\(\int \frac {1}{(d x)^{5/2} (a+b \log (c x^n))^2} \, dx\) [112]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (warning: unable to verify)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 98 \[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=-\frac {3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )}{2 b^2 d n^2 (d x)^{3/2}}-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )} \]

[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))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2343, 2347, 2209} \[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=-\frac {3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right )}{2 b^2 d n^2 (d x)^{3/2}}-\frac {1}{b d n (d x)^{3/2} \left (a+b \log \left (c x^n\right )\right )} \]

[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 2209

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

Rule 2343

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=-\frac {x \left (2 b n+3 e^{\frac {3 a}{2 b n}} \left (c x^n\right )^{\left .\frac {3}{2}\right /n} \operatorname {ExpIntegralEi}\left (-\frac {3 \left (a+b \log \left (c x^n\right )\right )}{2 b n}\right ) \left (a+b \log \left (c x^n\right )\right )\right )}{2 b^2 n^2 (d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )} \]

[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]))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.87 (sec) , antiderivative size = 432, normalized size of antiderivative = 4.41

method result size
risch \(-\frac {2}{b n x \sqrt {d x}\, \left (2 a +2 b \ln \left (c \right )+2 b \ln \left ({\mathrm e}^{n \ln \left (x \right )}\right )-i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )+i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}+i b \pi \,\operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}-i b \pi \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{3}\right ) d^{2}}+\frac {3 \,{\mathrm e}^{-\frac {3 i \left (b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )-b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}-b \pi \,\operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}+b \pi \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{3}+2 i b n \left (\ln \left (x \right )-\ln \left (d x \right )\right )+2 i b \ln \left (c \right )+2 i b \left (\ln \left ({\mathrm e}^{n \ln \left (x \right )}\right )-n \ln \left (x \right )\right )+2 i a \right )}{4 b n}} \operatorname {Ei}_{1}\left (\frac {3 \ln \left (d x \right )}{2}-\frac {3 i \left (b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )-b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}-b \pi \,\operatorname {csgn}\left (i {\mathrm e}^{n \ln \left (x \right )}\right ) \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}+b \pi \operatorname {csgn}\left (i c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{3}+2 i b n \left (\ln \left (x \right )-\ln \left (d x \right )\right )+2 i b \ln \left (c \right )+2 i b \left (\ln \left ({\mathrm e}^{n \ln \left (x \right )}\right )-n \ln \left (x \right )\right )+2 i a \right )}{4 b n}\right )}{2 d \,b^{2} n^{2}}\) \(432\)

[In]

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

[Out]

-2/b/n/x/(d*x)^(1/2)/(2*a+2*b*ln(c)+2*b*ln(exp(n*ln(x)))-I*b*Pi*csgn(I*c)*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*
ln(x)))+I*b*Pi*csgn(I*c)*csgn(I*c*exp(n*ln(x)))^2+I*b*Pi*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*ln(x)))^2-I*b*Pi*
csgn(I*c*exp(n*ln(x)))^3)/d^2+3/2/d/b^2/n^2*exp(-3/4*I*(b*Pi*csgn(I*c)*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*ln(
x)))-b*Pi*csgn(I*c)*csgn(I*c*exp(n*ln(x)))^2-b*Pi*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*ln(x)))^2+b*Pi*csgn(I*c*
exp(n*ln(x)))^3+2*I*b*n*(ln(x)-ln(d*x))+2*I*b*ln(c)+2*I*b*(ln(exp(n*ln(x)))-n*ln(x))+2*I*a)/b/n)*Ei(1,3/2*ln(d
*x)-3/4*I*(b*Pi*csgn(I*c)*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*ln(x)))-b*Pi*csgn(I*c)*csgn(I*c*exp(n*ln(x)))^2-
b*Pi*csgn(I*exp(n*ln(x)))*csgn(I*c*exp(n*ln(x)))^2+b*Pi*csgn(I*c*exp(n*ln(x)))^3+2*I*b*n*(ln(x)-ln(d*x))+2*I*b
*ln(c)+2*I*b*(ln(exp(n*ln(x)))-n*ln(x))+2*I*a)/b/n)

Fricas [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int { \frac {1}{\left (d x\right )^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}^{2}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int \frac {1}{\left (d x\right )^{\frac {5}{2}} \left (a + b \log {\left (c x^{n} \right )}\right )^{2}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int { \frac {1}{\left (d x\right )^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}^{2}} \,d x } \]

[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))

Giac [F]

\[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int { \frac {1}{\left (d x\right )^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}^{2}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d x)^{5/2} \left (a+b \log \left (c x^n\right )\right )^2} \, dx=\int \frac {1}{{\left (d\,x\right )}^{5/2}\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^2} \,d x \]

[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)