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

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

[Out]

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

________________________________________________________________________________________

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} \[ \frac {\sqrt {d x} e^{-\frac {a}{2 b n}} \left (c x^n\right )^{\left .-\frac {1}{2}\right /n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{2 b n}\right )}{2 b^2 d n^2}-\frac {\sqrt {d x}}{b d n \left (a+b \log \left (c x^n\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*x]*ExpIntegralEi[(a + b*Log[c*x^n])/(2*b*n)])/(2*b^2*d*E^(a/(2*b*n))*n^2*(c*x^n)^(1/(2*n))) - Sqrt[d*x
]/(b*d*n*(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}{\sqrt {d x} \left (a+b \log \left (c x^n\right )\right )^2} \, dx &=-\frac {\sqrt {d x}}{b d n \left (a+b \log \left (c x^n\right )\right )}+\frac {\int \frac {1}{\sqrt {d x} \left (a+b \log \left (c x^n\right )\right )} \, dx}{2 b n}\\ &=-\frac {\sqrt {d x}}{b d n \left (a+b \log \left (c x^n\right )\right )}+\frac {\left (\sqrt {d x} \left (c x^n\right )^{\left .-\frac {1}{2}\right /n}\right ) \operatorname {Subst}\left (\int \frac {e^{\frac {x}{2 n}}}{a+b x} \, dx,x,\log \left (c x^n\right )\right )}{2 b d n^2}\\ &=\frac {e^{-\frac {a}{2 b n}} \sqrt {d x} \left (c x^n\right )^{\left .-\frac {1}{2}\right /n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{2 b n}\right )}{2 b^2 d n^2}-\frac {\sqrt {d x}}{b d n \left (a+b \log \left (c x^n\right )\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.20, size = 83, normalized size = 0.85 \[ \frac {x \left (e^{-\frac {a}{2 b n}} \left (c x^n\right )^{\left .-\frac {1}{2}\right /n} \text {Ei}\left (\frac {a+b \log \left (c x^n\right )}{2 b n}\right )-\frac {2 b n}{a+b \log \left (c x^n\right )}\right )}{2 b^2 n^2 \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x}}{b^{2} d x \log \left (c x^{n}\right )^{2} + 2 \, a b d x \log \left (c x^{n}\right ) + a^{2} d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 2.00, size = 427, normalized size = 4.36 \[ -\frac {2 x}{\sqrt {d x}\, \left (-i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )+i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}+i \pi b \,\mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}-i \pi b \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{3}+2 b \ln \relax (c )+2 b \ln \left ({\mathrm e}^{n \ln \relax (x )}\right )+2 a \right ) b n}-\frac {\Ei \left (1, -\frac {\ln \left (d x \right )}{2}+\frac {i \left (\pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )-\pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}-\pi b \,\mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}+\pi b \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{3}+2 i \left (\ln \relax (x )-\ln \left (d x \right )\right ) b n +2 i b \ln \relax (c )+2 i a +2 i \left (-n \ln \relax (x )+\ln \left ({\mathrm e}^{n \ln \relax (x )}\right )\right ) b \right )}{4 b n}\right ) {\mathrm e}^{\frac {i \left (\pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )-\pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}-\pi b \,\mathrm {csgn}\left (i {\mathrm e}^{n \ln \relax (x )}\right ) \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{2}+\pi b \mathrm {csgn}\left (i c \,{\mathrm e}^{n \ln \relax (x )}\right )^{3}+2 i \left (\ln \relax (x )-\ln \left (d x \right )\right ) b n +2 i b \ln \relax (c )+2 i a +2 i \left (-n \ln \relax (x )+\ln \left ({\mathrm e}^{n \ln \relax (x )}\right )\right ) b \right )}{4 b n}}}{2 b^{2} d \,n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________