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

Optimal. Leaf size=84 \[ \frac {b n}{3 d e \sqrt {d+e x^2}}-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{3/2} e}-\frac {a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}} \]

[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)

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Rubi [A]
time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2376, 272, 53, 65, 214} \begin {gather*} -\frac {a+b \log \left (c x^n\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {b n \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \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) \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*}

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Mathematica [A]
time = 0.15, size = 97, normalized size = 1.15 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.30, size = 74, normalized size = 0.88 \begin {gather*} -\frac {1}{3} \, b n {\left (\frac {\operatorname {arsinh}\left (\frac {\sqrt {d} e^{\left (-\frac {1}{2}\right )}}{{\left | x \right |}}\right )}{d^{\frac {3}{2}}} - \frac {1}{\sqrt {x^{2} e + d} d}\right )} e^{\left (-1\right )} - \frac {b e^{\left (-1\right )} \log \left (c x^{n}\right )}{3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}}} - \frac {a e^{\left (-1\right )}}{3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*b*n*(arcsinh(sqrt(d)*e^(-1/2)/abs(x))/d^(3/2) - 1/(sqrt(x^2*e + d)*d))*e^(-1) - 1/3*b*e^(-1)*log(c*x^n)/(
x^2*e + d)^(3/2) - 1/3*a*e^(-1)/(x^2*e + d)^(3/2)

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Fricas [A]
time = 0.40, size = 272, normalized size = 3.24 \begin {gather*} \left [\frac {{\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {x^{2} e - 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b d n x^{2} e - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt {x^{2} e + d}}{6 \, {\left (d^{2} x^{4} e^{3} + 2 \, d^{3} x^{2} e^{2} + d^{4} e\right )}}, \frac {{\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (b d n x^{2} e - b d^{2} n \log \left (x\right ) + b d^{2} n - b d^{2} \log \left (c\right ) - a d^{2}\right )} \sqrt {x^{2} e + d}}{3 \, {\left (d^{2} x^{4} e^{3} + 2 \, d^{3} x^{2} e^{2} + d^{4} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (73) = 146\).
time = 16.92, size = 245, normalized size = 2.92 \begin {gather*} - \frac {a}{3 e \left (d + e x^{2}\right )^{\frac {3}{2}}} + \frac {2 b d^{3} n \sqrt {1 + \frac {e x^{2}}{d}}}{6 d^{\frac {9}{2}} e + 6 d^{\frac {7}{2}} e^{2} x^{2}} + \frac {b d^{3} n \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 b d^{3} n \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 {b d^{2} n 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 b d^{2} n 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}} - \frac {b \log {\left (c x^{n} \right )}}{3 e \left (d + e x^{2}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(3*e*(d + e*x**2)**(3/2)) + 2*b*d**3*n*sqrt(1 + e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) + b*d**3*n*
log(e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) - 2*b*d**3*n*log(sqrt(1 + e*x**2/d) + 1)/(6*d**(9/2)*e + 6
*d**(7/2)*e**2*x**2) + b*d**2*n*x**2*log(e*x**2/d)/(6*d**(9/2) + 6*d**(7/2)*e*x**2) - 2*b*d**2*n*x**2*log(sqrt
(1 + e*x**2/d) + 1)/(6*d**(9/2) + 6*d**(7/2)*e*x**2) - b*log(c*x**n)/(3*e*(d + e*x**2)**(3/2))

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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