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

Optimal. Leaf size=222 \[ -\frac{5 a^4}{4 b^5 \left (a+b \sqrt [5]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{20 a^3}{3 b^5 \left (a+b \sqrt [5]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}-\frac{15 a^2}{b^5 \left (a+b \sqrt [5]{x}\right ) \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{20 a}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{5 \left (a+b \sqrt [5]{x}\right ) \log \left (a+b \sqrt [5]{x}\right )}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}} \]

[Out]

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

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Rubi [A]  time = 0.125948, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1341, 646, 43} \[ -\frac{5 a^4}{4 b^5 \left (a+b \sqrt [5]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{20 a^3}{3 b^5 \left (a+b \sqrt [5]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}-\frac{15 a^2}{b^5 \left (a+b \sqrt [5]{x}\right ) \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{20 a}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{5 \left (a+b \sqrt [5]{x}\right ) \log \left (a+b \sqrt [5]{x}\right )}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1341

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 646

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

Rule 43

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])

Rubi steps

\begin{align*} \int \frac{1}{\left (a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}\right )^{5/2}} \, dx &=5 \operatorname{Subst}\left (\int \frac{x^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx,x,\sqrt [5]{x}\right )\\ &=\frac{\left (5 b^5 \left (a+b \sqrt [5]{x}\right )\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (a b+b^2 x\right )^5} \, dx,x,\sqrt [5]{x}\right )}{\sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}\\ &=\frac{\left (5 b^5 \left (a+b \sqrt [5]{x}\right )\right ) \operatorname{Subst}\left (\int \left (\frac{a^4}{b^9 (a+b x)^5}-\frac{4 a^3}{b^9 (a+b x)^4}+\frac{6 a^2}{b^9 (a+b x)^3}-\frac{4 a}{b^9 (a+b x)^2}+\frac{1}{b^9 (a+b x)}\right ) \, dx,x,\sqrt [5]{x}\right )}{\sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}\\ &=\frac{20 a}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}-\frac{5 a^4}{4 b^5 \left (a+b \sqrt [5]{x}\right )^3 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{20 a^3}{3 b^5 \left (a+b \sqrt [5]{x}\right )^2 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}-\frac{15 a^2}{b^5 \left (a+b \sqrt [5]{x}\right ) \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}+\frac{5 \left (a+b \sqrt [5]{x}\right ) \log \left (a+b \sqrt [5]{x}\right )}{b^5 \sqrt{a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5}}}\\ \end{align*}

Mathematica [A]  time = 0.0966441, size = 98, normalized size = 0.44 \[ \frac{5 a \left (88 a^2 b \sqrt [5]{x}+25 a^3+108 a b^2 x^{2/5}+48 b^3 x^{3/5}\right )+60 \left (a+b \sqrt [5]{x}\right )^4 \log \left (a+b \sqrt [5]{x}\right )}{12 b^5 \left (a+b \sqrt [5]{x}\right )^3 \sqrt{\left (a+b \sqrt [5]{x}\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*(25*a^3 + 88*a^2*b*x^(1/5) + 108*a*b^2*x^(2/5) + 48*b^3*x^(3/5)) + 60*(a + b*x^(1/5))^4*Log[a + b*x^(1/5)
])/(12*b^5*(a + b*x^(1/5))^3*Sqrt[(a + b*x^(1/5))^2])

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Maple [A]  time = 0.011, size = 152, normalized size = 0.7 \begin{align*}{\frac{5}{12\,{b}^{5}}\sqrt{{a}^{2}+2\,ab\sqrt [5]{x}+{b}^{2}{x}^{{\frac{2}{5}}}} \left ( 12\,{x}^{4/5}\ln \left ( a+b\sqrt [5]{x} \right ){b}^{4}+48\,{x}^{3/5}\ln \left ( a+b\sqrt [5]{x} \right ) a{b}^{3}+48\,{x}^{3/5}a{b}^{3}+72\,{x}^{2/5}\ln \left ( a+b\sqrt [5]{x} \right ){a}^{2}{b}^{2}+108\,{x}^{2/5}{a}^{2}{b}^{2}+48\,\sqrt [5]{x}\ln \left ( a+b\sqrt [5]{x} \right ){a}^{3}b+88\,\sqrt [5]{x}{a}^{3}b+12\,\ln \left ( a+b\sqrt [5]{x} \right ){a}^{4}+25\,{a}^{4} \right ) \left ( a+b\sqrt [5]{x} \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*(a^2+2*a*b*x^(1/5)+b^2*x^(2/5))^(1/2)*(12*x^(4/5)*ln(a+b*x^(1/5))*b^4+48*x^(3/5)*ln(a+b*x^(1/5))*a*b^3+48
*x^(3/5)*a*b^3+72*x^(2/5)*ln(a+b*x^(1/5))*a^2*b^2+108*x^(2/5)*a^2*b^2+48*x^(1/5)*ln(a+b*x^(1/5))*a^3*b+88*x^(1
/5)*a^3*b+12*ln(a+b*x^(1/5))*a^4+25*a^4)/(a+b*x^(1/5))^5/b^5

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Maxima [A]  time = 1.01338, size = 134, normalized size = 0.6 \begin{align*} \frac{5 \,{\left (48 \, a b^{3} x^{\frac{3}{5}} + 108 \, a^{2} b^{2} x^{\frac{2}{5}} + 88 \, a^{3} b x^{\frac{1}{5}} + 25 \, a^{4}\right )}}{12 \,{\left (b^{9} x^{\frac{4}{5}} + 4 \, a b^{8} x^{\frac{3}{5}} + 6 \, a^{2} b^{7} x^{\frac{2}{5}} + 4 \, a^{3} b^{6} x^{\frac{1}{5}} + a^{4} b^{5}\right )}} + \frac{5 \, \log \left (b x^{\frac{1}{5}} + a\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*(48*a*b^3*x^(3/5) + 108*a^2*b^2*x^(2/5) + 88*a^3*b*x^(1/5) + 25*a^4)/(b^9*x^(4/5) + 4*a*b^8*x^(3/5) + 6*a
^2*b^7*x^(2/5) + 4*a^3*b^6*x^(1/5) + a^4*b^5) + 5*log(b*x^(1/5) + a)/b^5

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Fricas [A]  time = 2.19283, size = 703, normalized size = 3.17 \begin{align*} \frac{5 \,{\left (300 \, a^{5} b^{15} x^{3} + 100 \, a^{15} b^{5} x + 25 \, a^{20} + 12 \,{\left (b^{20} x^{4} + 4 \, a^{5} b^{15} x^{3} + 6 \, a^{10} b^{10} x^{2} + 4 \, a^{15} b^{5} x + a^{20}\right )} \log \left (b x^{\frac{1}{5}} + a\right ) +{\left (48 \, a b^{19} x^{3} - 226 \, a^{6} b^{14} x^{2} + 104 \, a^{11} b^{9} x + 3 \, a^{16} b^{4}\right )} x^{\frac{4}{5}} -{\left (84 \, a^{2} b^{18} x^{3} - 228 \, a^{7} b^{13} x^{2} + 67 \, a^{12} b^{8} x + 4 \, a^{17} b^{3}\right )} x^{\frac{3}{5}} +{\left (136 \, a^{3} b^{17} x^{3} - 197 \, a^{8} b^{12} x^{2} + 48 \, a^{13} b^{7} x + 6 \, a^{18} b^{2}\right )} x^{\frac{2}{5}} -{\left (207 \, a^{4} b^{16} x^{3} - 124 \, a^{9} b^{11} x^{2} + 56 \, a^{14} b^{6} x + 12 \, a^{19} b\right )} x^{\frac{1}{5}}\right )}}{12 \,{\left (b^{25} x^{4} + 4 \, a^{5} b^{20} x^{3} + 6 \, a^{10} b^{15} x^{2} + 4 \, a^{15} b^{10} x + a^{20} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*(300*a^5*b^15*x^3 + 100*a^15*b^5*x + 25*a^20 + 12*(b^20*x^4 + 4*a^5*b^15*x^3 + 6*a^10*b^10*x^2 + 4*a^15*b
^5*x + a^20)*log(b*x^(1/5) + a) + (48*a*b^19*x^3 - 226*a^6*b^14*x^2 + 104*a^11*b^9*x + 3*a^16*b^4)*x^(4/5) - (
84*a^2*b^18*x^3 - 228*a^7*b^13*x^2 + 67*a^12*b^8*x + 4*a^17*b^3)*x^(3/5) + (136*a^3*b^17*x^3 - 197*a^8*b^12*x^
2 + 48*a^13*b^7*x + 6*a^18*b^2)*x^(2/5) - (207*a^4*b^16*x^3 - 124*a^9*b^11*x^2 + 56*a^14*b^6*x + 12*a^19*b)*x^
(1/5))/(b^25*x^4 + 4*a^5*b^20*x^3 + 6*a^10*b^15*x^2 + 4*a^15*b^10*x + a^20*b^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a**2+2*a*b*x**(1/5)+b**2*x**(2/5))**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x**(1/5) + b**2*x**(2/5))**(-5/2), x)

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Giac [A]  time = 1.13323, size = 113, normalized size = 0.51 \begin{align*} \frac{5 \, \log \left ({\left | b x^{\frac{1}{5}} + a \right |}\right )}{b^{5} \mathrm{sgn}\left (b x^{\frac{1}{5}} + a\right )} + \frac{5 \,{\left (48 \, a b^{2} x^{\frac{3}{5}} + 108 \, a^{2} b x^{\frac{2}{5}} + 88 \, a^{3} x^{\frac{1}{5}} + \frac{25 \, a^{4}}{b}\right )}}{12 \,{\left (b x^{\frac{1}{5}} + a\right )}^{4} b^{4} \mathrm{sgn}\left (b x^{\frac{1}{5}} + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*log(abs(b*x^(1/5) + a))/(b^5*sgn(b*x^(1/5) + a)) + 5/12*(48*a*b^2*x^(3/5) + 108*a^2*b*x^(2/5) + 88*a^3*x^(1/
5) + 25*a^4/b)/((b*x^(1/5) + a)^4*b^4*sgn(b*x^(1/5) + a))