3.4.16 \(\int \frac {a+b \log (c x^n)}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) [316]

Optimal. Leaf size=252 \[ -\frac {2 b e^2 n \left (d^2-e^2 x^2\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b e^3 n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )}{3 d^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}} \]

[Out]

-2/3*b*e^2*n*(-e^2*x^2+d^2)/d^4/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/9*b*n*(-e^2*x^2+d^2)^2/d^4/x^3/(-e*x+d)^(1/2)
/(e*x+d)^(1/2)-1/3*(-e^2*x^2+d^2)*(a+b*ln(c*x^n))/d^2/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-2/3*e^2*(-e^2*x^2+d^2)*
(a+b*ln(c*x^n))/d^4/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-2/3*b*e^3*n*arcsin(e*x/d)*(1-e^2*x^2/d^2)^(1/2)/d^3/(-e*x+d
)^(1/2)/(e*x+d)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2387, 277, 270, 2392, 12, 462, 283, 222} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b e^3 n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {ArcSin}\left (\frac {e x}{d}\right )}{3 d^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b e^2 n \left (d^2-e^2 x^2\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \sqrt {d-e x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^4*Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

(-2*b*e^2*n*(d^2 - e^2*x^2))/(3*d^4*x*Sqrt[d - e*x]*Sqrt[d + e*x]) - (b*n*(d^2 - e^2*x^2)^2)/(9*d^4*x^3*Sqrt[d
- e*x]*Sqrt[d + e*x]) - (2*b*e^3*n*Sqrt[1 - (e^2*x^2)/d^2]*ArcSin[(e*x)/d])/(3*d^3*Sqrt[d - e*x]*Sqrt[d + e*x
]) - ((d^2 - e^2*x^2)*(a + b*Log[c*x^n]))/(3*d^2*x^3*Sqrt[d - e*x]*Sqrt[d + e*x]) - (2*e^2*(d^2 - e^2*x^2)*(a
+ b*Log[c*x^n]))/(3*d^4*x*Sqrt[d - e*x]*Sqrt[d + e*x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 270

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

Rule 277

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

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 462

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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]

Rule 2392

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x^4 \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^4 \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 )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 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 {\left (-d^2-2 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{3 d^2 x^4} \, 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 )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 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 {\left (-d^2-2 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{x^4} \, dx}{3 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \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 )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (2 b e^2 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}{3 d^2 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {2 b e^2 n \left (d^2-e^2 x^2\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \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 )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (2 b e^4 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {1}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{3 d^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {2 b e^2 n \left (d^2-e^2 x^2\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b e^3 n \sqrt {1-\frac {e^2 x^2}{d^2}} \sin ^{-1}\left (\frac {e x}{d}\right )}{3 d^3 \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 )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 116, normalized size = 0.46 \begin {gather*} -\frac {6 b e^3 n x^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d-e x} \sqrt {d+e x}}\right )+\sqrt {d-e x} \sqrt {d+e x} \left (3 a \left (d^2+2 e^2 x^2\right )+b n \left (d^2+5 e^2 x^2\right )+3 b \left (d^2+2 e^2 x^2\right ) \log \left (c x^n\right )\right )}{9 d^4 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x^4*Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

-1/9*(6*b*e^3*n*x^3*ArcTan[(e*x)/(Sqrt[d - e*x]*Sqrt[d + e*x])] + Sqrt[d - e*x]*Sqrt[d + e*x]*(3*a*(d^2 + 2*e^
2*x^2) + b*n*(d^2 + 5*e^2*x^2) + 3*b*(d^2 + 2*e^2*x^2)*Log[c*x^n]))/(d^4*x^3)

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \sqrt {-e x +d}\, \sqrt {e x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

[Out]

int((a+b*ln(c*x^n))/x^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 131, normalized size = 0.52 \begin {gather*} \frac {12 \, b n x^{3} \arctan \left (\frac {{\left (\sqrt {x e + d} \sqrt {-x e + d} - d\right )} e^{\left (-1\right )}}{x}\right ) e^{3} - {\left (b d^{2} n + {\left (5 \, b n + 6 \, a\right )} x^{2} e^{2} + 3 \, a d^{2} + 3 \, {\left (2 \, b x^{2} e^{2} + b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (2 \, b n x^{2} e^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d} \sqrt {-x e + d}}{9 \, d^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

1/9*(12*b*n*x^3*arctan((sqrt(x*e + d)*sqrt(-x*e + d) - d)*e^(-1)/x)*e^3 - (b*d^2*n + (5*b*n + 6*a)*x^2*e^2 + 3
*a*d^2 + 3*(2*b*x^2*e^2 + b*d^2)*log(c) + 3*(2*b*n*x^2*e^2 + b*d^2*n)*log(x))*sqrt(x*e + d)*sqrt(-x*e + d))/(d
^4*x^3)

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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^{4} \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**4/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*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^4/(-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^4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\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^4*(d + e*x)^(1/2)*(d - e*x)^(1/2)),x)

[Out]

int((a + b*log(c*x^n))/(x^4*(d + e*x)^(1/2)*(d - e*x)^(1/2)), x)

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