3.2.29 \(\int \frac {1}{x^{5/2} (b \sqrt {x}+a x)^{3/2}} \, dx\) [129]

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

[Out]

4/b/x^2/(b*x^(1/2)+a*x)^(1/2)-40/9*(b*x^(1/2)+a*x)^(1/2)/b^2/x^(5/2)+320/63*a*(b*x^(1/2)+a*x)^(1/2)/b^3/x^2-12
8/21*a^2*(b*x^(1/2)+a*x)^(1/2)/b^4/x^(3/2)+512/63*a^3*(b*x^(1/2)+a*x)^(1/2)/b^5/x-1024/63*a^4*(b*x^(1/2)+a*x)^
(1/2)/b^6/x^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.17, antiderivative size = 165, normalized size of antiderivative = 1.00, 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 = {2040, 2041, 2039} \begin {gather*} -\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 {128 a^2 \sqrt {a x+b \sqrt {x}}}{21 b^4 x^{3/2}}+\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}}} \end {gather*}

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 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2040

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 2041

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^{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 = 0.19, size = 96, normalized size = 0.58 \begin {gather*} -\frac {4 \sqrt {b \sqrt {x}+a x} \left (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}\right )}{63 b^6 \left (b+a \sqrt {x}\right ) x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[b*Sqrt[x] + a*x]*(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*(b + a*Sqrt[x])*x^(5/2))

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 3 vs. order 2.
time = 0.40, size = 592, normalized size = 3.59

method	result	size
derivativedivides	\(-\frac {4}{9 b \,x^{2} \sqrt {b \sqrt {x}+a x}}-\frac {20 a \left (-\frac {2}{7 b \,x^{\frac {3}{2}} \sqrt {b \sqrt {x}+a x}}-\frac {8 a \left (-\frac {2}{5 b x \sqrt {b \sqrt {x}+a x}}-\frac {6 a \left (-\frac {2}{3 b \sqrt {x}\, \sqrt {b \sqrt {x}+a x}}+\frac {8 a \left (b +2 a \sqrt {x}\right )}{3 b^{3} \sqrt {b \sqrt {x}+a x}}\right )}{5 b}\right )}{7 b}\right )}{9 b}\)	\(124\)
default	\(\frac {4 \sqrt {b \sqrt {x}+a x}\, \left (126 x^{\frac {13}{2}} \sqrt {b \sqrt {x}+a x}\, a^{\frac {15}{2}}+126 x^{\frac {13}{2}} a^{\frac {15}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}-315 x^{\frac {11}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {13}{2}}+63 x^{\frac {11}{2}} a^{\frac {13}{2}} \left (\sqrt {x}\, \left (a \sqrt {x}+b \right )\right )^{\frac {3}{2}}+126 x^{\frac {11}{2}} \sqrt {b \sqrt {x}+a x}\, a^{\frac {11}{2}} b^{2}+126 x^{\frac {11}{2}} a^{\frac {11}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{2}-128 x^{\frac {9}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {9}{2}} b^{2}+63 x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{7} b -63 x^{\frac {13}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{7} b +252 x^{6} \sqrt {b \sqrt {x}+a x}\, a^{\frac {13}{2}} b +252 x^{6} a^{\frac {13}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b +63 x^{\frac {11}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{5} b^{3}-63 x^{\frac {11}{2}} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{5} b^{3}-508 x^{5} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {11}{2}} b -16 x^{\frac {7}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{4}+32 x^{4} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{3}-7 x^{\frac {5}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} \sqrt {a}\, b^{6}+126 x^{6} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{6} b^{2}-126 x^{6} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{6} b^{2}+10 x^{3} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{5}\right )}{63 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{7} x^{\frac {11}{2}} \sqrt {a}\, \left (a \sqrt {x}+b \right )^{2}}\)	\(592\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 2.40, size = 101, normalized size = 0.61 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {5}{2}} \left (a x + b \sqrt {x}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{5/2}\,{\left (a\,x+b\,\sqrt {x}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________