3.5.89 \(\int (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}})^{5/2} \, dx\) [489]

Optimal. Leaf size=289 \[ -\frac {4 b^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}}{\left (a+\frac {b}{\sqrt [4]{x}}\right ) \sqrt [4]{x}}+\frac {40 a^2 b^3 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt [4]{x}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a^3 b^2 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt {x}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a^4 b \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x^{3/4}}{3 \left (a+\frac {b}{\sqrt [4]{x}}\right )}+\frac {a^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a b^4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \log \left (\sqrt [4]{x}\right )}{a+\frac {b}{\sqrt [4]{x}}} \]

[Out]

-4*b^5*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))/x^(1/4)+40*a^2*b^3*x^(1/4)*(a^2+2*a*b/x^(1/4)+b^2/x
^(1/2))^(1/2)/(a+b/x^(1/4))+20/3*a^4*b*x^(3/4)*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+a^5*x*(a^2+
2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))+5*a*b^4*ln(x)*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4
))+20*a^3*b^2*x^(1/2)*(a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(1/2)/(a+b/x^(1/4))

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Rubi [A]
time = 0.09, antiderivative size = 289, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1355, 1369, 269, 45} \begin {gather*} -\frac {4 b^5 \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{\sqrt [4]{x} \left (a+\frac {b}{\sqrt [4]{x}}\right )}+\frac {20 a b^4 \log \left (\sqrt [4]{x}\right ) \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {40 a^2 b^3 \sqrt [4]{x} \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {a^5 x \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{a+\frac {b}{\sqrt [4]{x}}}+\frac {20 a^4 b x^{3/4} \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{3 \left (a+\frac {b}{\sqrt [4]{x}}\right )}+\frac {20 a^3 b^2 \sqrt {x} \sqrt {a^2+\frac {2 a b}{\sqrt [4]{x}}+\frac {b^2}{\sqrt {x}}}}{a+\frac {b}{\sqrt [4]{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4))^(5/2),x]

[Out]

(-4*b^5*Sqrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4)])/((a + b/x^(1/4))*x^(1/4)) + (40*a^2*b^3*Sqrt[a^2 + b^2/Sqrt
[x] + (2*a*b)/x^(1/4)]*x^(1/4))/(a + b/x^(1/4)) + (20*a^3*b^2*Sqrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4)]*Sqrt[x
])/(a + b/x^(1/4)) + (20*a^4*b*Sqrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4)]*x^(3/4))/(3*(a + b/x^(1/4))) + (a^5*S
qrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4)]*x)/(a + b/x^(1/4)) + (20*a*b^4*Sqrt[a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/
4)]*Log[x^(1/4)])/(a + b/x^(1/4))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 1355

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}\right )^{5/2} \, dx &=4 \text {Subst}\left (\int \left (a^2+\frac {b^2}{x^2}+\frac {2 a b}{x}\right )^{5/2} x^3 \, dx,x,\sqrt [4]{x}\right )\\ &=\frac {\left (4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}\right ) \text {Subst}\left (\int \left (a b+\frac {b^2}{x}\right )^5 x^3 \, dx,x,\sqrt [4]{x}\right )}{b^4 \left (a b+\frac {b^2}{\sqrt [4]{x}}\right )}\\ &=\frac {\left (4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}\right ) \text {Subst}\left (\int \frac {\left (b^2+a b x\right )^5}{x^2} \, dx,x,\sqrt [4]{x}\right )}{b^4 \left (a b+\frac {b^2}{\sqrt [4]{x}}\right )}\\ &=\frac {\left (4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}\right ) \text {Subst}\left (\int \left (10 a^2 b^8+\frac {b^{10}}{x^2}+\frac {5 a b^9}{x}+10 a^3 b^7 x+5 a^4 b^6 x^2+a^5 b^5 x^3\right ) \, dx,x,\sqrt [4]{x}\right )}{b^4 \left (a b+\frac {b^2}{\sqrt [4]{x}}\right )}\\ &=-\frac {4 b^6 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}}}{\left (a b+\frac {b^2}{\sqrt [4]{x}}\right ) \sqrt [4]{x}}+\frac {40 a^2 b^4 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt [4]{x}}{a b+\frac {b^2}{\sqrt [4]{x}}}+\frac {20 a^3 b^3 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \sqrt {x}}{a b+\frac {b^2}{\sqrt [4]{x}}}+\frac {20 a^4 b^2 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x^{3/4}}{3 \left (a b+\frac {b^2}{\sqrt [4]{x}}\right )}+\frac {a^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} x}{a+\frac {b}{\sqrt [4]{x}}}+\frac {5 a b^5 \sqrt {a^2+\frac {b^2}{\sqrt {x}}+\frac {2 a b}{\sqrt [4]{x}}} \log (x)}{a b+\frac {b^2}{\sqrt [4]{x}}}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 98, normalized size = 0.34 \begin {gather*} \frac {\sqrt {\frac {\left (b+a \sqrt [4]{x}\right )^2}{\sqrt {x}}} \left (-12 b^5+120 a^2 b^3 \sqrt {x}+60 a^3 b^2 x^{3/4}+20 a^4 b x+3 a^5 x^{5/4}+15 a b^4 \sqrt [4]{x} \log (x)\right )}{3 \left (b+a \sqrt [4]{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + b^2/Sqrt[x] + (2*a*b)/x^(1/4))^(5/2),x]

[Out]

(Sqrt[(b + a*x^(1/4))^2/Sqrt[x]]*(-12*b^5 + 120*a^2*b^3*Sqrt[x] + 60*a^3*b^2*x^(3/4) + 20*a^4*b*x + 3*a^5*x^(5
/4) + 15*a*b^4*x^(1/4)*Log[x]))/(3*(b + a*x^(1/4)))

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Maple [A]
time = 0.07, size = 94, normalized size = 0.33

method	result	size
derivativedivides	\(\frac {\left (\frac {a^{2} \sqrt {x}+2 a b \,x^{\frac {1}{4}}+b^{2}}{\sqrt {x}}\right )^{\frac {5}{2}} x \left (3 a^{5} x^{\frac {5}{4}}+20 b \,a^{4} x +60 b^{2} a^{3} x^{\frac {3}{4}}+15 b^{4} a \ln \left (x \right ) x^{\frac {1}{4}}+120 a^{2} b^{3} \sqrt {x}-12 b^{5}\right )}{3 \left (a \,x^{\frac {1}{4}}+b \right )^{5}}\)	\(91\)
default	\(\frac {\sqrt {\frac {a^{2} x^{\frac {3}{4}}+2 a b \sqrt {x}+b^{2} x^{\frac {1}{4}}}{x^{\frac {3}{4}}}}\, \left (3 a^{5} x^{\frac {5}{4}}+20 b \,a^{4} x +60 b^{2} a^{3} x^{\frac {3}{4}}+15 b^{4} a \ln \left (x \right ) x^{\frac {1}{4}}+120 a^{2} b^{3} \sqrt {x}-12 b^{5}\right )}{3 a \,x^{\frac {1}{4}}+3 b}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((a^2*x^(3/4)+2*a*b*x^(1/2)+b^2*x^(1/4))/x^(3/4))^(1/2)*(3*a^5*x^(5/4)+20*b*a^4*x+60*b^2*a^3*x^(3/4)+15*b^
4*a*ln(x)*x^(1/4)+120*a^2*b^3*x^(1/2)-12*b^5)/(a*x^(1/4)+b)

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Maxima [A]
time = 0.29, size = 57, normalized size = 0.20 \begin {gather*} 5 \, a b^{4} \log \left (x\right ) + \frac {3 \, a^{5} x^{\frac {5}{4}} + 20 \, a^{4} b x + 60 \, a^{3} b^{2} x^{\frac {3}{4}} + 120 \, a^{2} b^{3} \sqrt {x} - 12 \, b^{5}}{3 \, x^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(5/2),x, algorithm="maxima")

[Out]

5*a*b^4*log(x) + 1/3*(3*a^5*x^(5/4) + 20*a^4*b*x + 60*a^3*b^2*x^(3/4) + 120*a^2*b^3*sqrt(x) - 12*b^5)/x^(1/4)

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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((a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((a**2+2*a*b/x**(1/4)+b**2/x**(1/2))**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 5.21, size = 126, normalized size = 0.44 \begin {gather*} a^{5} x \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 5 \, a b^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + \frac {20}{3} \, a^{4} b x^{\frac {3}{4}} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 20 \, a^{3} b^{2} \sqrt {x} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) + 40 \, a^{2} b^{3} x^{\frac {1}{4}} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right ) - \frac {4 \, b^{5} \mathrm {sgn}\left (a x + b x^{\frac {3}{4}}\right ) \mathrm {sgn}\left (x\right )}{x^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2+2*a*b/x^(1/4)+b^2/x^(1/2))^(5/2),x, algorithm="giac")

[Out]

a^5*x*sgn(a*x + b*x^(3/4))*sgn(x) + 5*a*b^4*log(abs(x))*sgn(a*x + b*x^(3/4))*sgn(x) + 20/3*a^4*b*x^(3/4)*sgn(a
*x + b*x^(3/4))*sgn(x) + 20*a^3*b^2*sqrt(x)*sgn(a*x + b*x^(3/4))*sgn(x) + 40*a^2*b^3*x^(1/4)*sgn(a*x + b*x^(3/
4))*sgn(x) - 4*b^5*sgn(a*x + b*x^(3/4))*sgn(x)/x^(1/4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a^2+\frac {b^2}{\sqrt {x}}+\frac {2\,a\,b}{x^{1/4}}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2/x^(1/2) + (2*a*b)/x^(1/4))^(5/2),x)

[Out]

int((a^2 + b^2/x^(1/2) + (2*a*b)/x^(1/4))^(5/2), x)

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