt {x}+a x}}{33 b^3 x^{5/2}}-\frac {1280 a^2 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^2}-\frac {\left (640 a^3\right ) \int \frac {1}{x^2 \sqrt {b \sqrt {x}+a x}} \, dx}{77 b^4}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}+\frac {160 a \sqrt {b \sqrt {x}+a x}}{33 b^3 x^{5/2}}-\frac {1280 a^2 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^2}+\frac {512 a^3 \sqrt {b \sqrt {x}+a x}}{77 b^5 x^{3/2}}+\frac {\left (512 a^4\right ) \int \frac {1}{x^{3/2} \sqrt {b \sqrt {x}+a x}} \, dx}{77 b^5}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}+\frac {160 a \sqrt {b \sqrt {x}+a x}}{33 b^3 x^{5/2}}-\frac {1280 a^2 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^2}+\frac {512 a^3 \sqrt {b \sqrt {x}+a x}}{77 b^5 x^{3/2}}-\frac {2048 a^4 \sqrt {b \sqrt {x}+a x}}{231 b^6 x}-\frac {\left (1024 a^5\right ) \int \frac {1}{x \sqrt {b \sqrt {x}+a x}} \, dx}{231 b^6}\\ &=\frac {4}{b x^{5/2} \sqrt {b \sqrt {x}+a x}}-\frac {48 \sqrt {b \sqrt {x}+a x}}{11 b^2 x^3}+\frac {160 a \sqrt {b \sqrt {x}+a x}}{33 b^3 x^{5/2}}-\frac {1280 a^2 \sqrt {b \sqrt {x}+a x}}{231 b^4 x^2}+\frac {512 a^3 \sqrt {b \sqrt {x}+a x}}{77 b^5 x^{3/2}}-\frac {2048 a^4 \sqrt {b \sqrt {x}+a x}}{231 b^6 x}+\frac {4096 a^5 \sqrt {b \sqrt {x}+a x}}{231 b^7 \sqrt {x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 96, normalized size = 0.49 \[ \frac {4 \left (1024 a^6 x^3+512 a^5 b x^{5/2}-128 a^4 b^2 x^2+64 a^3 b^3 x^{3/2}-40 a^2 b^4 x+28 a b^5 \sqrt {x}-21 b^6\right )}{231 b^7 x^{5/2} \sqrt {a x+b \sqrt {x}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-21*b^6 + 28*a*b^5*Sqrt[x] - 40*a^2*b^4*x + 64*a^3*b^3*x^(3/2) - 128*a^4*b^2*x^2 + 512*a^5*b*x^(5/2) + 102
4*a^6*x^3))/(231*b^7*x^(5/2)*Sqrt[b*Sqrt[x] + a*x])

________________________________________________________________________________________

fricas [A]  time = 1.06, size = 109, normalized size = 0.56 \[ -\frac {4 \, {\left (512 \, a^{6} b x^{3} - 192 \, a^{4} b^{3} x^{2} - 68 \, a^{2} b^{5} x - 21 \, b^{7} - {\left (1024 \, a^{7} x^{3} - 640 \, a^{5} b^{2} x^{2} - 104 \, a^{3} b^{4} x - 49 \, a b^{6}\right )} \sqrt {x}\right )} \sqrt {a x + b \sqrt {x}}}{231 \, {\left (a^{2} b^{7} x^{4} - b^{9} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/231*(512*a^6*b*x^3 - 192*a^4*b^3*x^2 - 68*a^2*b^5*x - 21*b^7 - (1024*a^7*x^3 - 640*a^5*b^2*x^2 - 104*a^3*b^
4*x - 49*a*b^6)*sqrt(x))*sqrt(a*x + b*sqrt(x))/(a^2*b^7*x^4 - b^9*x^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.07, size = 614, normalized size = 3.15 \[ \frac {\sqrt {a x +b \sqrt {x}}\, \left (1155 a^{8} b \,x^{\frac {15}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-1155 a^{8} b \,x^{\frac {15}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+2310 a^{7} b^{2} x^{7} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-2310 a^{7} b^{2} x^{7} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )+1155 a^{6} b^{3} x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-1155 a^{6} b^{3} x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+b +2 \sqrt {a x +b \sqrt {x}}\, \sqrt {a}}{2 \sqrt {a}}\right )-2310 \sqrt {a x +b \sqrt {x}}\, a^{\frac {17}{2}} x^{\frac {15}{2}}-2310 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {17}{2}} x^{\frac {15}{2}}-4620 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {15}{2}} b \,x^{7}-4620 \sqrt {a x +b \sqrt {x}}\, a^{\frac {15}{2}} b \,x^{7}-2310 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, a^{\frac {13}{2}} b^{2} x^{\frac {13}{2}}-2310 \sqrt {a x +b \sqrt {x}}\, a^{\frac {13}{2}} b^{2} x^{\frac {13}{2}}+5544 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {15}{2}} x^{\frac {13}{2}}-924 \left (\left (a \sqrt {x}+b \right ) \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {15}{2}} x^{\frac {13}{2}}+8716 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {13}{2}} b \,x^{6}+2048 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {11}{2}} b^{2} x^{\frac {11}{2}}-512 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {9}{2}} b^{3} x^{5}+256 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{4} x^{\frac {9}{2}}-160 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{5} x^{4}+112 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{6} x^{\frac {7}{2}}-84 \left (a x +b \sqrt {x}\right )^{\frac {3}{2}} \sqrt {a}\, b^{7} x^{3}\right )}{231 \sqrt {\left (a \sqrt {x}+b \right ) \sqrt {x}}\, \left (a \sqrt {x}+b \right )^{2} \sqrt {a}\, b^{8} x^{\frac {13}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/231*(a*x+b*x^(1/2))^(1/2)*(-2310*a^(13/2)*x^(13/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)*b^2-2310*(a*x+b*x^(1/2))^(1
/2)*a^(13/2)*x^(13/2)*b^2-512*(a*x+b*x^(1/2))^(3/2)*a^(9/2)*x^5*b^3-4620*a^(15/2)*x^7*((a*x^(1/2)+b)*x^(1/2))^
(1/2)*b-4620*(a*x+b*x^(1/2))^(1/2)*a^(15/2)*x^7*b+2048*(a*x+b*x^(1/2))^(3/2)*a^(11/2)*x^(11/2)*b^2+8716*(a*x+b
*x^(1/2))^(3/2)*a^(13/2)*x^6*b-84*(a*x+b*x^(1/2))^(3/2)*a^(1/2)*x^3*b^7+5544*(a*x+b*x^(1/2))^(3/2)*a^(15/2)*x^
(13/2)-2310*(a*x+b*x^(1/2))^(1/2)*a^(17/2)*x^(15/2)+1155*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)
*a^(1/2))/a^(1/2))*x^(15/2)*a^8*b-2310*a^(17/2)*x^(15/2)*((a*x^(1/2)+b)*x^(1/2))^(1/2)-1155*ln(1/2*(2*a*x^(1/2
)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^(15/2)*a^8*b+2310*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/
2))^(1/2)*a^(1/2))/a^(1/2))*x^7*a^7*b^2-924*a^(15/2)*x^(13/2)*((a*x^(1/2)+b)*x^(1/2))^(3/2)-2310*ln(1/2*(2*a*x
^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^7*a^7*b^2+256*(a*x+b*x^(1/2))^(3/2)*a^(7/2)*x^(9/2)*b^4+1
155*ln(1/2*(2*a*x^(1/2)+b+2*((a*x^(1/2)+b)*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^(13/2)*a^6*b^3-1155*ln(1/2*(2*a*
x^(1/2)+b+2*(a*x+b*x^(1/2))^(1/2)*a^(1/2))/a^(1/2))*x^(13/2)*a^6*b^3-160*(a*x+b*x^(1/2))^(3/2)*a^(5/2)*x^4*b^5
+112*(a*x+b*x^(1/2))^(3/2)*a^(3/2)*x^(7/2)*b^6)/((a*x^(1/2)+b)*x^(1/2))^(1/2)/b^8/x^(13/2)/a^(1/2)/(a*x^(1/2)+
b)^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a x + b \sqrt {x}\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________