3.112 \(\int \frac {1}{(d x)^{5/2} (a+b \log (c x^n))^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac {3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \text {Ei}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )}{2 b^2 d n^2 (d x)^{3/2}}-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )} \]

[Out]

-3/2*exp(3/2*a/b/n)*(c*x^n)^(3/2/n)*Ei(-3/2*(a+b*ln(c*x^n))/b/n)/b^2/d/n^2/(d*x)^(3/2)-1/b/d/n/(d*x)^(3/2)/(a+
b*ln(c*x^n))

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Rubi [A]  time = 0.09, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2306, 2310, 2178} \[ -\frac {3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \text {Ei}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )}{2 b^2 d n^2 (d x)^{3/2}}-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*x)^(5/2)*(a + b*Log[c*x^n])^2),x]

[Out]

(-3*E^((3*a)/(2*b*n))*(c*x^n)^(3/(2*n))*ExpIntegralEi[(-3*(a + b*Log[c*x^n]))/(2*b*n)])/(2*b^2*d*n^2*(d*x)^(3/
2)) - 1/(b*d*n*(d*x)^(3/2)*(a + b*Log[c*x^n]))

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p + 1), x]
, x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && LtQ[p, -1]

Rule 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps

\begin {align*} \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx &=-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}-\frac {3 \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )} \, dx}{2 b n}\\ &=-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}-\frac {\left (3 \left (c x^n\right )^{\left .\frac {3}{2}\right /n}\right ) \operatorname {Subst}\left (\int \frac {e^{-\frac {3 x}{2 n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )}{2 b d n^2 (d x)^{3/2}}\\ &=-\frac {3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \text {Ei}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )}{2 b^2 d n^2 (d x)^{3/2}}-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 94, normalized size = 0.96 \[ -\frac {x \left (3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \left (a+b \log \left (c x^n\right )\right ) \text {Ei}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )+2 b n\right )}{2 b^2 n^2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*x)^(5/2)*(a + b*Log[c*x^n])^2),x]

[Out]

-1/2*(x*(2*b*n + 3*E^((3*a)/(2*b*n))*(c*x^n)^(3/(2*n))*ExpIntegralEi[(-3*(a + b*Log[c*x^n]))/(2*b*n)]*(a + b*L
og[c*x^n])))/(b^2*n^2*(d*x)^(5/2)*(a + b*Log[c*x^n]))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x}}{b^{2} d^{3} x^{3} \log \left (c x^{n}\right )^{2} + 2 \, a b d^{3} x^{3} \log \left (c x^{n}\right ) + a^{2} d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*x)/(b^2*d^3*x^3*log(c*x^n)^2 + 2*a*b*d^3*x^3*log(c*x^n) + a^2*d^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d x\right )^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((d*x)^(5/2)*(b*log(c*x^n) + a)^2), x)

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maple [F]  time = 5.16, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d x \right )^{\frac {5}{2}} \left (b \ln \left (c \,x^{n}\right )+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*x)^(5/2)/(b*ln(c*x^n)+a)^2,x)

[Out]

int(1/(d*x)^(5/2)/(b*ln(c*x^n)+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -4 \, b n \int \frac {1}{3 \, {\left (b^{3} d^{\frac {5}{2}} \log \relax (c)^{3} + b^{3} d^{\frac {5}{2}} \log \left (x^{n}\right )^{3} + 3 \, a b^{2} d^{\frac {5}{2}} \log \relax (c)^{2} + 3 \, a^{2} b d^{\frac {5}{2}} \log \relax (c) + a^{3} d^{\frac {5}{2}} + 3 \, {\left (b^{3} d^{\frac {5}{2}} \log \relax (c) + a b^{2} d^{\frac {5}{2}}\right )} \log \left (x^{n}\right )^{2} + 3 \, {\left (b^{3} d^{\frac {5}{2}} \log \relax (c)^{2} + 2 \, a b^{2} d^{\frac {5}{2}} \log \relax (c) + a^{2} b d^{\frac {5}{2}}\right )} \log \left (x^{n}\right )\right )} x^{\frac {5}{2}}}\,{d x} - \frac {2}{3 \, {\left (b^{2} d^{\frac {5}{2}} \log \relax (c)^{2} + b^{2} d^{\frac {5}{2}} \log \left (x^{n}\right )^{2} + 2 \, a b d^{\frac {5}{2}} \log \relax (c) + a^{2} d^{\frac {5}{2}} + 2 \, {\left (b^{2} d^{\frac {5}{2}} \log \relax (c) + a b d^{\frac {5}{2}}\right )} \log \left (x^{n}\right )\right )} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*b*n*integrate(1/3/((b^3*d^(5/2)*log(c)^3 + b^3*d^(5/2)*log(x^n)^3 + 3*a*b^2*d^(5/2)*log(c)^2 + 3*a^2*b*d^(5
/2)*log(c) + a^3*d^(5/2) + 3*(b^3*d^(5/2)*log(c) + a*b^2*d^(5/2))*log(x^n)^2 + 3*(b^3*d^(5/2)*log(c)^2 + 2*a*b
^2*d^(5/2)*log(c) + a^2*b*d^(5/2))*log(x^n))*x^(5/2)), x) - 2/3/((b^2*d^(5/2)*log(c)^2 + b^2*d^(5/2)*log(x^n)^
2 + 2*a*b*d^(5/2)*log(c) + a^2*d^(5/2) + 2*(b^2*d^(5/2)*log(c) + a*b*d^(5/2))*log(x^n))*x^(3/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (d\,x\right )}^{5/2}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d*x)^(5/2)*(a + b*log(c*x^n))^2),x)

[Out]

int(1/((d*x)^(5/2)*(a + b*log(c*x^n))^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (d x\right )^{\frac {5}{2}} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*x)**(5/2)/(a+b*ln(c*x**n))**2,x)

[Out]

Integral(1/((d*x)**(5/2)*(a + b*log(c*x**n))**2), x)

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