\(\int \frac {x (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\) [300]

Optimal result

Integrand size = 23, antiderivative size = 84 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {b n}{3 d e \sqrt {d+e x^2}}-\frac {b n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{3/2} e}-\frac {a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2376, 272, 53, 65, 214} \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {b n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{3/2} e}+\frac {b n}{3 d e \sqrt {d+e x^2}} \]

[In]

Int[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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 2376

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

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.15 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {\frac {a}{\left (d+e x^2\right )^{3/2}}-\frac {b n}{d \sqrt {d+e x^2}}-\frac {b n \log (x)}{d^{3/2}}+\frac {b \log \left (c x^n\right )}{\left (d+e x^2\right )^{3/2}}+\frac {b n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{d^{3/2}}}{3 e} \]

[In]

Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Maple [F]

\[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]

[In]

int(x*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

int(x*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.18 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b d e n x^{2} - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}, \frac {{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b d e n x^{2} - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}\right ] \]

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 13.69 (sec) , antiderivative size = 272, normalized size of antiderivative = 3.24 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=a \left (\begin {cases} - \frac {1}{3 e \left (d + e x^{2}\right )^{\frac {3}{2}}} & \text {for}\: e \neq 0 \\\frac {x^{2}}{2 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {2 d^{3} \sqrt {1 + \frac {e x^{2}}{d}}}{6 d^{\frac {9}{2}} e + 6 d^{\frac {7}{2}} e^{2} x^{2}} - \frac {d^{3} \log {\left (\frac {e x^{2}}{d} \right )}}{6 d^{\frac {9}{2}} e + 6 d^{\frac {7}{2}} e^{2} x^{2}} + \frac {2 d^{3} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{6 d^{\frac {9}{2}} e + 6 d^{\frac {7}{2}} e^{2} x^{2}} - \frac {d^{2} x^{2} \log {\left (\frac {e x^{2}}{d} \right )}}{6 d^{\frac {9}{2}} + 6 d^{\frac {7}{2}} e x^{2}} + \frac {2 d^{2} x^{2} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{6 d^{\frac {9}{2}} + 6 d^{\frac {7}{2}} e x^{2}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{2}}{4 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {1}{3 e \left (d + e x^{2}\right )^{\frac {3}{2}}} & \text {for}\: e \neq 0 \\\frac {x^{2}}{2 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

[In]

integrate(x*(a+b*ln(c*x**n))/(e*x**2+d)**(5/2),x)

[Out]

a*Piecewise((-1/(3*e*(d + e*x**2)**(3/2)), Ne(e, 0)), (x**2/(2*d**(5/2)), True)) - b*n*Piecewise((-2*d**3*sqrt
(1 + e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) - d**3*log(e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2
) + 2*d**3*log(sqrt(1 + e*x**2/d) + 1)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) - d**2*x**2*log(e*x**2/d)/(6*d**(
9/2) + 6*d**(7/2)*e*x**2) + 2*d**2*x**2*log(sqrt(1 + e*x**2/d) + 1)/(6*d**(9/2) + 6*d**(7/2)*e*x**2), (e > -oo
) & (e < oo) & Ne(e, 0)), (x**2/(4*d**(5/2)), True)) + b*Piecewise((-1/(3*e*(d + e*x**2)**(3/2)), Ne(e, 0)), (
x**2/(2*d**(5/2)), True))*log(c*x**n)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),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]

\[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x/(e*x^2 + d)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

[In]

int((x*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2),x)

[Out]

int((x*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2), x)