3.4.15 \(\int \frac {a+b \log (c x^n)}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) [315]

Optimal. Leaf size=142 \[ -\frac {b n \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b e n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )}{d \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2387, 2373, 283, 222} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b e n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {ArcSin}\left (\frac {e x}{d}\right )}{d \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 222

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

Rule 283

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 2373

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]

Rule 2387

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^(q_)*((d2_) + (e2_.)*(x_))^(q_), x_
Symbol] :> Dist[(d1 + e1*x)^q*((d2 + e2*x)^q/(1 + e1*(e2/(d1*d2))*x^2)^q), Int[x^m*(1 + e1*(e2/(d1*d2))*x^2)^q
*(a + b*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[d2*e1 + d1*e2, 0] && IntegerQ[m]
&& IntegerQ[q - 1/2]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 70, normalized size = 0.49 \begin {gather*} -\frac {b e n x \tan ^{-1}\left (\frac {e x}{\sqrt {d-e x} \sqrt {d+e x}}\right )+\sqrt {d-e x} \sqrt {d+e x} \left (a+b n+b \log \left (c x^n\right )\right )}{d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{2} \sqrt {-e x +d}\, \sqrt {e x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 87, normalized size = 0.61 \begin {gather*} -\frac {{\left (\arcsin \left (\frac {x e}{d}\right ) e + \frac {\sqrt {-x^{2} e^{2} + d^{2}}}{x}\right )} b n}{d^{2}} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} b \log \left (c x^{n}\right )}{d^{2} x} - \frac {\sqrt {-x^{2} e^{2} + d^{2}} a}{d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 77, normalized size = 0.54 \begin {gather*} \frac {2 \, b n x \arctan \left (\frac {{\left (\sqrt {x e + d} \sqrt {-x e + d} - d\right )} e^{\left (-1\right )}}{x}\right ) e - {\left (b n \log \left (x\right ) + b n + b \log \left (c\right ) + a\right )} \sqrt {x e + d} \sqrt {-x e + d}}{d^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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