3.3.78 \(\int \frac {x (a+b \log (c x^n))}{\sqrt {d+e x^2}} \, dx\) [278]

Optimal. Leaf size=73 \[ -\frac {b n \sqrt {d+e x^2}}{e}+\frac {b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 73, 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, 52, 65, 214} \begin {gather*} \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {b n \sqrt {d+e x^2}}{e}+\frac {b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 )}{\sqrt {d+e x^2}} \, dx &=\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {(b n) \int \frac {\sqrt {d+e x^2}}{x} \, dx}{e}\\ &=\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {(b n) \text {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )}{2 e}\\ &=-\frac {b n \sqrt {d+e x^2}}{e}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {(b d n) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{2 e}\\ &=-\frac {b n \sqrt {d+e x^2}}{e}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}-\frac {(b d 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 \sqrt {d+e x^2}}{e}+\frac {b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 91, normalized size = 1.25 \begin {gather*} \frac {a \sqrt {d+e x^2}-b n \sqrt {d+e x^2}-b \sqrt {d} n \log (x)+b \sqrt {d+e x^2} \log \left (c x^n\right )+b \sqrt {d} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 68, normalized size = 0.93 \begin {gather*} {\left (\sqrt {d} \operatorname {arsinh}\left (\frac {\sqrt {d} e^{\left (-\frac {1}{2}\right )}}{{\left | x \right |}}\right ) - \sqrt {x^{2} e + d}\right )} b n e^{\left (-1\right )} + \sqrt {x^{2} e + d} b e^{\left (-1\right )} \log \left (c x^{n}\right ) + \sqrt {x^{2} e + d} a e^{\left (-1\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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.42, size = 127, normalized size = 1.74 \begin {gather*} \left [\frac {1}{2} \, {\left (b \sqrt {d} n \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, \sqrt {x^{2} e + d} {\left (b n \log \left (x\right ) - b n + b \log \left (c\right ) + a\right )}\right )} e^{\left (-1\right )}, -{\left (b \sqrt {-d} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) - \sqrt {x^{2} e + d} {\left (b n \log \left (x\right ) - b n + b \log \left (c\right ) + a\right )}\right )} e^{\left (-1\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)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 2.67, size = 126, normalized size = 1.73 \begin {gather*} a \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d}} & \text {for}\: e = 0 \\\frac {\sqrt {d + e x^{2}}}{e} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {x^{2}}{4 \sqrt {d}} & \text {for}\: e = 0 \\- \frac {\sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e} + \frac {d}{e^{\frac {3}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {x}{\sqrt {e} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d}} & \text {for}\: e = 0 \\\frac {\sqrt {d + e x^{2}}}{e} & \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)**(1/2),x)

[Out]

a*Piecewise((x**2/(2*sqrt(d)), Eq(e, 0)), (sqrt(d + e*x**2)/e, True)) - b*n*Piecewise((x**2/(4*sqrt(d)), Eq(e,
0)), (-sqrt(d)*asinh(sqrt(d)/(sqrt(e)*x))/e + d/(e**(3/2)*x*sqrt(d/(e*x**2) + 1)) + x/(sqrt(e)*sqrt(d/(e*x**2
) + 1)), True)) + b*Piecewise((x**2/(2*sqrt(d)), Eq(e, 0)), (sqrt(d + e*x**2)/e, 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)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x/sqrt(x^2*e + d), 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 )}{\sqrt {e\,x^2+d}} \,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)^(1/2),x)

[Out]

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

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