3.2.3 \(\int \frac {x^2}{\sqrt {b \sqrt {x}+a x}} \, dx\) [103]

Optimal. Leaf size=174 \[ \frac {63 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^5}-\frac {21 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {63 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{64 a^{11/2}} \]

[Out]

-63/64*b^5*arctanh(a^(1/2)*x^(1/2)/(b*x^(1/2)+a*x)^(1/2))/a^(11/2)+63/64*b^4*(b*x^(1/2)+a*x)^(1/2)/a^5+21/40*b
^2*x*(b*x^(1/2)+a*x)^(1/2)/a^3-9/20*b*x^(3/2)*(b*x^(1/2)+a*x)^(1/2)/a^2+2/5*x^2*(b*x^(1/2)+a*x)^(1/2)/a-21/32*
b^3*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)/a^4

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Rubi [A]
time = 0.11, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2043, 684, 654, 634, 212} \begin {gather*} -\frac {63 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {a x+b \sqrt {x}}}\right )}{64 a^{11/2}}+\frac {63 b^4 \sqrt {a x+b \sqrt {x}}}{64 a^5}-\frac {21 b^3 \sqrt {x} \sqrt {a x+b \sqrt {x}}}{32 a^4}+\frac {21 b^2 x \sqrt {a x+b \sqrt {x}}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {a x+b \sqrt {x}}}{20 a^2}+\frac {2 x^2 \sqrt {a x+b \sqrt {x}}}{5 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(63*b^4*Sqrt[b*Sqrt[x] + a*x])/(64*a^5) - (21*b^3*Sqrt[x]*Sqrt[b*Sqrt[x] + a*x])/(32*a^4) + (21*b^2*x*Sqrt[b*S
qrt[x] + a*x])/(40*a^3) - (9*b*x^(3/2)*Sqrt[b*Sqrt[x] + a*x])/(20*a^2) + (2*x^2*Sqrt[b*Sqrt[x] + a*x])/(5*a) -
(63*b^5*ArcTanh[(Sqrt[a]*Sqrt[x])/Sqrt[b*Sqrt[x] + a*x]])/(64*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 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^2}{\sqrt {b \sqrt {x}+a x}} \, dx &=2 \text {Subst}\left (\int \frac {x^5}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {(9 b) \text {Subst}\left (\int \frac {x^4}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{5 a}\\ &=-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}+\frac {\left (63 b^2\right ) \text {Subst}\left (\int \frac {x^3}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{40 a^2}\\ &=\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {\left (21 b^3\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{16 a^3}\\ &=-\frac {21 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}+\frac {\left (63 b^4\right ) \text {Subst}\left (\int \frac {x}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{64 a^4}\\ &=\frac {63 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^5}-\frac {21 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {\left (63 b^5\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b x+a x^2}} \, dx,x,\sqrt {x}\right )}{128 a^5}\\ &=\frac {63 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^5}-\frac {21 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {\left (63 b^5\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{64 a^5}\\ &=\frac {63 b^4 \sqrt {b \sqrt {x}+a x}}{64 a^5}-\frac {21 b^3 \sqrt {x} \sqrt {b \sqrt {x}+a x}}{32 a^4}+\frac {21 b^2 x \sqrt {b \sqrt {x}+a x}}{40 a^3}-\frac {9 b x^{3/2} \sqrt {b \sqrt {x}+a x}}{20 a^2}+\frac {2 x^2 \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {63 b^5 \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b \sqrt {x}+a x}}\right )}{64 a^{11/2}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 113, normalized size = 0.65 \begin {gather*} \frac {\sqrt {b \sqrt {x}+a x} \left (315 b^4-210 a b^3 \sqrt {x}+168 a^2 b^2 x-144 a^3 b x^{3/2}+128 a^4 x^2\right )}{320 a^5}+\frac {63 b^5 \log \left (b+2 a \sqrt {x}-2 \sqrt {a} \sqrt {b \sqrt {x}+a x}\right )}{128 a^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[b*Sqrt[x] + a*x],x]

[Out]

(Sqrt[b*Sqrt[x] + a*x]*(315*b^4 - 210*a*b^3*Sqrt[x] + 168*a^2*b^2*x - 144*a^3*b*x^(3/2) + 128*a^4*x^2))/(320*a
^5) + (63*b^5*Log[b + 2*a*Sqrt[x] - 2*Sqrt[a]*Sqrt[b*Sqrt[x] + a*x]])/(128*a^(11/2))

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Maple [A]
time = 0.39, size = 223, normalized size = 1.28

method	result	size
derivativedivides	\(\frac {2 x^{2} \sqrt {b \sqrt {x}+a x}}{5 a}-\frac {9 b \left (\frac {x^{\frac {3}{2}} \sqrt {b \sqrt {x}+a x}}{4 a}-\frac {7 b \left (\frac {x \sqrt {b \sqrt {x}+a x}}{3 a}-\frac {5 b \left (\frac {\sqrt {x}\, \sqrt {b \sqrt {x}+a x}}{2 a}-\frac {3 b \left (\frac {\sqrt {b \sqrt {x}+a x}}{a}-\frac {b \ln \left (\frac {\frac {b}{2}+a \sqrt {x}}{\sqrt {a}}+\sqrt {b \sqrt {x}+a x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )}{6 a}\right )}{8 a}\right )}{5 a}\)	\(151\)
default	\(-\frac {\sqrt {b \sqrt {x}+a x}\, \left (544 \sqrt {x}\, \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {7}{2}} b -256 x \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {9}{2}}+1300 \sqrt {x}\, \sqrt {b \sqrt {x}+a x}\, a^{\frac {5}{2}} b^{3}-880 \left (b \sqrt {x}+a x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b^{2}+650 \sqrt {b \sqrt {x}+a x}\, a^{\frac {3}{2}} b^{4}+640 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{5}-1280 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {3}{2}} b^{4}-325 \ln \left (\frac {2 a \sqrt {x}+2 \sqrt {b \sqrt {x}+a x}\, \sqrt {a}+b}{2 \sqrt {a}}\right ) a \,b^{5}\right )}{640 \sqrt {\sqrt {x}\, \left (a \sqrt {x}+b \right )}\, a^{\frac {13}{2}}}\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/640*(b*x^(1/2)+a*x)^(1/2)*(544*x^(1/2)*(b*x^(1/2)+a*x)^(3/2)*a^(7/2)*b-256*x*(b*x^(1/2)+a*x)^(3/2)*a^(9/2)+
1300*x^(1/2)*(b*x^(1/2)+a*x)^(1/2)*a^(5/2)*b^3-880*(b*x^(1/2)+a*x)^(3/2)*a^(5/2)*b^2+650*(b*x^(1/2)+a*x)^(1/2)
*a^(3/2)*b^4+640*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*b^5-1280*(x^(1/2)*(
a*x^(1/2)+b))^(1/2)*a^(3/2)*b^4-325*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*b^5)/(x^
(1/2)*(a*x^(1/2)+b))^(1/2)/a^(13/2)

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.82, size = 111, normalized size = 0.64 \begin {gather*} \frac {1}{320} \, \sqrt {a x + b \sqrt {x}} {\left (2 \, {\left (4 \, {\left (2 \, \sqrt {x} {\left (\frac {8 \, \sqrt {x}}{a} - \frac {9 \, b}{a^{2}}\right )} + \frac {21 \, b^{2}}{a^{3}}\right )} \sqrt {x} - \frac {105 \, b^{3}}{a^{4}}\right )} \sqrt {x} + \frac {315 \, b^{4}}{a^{5}}\right )} + \frac {63 \, b^{5} \log \left ({\left | -2 \, \sqrt {a} {\left (\sqrt {a} \sqrt {x} - \sqrt {a x + b \sqrt {x}}\right )} - b \right |}\right )}{128 \, a^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/320*sqrt(a*x + b*sqrt(x))*(2*(4*(2*sqrt(x)*(8*sqrt(x)/a - 9*b/a^2) + 21*b^2/a^3)*sqrt(x) - 105*b^3/a^4)*sqrt
(x) + 315*b^4/a^5) + 63/128*b^5*log(abs(-2*sqrt(a)*(sqrt(a)*sqrt(x) - sqrt(a*x + b*sqrt(x))) - b))/a^(11/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x + b*x^(1/2))^(1/2),x)

[Out]

int(x^2/(a*x + b*x^(1/2))^(1/2), x)

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