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

Optimal. Leaf size=129 \[ -\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^2}+\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2} \]

[Out]

-1/9*b*n*(e*x^2+d)^(3/2)/e^2-2/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^2+1/3*(e*x^2+d)^(3/2)*(a+b*ln(
c*x^n))/e^2+2/3*b*d*n*(e*x^2+d)^(1/2)/e^2-d*(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 = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {266, 43, 2350, 12, 446, 80, 50, 63, 208} \[ -\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^2}+\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*d*n*Sqrt[d + e*x^2])/(3*e^2) - (b*n*(d + e*x^2)^(3/2))/(9*e^2) - (2*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x^2]/S
qrt[d]])/(3*e^2) - (d*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^2 + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*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 50

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 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 )}{\sqrt {d+e x^2}} \, dx &=-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-(b n) \int \frac {\left (-2 d+e x^2\right ) \sqrt {d+e x^2}}{3 e^2 x} \, dx\\ &=-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {(b n) \int \frac {\left (-2 d+e x^2\right ) \sqrt {d+e x^2}}{x} \, dx}{3 e^2}\\ &=-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {(b n) \operatorname {Subst}\left (\int \frac {(-2 d+e x) \sqrt {d+e x}}{x} \, dx,x,x^2\right )}{6 e^2}\\ &=-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {(b d n) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )}{3 e^2}\\ &=\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^2}\\ &=\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (2 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 e^3}\\ &=\frac {2 b d n \sqrt {d+e x^2}}{3 e^2}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^2}-\frac {2 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 145, normalized size = 1.12 \[ \frac {3 a e x^2 \sqrt {d+e x^2}-6 a d \sqrt {d+e x^2}+3 b \left (e x^2-2 d\right ) \sqrt {d+e x^2} \log \left (c x^n\right )-6 b d^{3/2} n \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+6 b d^{3/2} n \log (x)-b e n x^2 \sqrt {d+e x^2}+5 b d n \sqrt {d+e x^2}}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*d*Sqrt[d + e*x^2] + 5*b*d*n*Sqrt[d + e*x^2] + 3*a*e*x^2*Sqrt[d + e*x^2] - b*e*n*x^2*Sqrt[d + e*x^2] + 6*
b*d^(3/2)*n*Log[x] + 3*b*(-2*d + e*x^2)*Sqrt[d + e*x^2]*Log[c*x^n] - 6*b*d^(3/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*
x^2]])/(9*e^2)

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fricas [A]  time = 0.46, size = 207, normalized size = 1.60 \[ \left [\frac {3 \, b d^{\frac {3}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + {\left (5 \, b d n - {\left (b e n - 3 \, a e\right )} x^{2} - 6 \, a d + 3 \, {\left (b e x^{2} - 2 \, b d\right )} \log \relax (c) + 3 \, {\left (b e n x^{2} - 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{9 \, e^{2}}, \frac {6 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (5 \, b d n - {\left (b e n - 3 \, a e\right )} x^{2} - 6 \, a d + 3 \, {\left (b e x^{2} - 2 \, b d\right )} \log \relax (c) + 3 \, {\left (b e n x^{2} - 2 \, b d n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{9 \, e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/9*(3*b*d^(3/2)*n*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + (5*b*d*n - (b*e*n - 3*a*e)*x^2 - 6*a
*d + 3*(b*e*x^2 - 2*b*d)*log(c) + 3*(b*e*n*x^2 - 2*b*d*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/9*(6*b*sqrt(-d)*d*n*
arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (5*b*d*n - (b*e*n - 3*a*e)*x^2 - 6*a*d + 3*(b*e*x^2 - 2*b*d)*log(c) + 3*(b*
e*n*x^2 - 2*b*d*n)*log(x))*sqrt(e*x^2 + d))/e^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{3}}{\sqrt {e \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.53, size = 149, normalized size = 1.16 \[ \frac {1}{9} \, b n {\left (\frac {3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{2}} - \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} - 6 \, \sqrt {e x^{2} + d} d}{e^{2}}\right )} + \frac {1}{3} \, {\left (\frac {\sqrt {e x^{2} + d} x^{2}}{e} - \frac {2 \, \sqrt {e x^{2} + d} d}{e^{2}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{3} \, {\left (\frac {\sqrt {e x^{2} + d} x^{2}}{e} - \frac {2 \, \sqrt {e x^{2} + d} d}{e^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^(1/2),x)

[Out]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d)**(1/2),x)

[Out]

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

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