\(\int \frac {(d+e x^2)^{3/2} (a+b \log (c x^n))}{x^6} \, dx\) [272]

Optimal result

Integrand size = 25, antiderivative size = 138 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b e^2 n \sqrt {d+e x^2}}{5 d x}-\frac {b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 d x^5}+\frac {b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 d}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2373, 283, 223, 212} \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{5 d}-\frac {b e^2 n \sqrt {d+e x^2}}{5 d x}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 d x^5}-\frac {b e n \left (d+e x^2\right )^{3/2}}{15 d x^3} \]

[In]

Int[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-1/5*(b*e^2*n*Sqrt[d + e*x^2])/(d*x) - (b*e*n*(d + e*x^2)^(3/2))/(15*d*x^3) - (b*n*(d + e*x^2)^(5/2))/(25*d*x^
5) + (b*e^(5/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(5*d) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*d*x^
5)

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 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]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {\sqrt {d+e x^2} \left (15 a \left (d+e x^2\right )^2+b n \left (3 d^2+11 d e x^2+23 e^2 x^4\right )\right )+15 b \left (d+e x^2\right )^{5/2} \log \left (c x^n\right )-15 b e^{5/2} n x^5 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{75 d x^5} \]

[In]

Integrate[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-1/75*(Sqrt[d + e*x^2]*(15*a*(d + e*x^2)^2 + b*n*(3*d^2 + 11*d*e*x^2 + 23*e^2*x^4)) + 15*b*(d + e*x^2)^(5/2)*L
og[c*x^n] - 15*b*e^(5/2)*n*x^5*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(d*x^5)

Maple [F]

\[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{6}}d x\]

[In]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^6,x)

[Out]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^6,x)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.28 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\left [\frac {15 \, b e^{\frac {5}{2}} n x^{5} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - 2 \, {\left ({\left (23 \, b e^{2} n + 15 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 15 \, a d^{2} + {\left (11 \, b d e n + 30 \, a d e\right )} x^{2} + 15 \, {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{150 \, d x^{5}}, -\frac {15 \, b \sqrt {-e} e^{2} n x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left ({\left (23 \, b e^{2} n + 15 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 15 \, a d^{2} + {\left (11 \, b d e n + 30 \, a d e\right )} x^{2} + 15 \, {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{75 \, d x^{5}}\right ] \]

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,x, algorithm="fricas")

[Out]

[1/150*(15*b*e^(5/2)*n*x^5*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - 2*((23*b*e^2*n + 15*a*e^2)*x^4 +
3*b*d^2*n + 15*a*d^2 + (11*b*d*e*n + 30*a*d*e)*x^2 + 15*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*log(c) + 15*(b*e^2*n
*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d*x^5), -1/75*(15*b*sqrt(-e)*e^2*n*x^5*arctan(sqrt(-
e)*x/sqrt(e*x^2 + d)) + ((23*b*e^2*n + 15*a*e^2)*x^4 + 3*b*d^2*n + 15*a*d^2 + (11*b*d*e*n + 30*a*d*e)*x^2 + 15
*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*log(c) + 15*(b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x^2 + d)
)/(d*x^5)]

Sympy [F]

\[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{6}}\, dx \]

[In]

integrate((e*x**2+d)**(3/2)*(a+b*ln(c*x**n))/x**6,x)

[Out]

Integral((a + b*log(c*x**n))*(d + e*x**2)**(3/2)/x**6, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,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]

\[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{6}} \,d x } \]

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*log(c*x^n) + a)/x^6, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \]

[In]

int(((d + e*x^2)^(3/2)*(a + b*log(c*x^n)))/x^6,x)

[Out]

int(((d + e*x^2)^(3/2)*(a + b*log(c*x^n)))/x^6, x)