3.2.24 \(\int \frac {x^{5/2}}{(b \sqrt {x}+a x)^{3/2}} \, dx\) [124]

Optimal. Leaf size=171 \[ -\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^{11/2}} \]

[Out]

315/32*b^4*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(11/2)-4*x^(5/2)/a/(b*x^(1/2)+a*x)^(1/2)-315/32*b^
3*(b*x^(1/2)+a*x)^(1/2)/a^5-21/4*b*x*(b*x^(1/2)+a*x)^(1/2)/a^3+9/2*x^(3/2)*(b*x^(1/2)+a*x)^(1/2)/a^2+105/16*b^
2*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)/a^4

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2043, 682, 684, 654, 634, 212} \begin {gather*} \frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{32 a^{11/2}}-\frac {315 b^3 \sqrt {a x+b \sqrt {x}}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{16 a^4}-\frac {21 b x \sqrt {a x+b \sqrt {x}}}{4 a^3}+\frac {9 x^{3/2} \sqrt {a x+b \sqrt {x}}}{2 a^2}-\frac {4 x^{5/2}}{a \sqrt {a x+b \sqrt {x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x^(5/2))/(a*Sqrt[b*Sqrt[x] + a*x]) - (315*b^3*Sqrt[b*Sqrt[x] + a*x])/(32*a^5) + (105*b^2*Sqrt[x]*Sqrt[b*Sq
rt[x] + a*x])/(16*a^4) - (21*b*x*Sqrt[b*Sqrt[x] + a*x])/(4*a^3) + (9*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(2*a^2) +
(315*b^4*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(32*a^(11/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
 1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 682

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Dist[e^2*((m + p)/(c*(p + 1))), Int[(d + e*x)^(m - 2)*(a + b*x +
 c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]

Rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 2043

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \frac {x^{5/2}}{\left (b \sqrt {x}+a x\right )^{3/2}} \, dx &=2 \text {Subst}\left (\int \frac {x^6}{\left (b x+a x^2\right )^{3/2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {18 \text {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{a}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}-\frac {(63 b) \text {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{4 a^2}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (105 b^2\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{8 a^3}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}-\frac {\left (315 b^3\right ) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{32 a^4}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (315 b^4\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{64 a^5}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {\left (315 b^4\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^5}\\ &=-\frac {4 x^{5/2}}{a \sqrt {b \sqrt {x}+a x}}-\frac {315 b^3 \sqrt {b \sqrt {x}+a x}}{32 a^5}+\frac {105 b^2 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{16 a^4}-\frac {21 b x \sqrt {b \sqrt {x}+a x}}{4 a^3}+\frac {9 x^{3/2} \sqrt {b \sqrt {x}+a x}}{2 a^2}+\frac {315 b^4 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{32 a^{11/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.29, size = 124, normalized size = 0.73 \begin {gather*} \frac {\sqrt {b \sqrt {x}+a x} \left (-315 b^4-105 a b^3 \sqrt {x}+42 a^2 b^2 x-24 a^3 b x^{3/2}+16 a^4 x^2\right )}{32 a^5 \left (b+a \sqrt {x}\right )}-\frac {315 b^4 \log \left (b+2 a \sqrt {x}-2 \sqrt {a} \sqrt {b \sqrt {x}+a x}\right )}{64 a^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(-315*b^4 - 105*a*b^3*Sqrt[x] + 42*a^2*b^2*x - 24*a^3*b*x^(3/2) + 16*a^4*x^2))/(32*a^5*
(b + a*Sqrt[x])) - (315*b^4*Log[b + 2*a*Sqrt[x] - 2*Sqrt[a]*Sqrt[b*Sqrt[x] + a*x]])/(64*a^(11/2))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(526\) vs. \(2(125)=250\).
time = 0.39, size = 527, normalized size = 3.08

method result size
derivativedivides \(\frac {x^{\frac {5}{2}}}{2 a \sqrt {b \sqrt {x}+a x}}-\frac {9 b \left (\frac {x^{2}}{3 a \sqrt {b \sqrt {x}+a x}}-\frac {7 b \left (\frac {x^{\frac {3}{2}}}{2 a \sqrt {b \sqrt {x}+a x}}-\frac {5 b \left (\frac {x}{a \sqrt {b \sqrt {x}+a x}}-\frac {3 b \left (-\frac {\sqrt {x}}{a \sqrt {b \sqrt {x}+a x}}-\frac {b \left (-\frac {1}{a \sqrt {b \sqrt {x}+a x}}+\frac {b +2 a \sqrt {x}}{b a \sqrt {b \sqrt {x}+a x}}\right )}{2 a}+\frac {\ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )}{4 a}\right )}{6 a}\right )}{4 a}\) \(201\)
default \(\frac {\sqrt {b \sqrt {x}+a x}\, \left (32 x^{\frac {3}{2}} \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {11}{2}}+276 x^{\frac {3}{2}} \sqrt {b \sqrt {x}+a x}\, a^{\frac {9}{2}} b^{2}-48 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {9}{2}} b x +690 x \sqrt {b \sqrt {x}+a x}\, a^{\frac {7}{2}} b^{3}-768 x \,a^{\frac {7}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{3}+384 x \,a^{3} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{4}-192 \sqrt {x}\, \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b^{2}-69 x \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{3} b^{4}+552 \sqrt {x}\, \sqrt {b \sqrt {x}+a x}\, a^{\frac {5}{2}} b^{4}-1536 \sqrt {x}\, a^{\frac {5}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, b^{4}+768 \sqrt {x}\, a^{2} \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{5}-112 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{3}+256 a^{\frac {5}{2}} \left (\sqrt {x}\, \left (a \sqrt {x}+b \right )\right )^{\frac {3}{2}} b^{3}-138 \sqrt {x}\, \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a^{2} b^{5}+138 \sqrt {b \sqrt {x}+a x}\, a^{\frac {3}{2}} b^{5}-768 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}} b^{5}+384 a \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) b^{6}-69 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{6}\right )}{64 a^{\frac {13}{2}} \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \left (a \sqrt {x}+b \right )^{2}}\) \(527\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b*x^(1/2)+a*x)^(1/2)/a^(13/2)*(32*x^(3/2)*(b*x^(1/2)+a*x)^(3/2)*a^(11/2)+276*x^(3/2)*(b*x^(1/2)+a*x)^(1/
2)*a^(9/2)*b^2-48*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)*b*x+690*x*(b*x^(1/2)+a*x)^(1/2)*a^(7/2)*b^3-768*x*a^(7/2)*(x^(
1/2)*(a*x^(1/2)+b))^(1/2)*b^3+384*x*a^3*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)
)*b^4-192*x^(1/2)*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*b^2-69*x*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^3*b^4+552*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*b^4-1536*x^(1/2)*a^(5/2)*(x^(1/2)*(a*x^(1/2)+b))^
(1/2)*b^4+768*x^(1/2)*a^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))*b^5-112*(b*x
^(1/2)+a*x)^(3/2)*a^(5/2)*b^3+256*a^(5/2)*(x^(1/2)*(a*x^(1/2)+b))^(3/2)*b^3-138*x^(1/2)*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^2*b^5+138*(b*x^(1/2)+a*x)^(1/2)*a^(3/2)*b^5-768*(x^(1/2)*(a*x^(1/2
)+b))^(1/2)*a^(3/2)*b^5+384*a*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))*b^6-69*l
n(1/2*(2*a*x^(1/2)+2*(b*x^(1/2)+a*x)^(1/2)*a^(1/2)+b)/a^(1/2))*a*b^6)/(x^(1/2)*(a*x^(1/2)+b))^(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(x^(5/2)/(b*x^(1/2)+a*x)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.64, size = 136, normalized size = 0.80 \begin {gather*} \frac {1}{32} \, \sqrt {a x + b \sqrt {x}} {\left (2 \, {\left (4 \, \sqrt {x} {\left (\frac {2 \, \sqrt {x}}{a^{2}} - \frac {5 \, b}{a^{3}}\right )} + \frac {41 \, b^{2}}{a^{4}}\right )} \sqrt {x} - \frac {187 \, b^{3}}{a^{5}}\right )} - \frac {315 \, b^{4} \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{64 \, a^{\frac {11}{2}}} - \frac {4 \, b^{5}}{{\left (a {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} + \sqrt {a} b\right )} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(a*x + b*sqrt(x))*(2*(4*sqrt(x)*(2*sqrt(x)/a^2 - 5*b/a^3) + 41*b^2/a^4)*sqrt(x) - 187*b^3/a^5) - 315/
64*b^4*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(11/2) - 4*b^5/((a*(sqrt(a)*sqrt(x
) - sqrt(a*x + b*sqrt(x))) + sqrt(a)*b)*a^5)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________