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

Optimal. Leaf size=100 \[ \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^2}+\frac {2 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e^2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {266, 43, 2350, 12, 446, 80, 63, 208} \[ \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {d \left (a+b \log \left (c x^n\right )\right )}{e^2 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^2}+\frac {2 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 63

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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 118, normalized size = 1.18 \[ \frac {2 a d+a e x^2+b \left (2 d+e x^2\right ) \log \left (c x^n\right )-2 b \sqrt {d} n \log (x) \sqrt {d+e x^2}+2 b \sqrt {d} n \sqrt {d+e x^2} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )-b d n-b e n x^2}{e^2 \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 245, normalized size = 2.45 \[ \left [\frac {{\left (b e n x^{2} + b d n\right )} \sqrt {d} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \relax (c) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}, -\frac {2 \, {\left (b e n x^{2} + b d n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b d n + {\left (b e n - a e\right )} x^{2} - 2 \, a d - {\left (b e x^{2} + 2 \, b d\right )} \log \relax (c) - {\left (b e n x^{2} + 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{e^{3} x^{2} + d e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{3}}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.68, size = 132, normalized size = 1.32 \[ -b n {\left (\frac {\sqrt {d} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{2}} + \frac {\sqrt {e x^{2} + d}}{e^{2}}\right )} + b {\left (\frac {x^{2}}{\sqrt {e x^{2} + d} e} + \frac {2 \, d}{\sqrt {e x^{2} + d} e^{2}}\right )} \log \left (c x^{n}\right ) + a {\left (\frac {x^{2}}{\sqrt {e x^{2} + d} e} + \frac {2 \, d}{\sqrt {e x^{2} + d} e^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 47.87, size = 163, normalized size = 1.63 \[ a \left (\begin {cases} \frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {for}\: e = 0 \\\frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {x^{4}}{16 d^{\frac {3}{2}}} & \text {for}\: e = 0 \\- \frac {2 \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{2}} + \frac {d}{e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {x}{e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {for}\: e = 0 \\\frac {d}{e^{2} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((x**4/(4*d**(3/2)), Eq(e, 0)), (d/(e**2*sqrt(d + e*x**2)) + sqrt(d + e*x**2)/e**2, True)) - b*n*Pi
ecewise((x**4/(16*d**(3/2)), Eq(e, 0)), (-2*sqrt(d)*asinh(sqrt(d)/(sqrt(e)*x))/e**2 + d/(e**(5/2)*x*sqrt(d/(e*
x**2) + 1)) + x/(e**(3/2)*sqrt(d/(e*x**2) + 1)), True)) + b*Piecewise((x**4/(4*d**(3/2)), Eq(e, 0)), (d/(e**2*
sqrt(d + e*x**2)) + sqrt(d + e*x**2)/e**2, True))*log(c*x**n)

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