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

Optimal. Leaf size=81 \[ -\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{d x}-\frac {b n \sqrt {d+e x^2}}{d x}+\frac {b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2335, 277, 217, 206} \[ -\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{d x}-\frac {b n \sqrt {d+e x^2}}{d x}+\frac {b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n*Sqrt[d + e*x^2])/(d*x)) + (b*Sqrt[e]*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/d - (Sqrt[d + e*x^2]*(a +
b*Log[c*x^n]))/(d*x)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 77, normalized size = 0.95 \[ \frac {(a+b n) \left (-\sqrt {d+e x^2}\right )-b \sqrt {d+e x^2} \log \left (c x^n\right )+b \sqrt {e} n x \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 127, normalized size = 1.57 \[ \left [\frac {b \sqrt {e} n x \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, \sqrt {e x^{2} + d} {\left (b n \log \relax (x) + b n + b \log \relax (c) + a\right )}}{2 \, d x}, -\frac {b \sqrt {-e} n x \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + \sqrt {e x^{2} + d} {\left (b n \log \relax (x) + b n + b \log \relax (c) + a\right )}}{d x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(b*sqrt(e)*n*x*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - 2*sqrt(e*x^2 + d)*(b*n*log(x) + b*n + b*
log(c) + a))/(d*x), -(b*sqrt(-e)*n*x*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + sqrt(e*x^2 + d)*(b*n*log(x) + b*n +
b*log(c) + a))/(d*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\sqrt {e \,x^{2}+d}\, x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 77, normalized size = 0.95 \[ \frac {{\left (\sqrt {e} \operatorname {arsinh}\left (\frac {e x}{\sqrt {d e}}\right ) - \frac {\sqrt {e x^{2} + d}}{x}\right )} b n}{d} - \frac {\sqrt {e x^{2} + d} b \log \left (c x^{n}\right )}{d x} - \frac {\sqrt {e x^{2} + d} a}{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(e)*arcsinh(e*x/sqrt(d*e)) - sqrt(e*x^2 + d)/x)*b*n/d - sqrt(e*x^2 + d)*b*log(c*x^n)/(d*x) - sqrt(e*x^2 +
d)*a/(d*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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