3.5.91 \(\int \frac {1}{(a^2+2 a b \sqrt [5]{x}+b^2 x^{2/5})^{5/2}} \, dx\) [491]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 222, normalized size of antiderivative = 1.00, 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 = {1355, 660, 45} \begin {gather*} -\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}}}-\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}}} \end {gather*}

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 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 660

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 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]

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 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 98, normalized size = 0.44 \begin {gather*} \frac {5 a \left (25 a^3+88 a^2 b \sqrt [5]{x}+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}} \end {gather*}

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 = 0.05, size = 152, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {5 \left (12 \ln \left (a +b \,x^{\frac {1}{5}}\right ) b^{4} x^{\frac {4}{5}}+48 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a \,b^{3} x^{\frac {3}{5}}+72 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{2} b^{2} x^{\frac {2}{5}}+48 a \,b^{3} x^{\frac {3}{5}}+48 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{3} b \,x^{\frac {1}{5}}+108 a^{2} b^{2} x^{\frac {2}{5}}+12 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{4}+88 a^{3} b \,x^{\frac {1}{5}}+25 a^{4}\right ) \left (a +b \,x^{\frac {1}{5}}\right )}{12 b^{5} \left (\left (a +b \,x^{\frac {1}{5}}\right )^{2}\right )^{\frac {5}{2}}}\)	\(141\)
default	\(\frac {5 \sqrt {a^{2}+2 a b \,x^{\frac {1}{5}}+b^{2} x^{\frac {2}{5}}}\, \left (12 \ln \left (a +b \,x^{\frac {1}{5}}\right ) b^{4} x^{\frac {4}{5}}+48 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a \,b^{3} x^{\frac {3}{5}}+72 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{2} b^{2} x^{\frac {2}{5}}+48 a \,b^{3} x^{\frac {3}{5}}+48 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{3} b \,x^{\frac {1}{5}}+108 a^{2} b^{2} x^{\frac {2}{5}}+12 \ln \left (a +b \,x^{\frac {1}{5}}\right ) a^{4}+88 a^{3} b \,x^{\frac {1}{5}}+25 a^{4}\right )}{12 \left (a +b \,x^{\frac {1}{5}}\right )^{5} b^{5}}\)	\(152\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 99, normalized size = 0.45 \begin {gather*} \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 {gather*}

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 = 0.45, size = 302, normalized size = 1.36 \begin {gather*} \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 {gather*}

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

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 = 6.02, size = 84, normalized size = 0.38 \begin {gather*} \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 {gather*}

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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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