3.4.9 \(\int \frac {x^3 (a+b \log (c x^n))}{\sqrt {d-e x} \sqrt {d+e x}} \, dx\) [309]

Optimal. Leaf size=251 \[ \frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}} \]

[Out]

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

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Rubi [A]
time = 0.34, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {2387, 272, 45, 2392, 12, 457, 81, 52, 65, 214} \begin {gather*} -\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*x^n]))/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

(2*b*d^2*n*(d^2 - e^2*x^2))/(3*e^4*Sqrt[d - e*x]*Sqrt[d + e*x]) - (b*n*(d^2 - e^2*x^2)^2)/(9*e^4*Sqrt[d - e*x]
*Sqrt[d + e*x]) - (2*b*d^4*n*Sqrt[1 - (e^2*x^2)/d^2]*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])/(3*e^4*Sqrt[d - e*x]*Sq
rt[d + e*x]) - (d^2*(d^2 - e^2*x^2)*(a + b*Log[c*x^n]))/(e^4*Sqrt[d - e*x]*Sqrt[d + e*x]) + ((d^2 - e^2*x^2)^2
*(a + b*Log[c*x^n]))/(3*e^4*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 81

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 214

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

Rule 272

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

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

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 {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx &=\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {1-\frac {e^2 x^2}{d^2}}} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {d^2 \left (-2 d^2-e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{3 e^4 x} \, dx}{\sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \int \frac {\left (-2 d^2-e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{x} \, dx}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (b d^2 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \text {Subst}\left (\int \frac {\left (-2 d^2-e^2 x\right ) \sqrt {1-\frac {e^2 x}{d^2}}}{x} \, dx,x,x^2\right )}{6 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \text {Subst}\left (\int \frac {\sqrt {1-\frac {e^2 x}{d^2}}}{x} \, dx,x,x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {e^2 x}{d^2}}} \, dx,x,x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (2 b d^6 n \sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {d^2 x^2}{e^2}} \, dx,x,\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{3 e^6 \sqrt {d-e x} \sqrt {d+e x}}\\ &=\frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}} \tanh ^{-1}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 163, normalized size = 0.65 \begin {gather*} -\frac {-6 b d^3 n \log (x)+3 b n \sqrt {d-e x} \sqrt {d+e x} \left (2 d^2+e^2 x^2\right ) \log (x)+\sqrt {d-e x} \sqrt {d+e x} \left (e^2 x^2 \left (3 a-b n-3 b n \log (x)+3 b \log \left (c x^n\right )\right )+d^2 \left (6 a-5 b n-6 b n \log (x)+6 b \log \left (c x^n\right )\right )\right )+6 b d^3 n \log \left (d+\sqrt {d-e x} \sqrt {d+e x}\right )}{9 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*x^n]))/(Sqrt[d - e*x]*Sqrt[d + e*x]),x]

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {-e x +d}\, \sqrt {e x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

[Out]

int(x^3*(a+b*ln(c*x^n))/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x)

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Maxima [A]
time = 0.55, size = 184, normalized size = 0.73 \begin {gather*} -\frac {1}{9} \, {\left (3 \, d^{3} e^{\left (-4\right )} \log \left (d + \sqrt {-x^{2} e^{2} + d^{2}}\right ) - 3 \, d^{3} e^{\left (-4\right )} \log \left (-d + \sqrt {-x^{2} e^{2} + d^{2}}\right ) - {\left (6 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} - {\left (-x^{2} e^{2} + d^{2}\right )}^{\frac {3}{2}}\right )} e^{\left (-4\right )}\right )} b n - \frac {1}{3} \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{\left (-2\right )} + 2 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-4\right )}\right )} b \log \left (c x^{n}\right ) - \frac {1}{3} \, {\left (\sqrt {-x^{2} e^{2} + d^{2}} x^{2} e^{\left (-2\right )} + 2 \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-4\right )}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.40, size = 122, normalized size = 0.49 \begin {gather*} \frac {1}{9} \, {\left (6 \, b d^{3} n \log \left (\frac {\sqrt {x e + d} \sqrt {-x e + d} - d}{x}\right ) + {\left (5 \, b d^{2} n + {\left (b n - 3 \, a\right )} x^{2} e^{2} - 6 \, a d^{2} - 3 \, {\left (b x^{2} e^{2} + 2 \, b d^{2}\right )} \log \left (c\right ) - 3 \, {\left (b n x^{2} e^{2} + 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d} \sqrt {-x e + d}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )}{\sqrt {d - e x} \sqrt {d + e x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(-e*x+d)**(1/2)/(e*x+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*log(c*x**n))/(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(x^3*(a+b*log(c*x^n))/(-e*x+d)^(1/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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