\(\int \frac {a+b \log (c x^n)}{x^6 \sqrt {d+e x^2}} \, dx\) [285]

Optimal result

Integrand size = 25, antiderivative size = 204 \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=-\frac {8 b e^2 n \sqrt {d+e x^2}}{15 d^3 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}+\frac {8 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{15 d^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x} \]

[Out]

-1/25*b*n*(e*x^2+d)^(3/2)/d^2/x^5+26/225*b*e*n*(e*x^2+d)^(3/2)/d^3/x^3+8/15*b*e^(5/2)*n*arctanh(x*e^(1/2)/(e*x
^2+d)^(1/2))/d^3-8/15*b*e^2*n*(e*x^2+d)^(1/2)/d^3/x-1/5*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d/x^5+4/15*e*(a+b*ln(c
*x^n))*(e*x^2+d)^(1/2)/d^2/x^3-8/15*e^2*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d^3/x

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {277, 270, 2392, 12, 1279, 462, 283, 223, 212} \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {8 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{15 d^3}-\frac {8 b e^2 n \sqrt {d+e x^2}}{15 d^3 x}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}-\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5} \]

[In]

Int[(a + b*Log[c*x^n])/(x^6*Sqrt[d + e*x^2]),x]

[Out]

(-8*b*e^2*n*Sqrt[d + e*x^2])/(15*d^3*x) - (b*n*(d + e*x^2)^(3/2))/(25*d^2*x^5) + (26*b*e*n*(d + e*x^2)^(3/2))/
(225*d^3*x^3) + (8*b*e^(5/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(15*d^3) - (Sqrt[d + e*x^2]*(a + b*Log[c*
x^n]))/(5*d*x^5) + (4*e*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(15*d^2*x^3) - (8*e^2*Sqrt[d + e*x^2]*(a + b*Log[c
*x^n]))/(15*d^3*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 212

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

Rule 223

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 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 1279

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

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*} \text {integral}& = -\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}-(b n) \int \frac {\sqrt {d+e x^2} \left (-3 d^2+4 d e x^2-8 e^2 x^4\right )}{15 d^3 x^6} \, dx \\ & = -\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}-\frac {(b n) \int \frac {\sqrt {d+e x^2} \left (-3 d^2+4 d e x^2-8 e^2 x^4\right )}{x^6} \, dx}{15 d^3} \\ & = -\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}+\frac {(b n) \int \frac {\sqrt {d+e x^2} \left (-26 d^2 e+40 d e^2 x^2\right )}{x^4} \, dx}{75 d^4} \\ & = -\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}+\frac {\left (8 b e^2 n\right ) \int \frac {\sqrt {d+e x^2}}{x^2} \, dx}{15 d^3} \\ & = -\frac {8 b e^2 n \sqrt {d+e x^2}}{15 d^3 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}+\frac {\left (8 b e^3 n\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{15 d^3} \\ & = -\frac {8 b e^2 n \sqrt {d+e x^2}}{15 d^3 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x}+\frac {\left (8 b e^3 n\right ) \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{15 d^3} \\ & = -\frac {8 b e^2 n \sqrt {d+e x^2}}{15 d^3 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{25 d^2 x^5}+\frac {26 b e n \left (d+e x^2\right )^{3/2}}{225 d^3 x^3}+\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{15 d^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {4 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {8 e^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{15 d^3 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=-\frac {\sqrt {d+e x^2} \left (15 a \left (3 d^2-4 d e x^2+8 e^2 x^4\right )+b n \left (9 d^2-17 d e x^2+94 e^2 x^4\right )\right )+15 b \sqrt {d+e x^2} \left (3 d^2-4 d e x^2+8 e^2 x^4\right ) \log \left (c x^n\right )-120 b e^{5/2} n x^5 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{225 d^3 x^5} \]

[In]

Integrate[(a + b*Log[c*x^n])/(x^6*Sqrt[d + e*x^2]),x]

[Out]

-1/225*(Sqrt[d + e*x^2]*(15*a*(3*d^2 - 4*d*e*x^2 + 8*e^2*x^4) + b*n*(9*d^2 - 17*d*e*x^2 + 94*e^2*x^4)) + 15*b*
Sqrt[d + e*x^2]*(3*d^2 - 4*d*e*x^2 + 8*e^2*x^4)*Log[c*x^n] - 120*b*e^(5/2)*n*x^5*Log[e*x + Sqrt[e]*Sqrt[d + e*
x^2]])/(d^3*x^5)

Maple [F]

\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{6} \sqrt {e \,x^{2}+d}}d x\]

[In]

int((a+b*ln(c*x^n))/x^6/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*ln(c*x^n))/x^6/(e*x^2+d)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.60 \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=\left [\frac {60 \, b e^{\frac {5}{2}} n x^{5} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left (2 \, {\left (47 \, b e^{2} n + 60 \, a e^{2}\right )} x^{4} + 9 \, b d^{2} n + 45 \, a d^{2} - {\left (17 \, b d e n + 60 \, a d e\right )} x^{2} + 15 \, {\left (8 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (8 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, d^{3} x^{5}}, -\frac {120 \, b \sqrt {-e} e^{2} n x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, {\left (47 \, b e^{2} n + 60 \, a e^{2}\right )} x^{4} + 9 \, b d^{2} n + 45 \, a d^{2} - {\left (17 \, b d e n + 60 \, a d e\right )} x^{2} + 15 \, {\left (8 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (8 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, d^{3} x^{5}}\right ] \]

[In]

integrate((a+b*log(c*x^n))/x^6/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/225*(60*b*e^(5/2)*n*x^5*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2*(47*b*e^2*n + 60*a*e^2)*x^4 +
9*b*d^2*n + 45*a*d^2 - (17*b*d*e*n + 60*a*d*e)*x^2 + 15*(8*b*e^2*x^4 - 4*b*d*e*x^2 + 3*b*d^2)*log(c) + 15*(8*b
*e^2*n*x^4 - 4*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*x^5), -1/225*(120*b*sqrt(-e)*e^2*n*x^5*a
rctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (2*(47*b*e^2*n + 60*a*e^2)*x^4 + 9*b*d^2*n + 45*a*d^2 - (17*b*d*e*n + 60*a
*d*e)*x^2 + 15*(8*b*e^2*x^4 - 4*b*d*e*x^2 + 3*b*d^2)*log(c) + 15*(8*b*e^2*n*x^4 - 4*b*d*e*n*x^2 + 3*b*d^2*n)*l
og(x))*sqrt(e*x^2 + d))/(d^3*x^5)]

Sympy [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{6} \sqrt {d + e x^{2}}}\, dx \]

[In]

integrate((a+b*ln(c*x**n))/x**6/(e*x**2+d)**(1/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**6*sqrt(d + e*x**2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*x^n))/x^6/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x^{2} + d} x^{6}} \,d x } \]

[In]

integrate((a+b*log(c*x^n))/x^6/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/(sqrt(e*x^2 + d)*x^6), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^6 \sqrt {d+e x^2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^6\,\sqrt {e\,x^2+d}} \,d x \]

[In]

int((a + b*log(c*x^n))/(x^6*(d + e*x^2)^(1/2)),x)

[Out]

int((a + b*log(c*x^n))/(x^6*(d + e*x^2)^(1/2)), x)