3.3.89 \(\int \frac {x (a+b \log (c x^n))}{(d+e x^2)^{3/2}} \, dx\) [289]

Optimal. Leaf size=57 \[ -\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{\sqrt {d} e}-\frac {a+b \log \left (c x^n\right )}{e \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2376, 272, 65, 214} \begin {gather*} -\frac {a+b \log \left (c x^n\right )}{e \sqrt {d+e x^2}}-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{\sqrt {d} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.09, size = 77, normalized size = 1.35 \begin {gather*} -\frac {\frac {a}{\sqrt {d+e x^2}}-\frac {b n \log (x)}{\sqrt {d}}+\frac {b \log \left (c x^n\right )}{\sqrt {d+e x^2}}+\frac {b n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{\sqrt {d}}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 57, normalized size = 1.00 \begin {gather*} -\frac {b n \operatorname {arsinh}\left (\frac {\sqrt {d} e^{\left (-\frac {1}{2}\right )}}{{\left | x \right |}}\right ) e^{\left (-1\right )}}{\sqrt {d}} - \frac {b e^{\left (-1\right )} \log \left (c x^{n}\right )}{\sqrt {x^{2} e + d}} - \frac {a e^{\left (-1\right )}}{\sqrt {x^{2} e + d}} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 176, normalized size = 3.09 \begin {gather*} \left [\frac {{\left (b n x^{2} e + b d 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 \log \left (x\right ) + b d \log \left (c\right ) + a d\right )} \sqrt {x^{2} e + d}}{2 \, {\left (d x^{2} e^{2} + d^{2} e\right )}}, \frac {{\left (b n x^{2} e + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) - {\left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )} \sqrt {x^{2} e + d}}{d x^{2} e^{2} + d^{2} e}\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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 6.14, size = 80, normalized size = 1.40 \begin {gather*} - \frac {a}{e \sqrt {d + e x^{2}}} - b n \left (\begin {cases} \frac {x^{2}}{4 d^{\frac {3}{2}}} & \text {for}\: e = 0 \\\frac {\operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{\sqrt {d} e} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {x^{2}}{2 d^{\frac {3}{2}}} & \text {for}\: e = 0 \\- \frac {1}{e \sqrt {d + e x^{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \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)**(3/2),x)

[Out]

-a/(e*sqrt(d + e*x**2)) - b*n*Piecewise((x**2/(4*d**(3/2)), Eq(e, 0)), (asinh(sqrt(d)/(sqrt(e)*x))/(sqrt(d)*e)
, True)) + b*Piecewise((x**2/(2*d**(3/2)), Eq(e, 0)), (-1/(e*sqrt(d + e*x**2)), True))*log(c*x**n)

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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)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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