3.276 \(\int \frac {x^5 (a+b \log (c x^n))}{\sqrt {d+e x^2}} \, dx\)

Optimal. Leaf size=182 \[ \frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^3}-\frac {8 b d^2 n \sqrt {d+e x^2}}{15 e^3}+\frac {7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^3} \]

[Out]

7/45*b*d*n*(e*x^2+d)^(3/2)/e^3-1/25*b*n*(e*x^2+d)^(5/2)/e^3+8/15*b*d^(5/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/
e^3-2/3*d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^3+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^3-8/15*b*d^2*n*(e*x^2+d)^(
1/2)/e^3+d^2*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^3

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Rubi [A]  time = 0.23, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {266, 43, 2350, 12, 1251, 897, 1261, 208} \[ \frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {8 b d^2 n \sqrt {d+e x^2}}{15 e^3}+\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^3}+\frac {7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*x^n]))/Sqrt[d + e*x^2],x]

[Out]

(-8*b*d^2*n*Sqrt[d + e*x^2])/(15*e^3) + (7*b*d*n*(d + e*x^2)^(3/2))/(45*e^3) - (b*n*(d + e*x^2)^(5/2))/(25*e^3
) + (8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(15*e^3) + (d^2*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3 -
(2*d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)

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 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 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-(b n) \int \frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{15 e^3 x} \, dx\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(b n) \int \frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{x} \, dx}{15 e^3}\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x} \, dx,x,x^2\right )}{30 e^3}\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \frac {x^2 \left (15 d^2-10 d x^2+3 x^4\right )}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{15 e^4}\\ &=\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {(b n) \operatorname {Subst}\left (\int \left (8 d^2 e-7 d e x^2+3 e x^4+\frac {8 d^3}{-\frac {d}{e}+\frac {x^2}{e}}\right ) \, dx,x,\sqrt {d+e x^2}\right )}{15 e^4}\\ &=-\frac {8 b d^2 n \sqrt {d+e x^2}}{15 e^3}+\frac {7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^3}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {\left (8 b d^3 n\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{15 e^4}\\ &=-\frac {8 b d^2 n \sqrt {d+e x^2}}{15 e^3}+\frac {7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^3}+\frac {8 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^3}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 204, normalized size = 1.12 \[ \frac {120 a d^2 \sqrt {d+e x^2}+45 a e^2 x^4 \sqrt {d+e x^2}-60 a d e x^2 \sqrt {d+e x^2}+15 b \sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right ) \log \left (c x^n\right )+120 b d^{5/2} n \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )-120 b d^{5/2} n \log (x)-94 b d^2 n \sqrt {d+e x^2}-9 b e^2 n x^4 \sqrt {d+e x^2}+17 b d e n x^2 \sqrt {d+e x^2}}{225 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*x^n]))/Sqrt[d + e*x^2],x]

[Out]

(120*a*d^2*Sqrt[d + e*x^2] - 94*b*d^2*n*Sqrt[d + e*x^2] - 60*a*d*e*x^2*Sqrt[d + e*x^2] + 17*b*d*e*n*x^2*Sqrt[d
+ e*x^2] + 45*a*e^2*x^4*Sqrt[d + e*x^2] - 9*b*e^2*n*x^4*Sqrt[d + e*x^2] - 120*b*d^(5/2)*n*Log[x] + 15*b*Sqrt[
d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4)*Log[c*x^n] + 120*b*d^(5/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(225
*e^3)

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fricas [A]  time = 0.47, size = 314, normalized size = 1.73 \[ \left [\frac {60 \, b d^{\frac {5}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} - {\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \relax (c) - 15 \, {\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{225 \, e^{3}}, -\frac {120 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} - {\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \relax (c) - 15 \, {\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{225 \, e^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/225*(60*b*d^(5/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (9*(b*e^2*n - 5*a*e^2)*x^4 + 94*b
*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60*a*d*e)*x^2 - 15*(3*b*e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(3*b*e
^2*n*x^4 - 4*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e^3, -1/225*(120*b*sqrt(-d)*d^2*n*arctan(sqrt(-
d)/sqrt(e*x^2 + d)) + (9*(b*e^2*n - 5*a*e^2)*x^4 + 94*b*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60*a*d*e)*x^2 - 15*(
3*b*e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^4 - 4*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^
2 + d))/e^3]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{5}}{\sqrt {e \,x^{2}+d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*ln(c*x^n)+a)/(e*x^2+d)^(1/2),x)

[Out]

int(x^5*(b*ln(c*x^n)+a)/(e*x^2+d)^(1/2),x)

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maxima [A]  time = 1.34, size = 206, normalized size = 1.13 \[ -\frac {1}{225} \, b n {\left (\frac {60 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{3}} + \frac {9 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {e x^{2} + d} d^{2}}{e^{3}}\right )} + \frac {1}{15} \, {\left (\frac {3 \, \sqrt {e x^{2} + d} x^{4}}{e} - \frac {4 \, \sqrt {e x^{2} + d} d x^{2}}{e^{2}} + \frac {8 \, \sqrt {e x^{2} + d} d^{2}}{e^{3}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{15} \, {\left (\frac {3 \, \sqrt {e x^{2} + d} x^{4}}{e} - \frac {4 \, \sqrt {e x^{2} + d} d x^{2}}{e^{2}} + \frac {8 \, \sqrt {e x^{2} + d} d^{2}}{e^{3}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/225*b*n*(60*d^(5/2)*log((sqrt(e*x^2 + d) - sqrt(d))/(sqrt(e*x^2 + d) + sqrt(d)))/e^3 + (9*(e*x^2 + d)^(5/2)
- 35*(e*x^2 + d)^(3/2)*d + 120*sqrt(e*x^2 + d)*d^2)/e^3) + 1/15*(3*sqrt(e*x^2 + d)*x^4/e - 4*sqrt(e*x^2 + d)*
d*x^2/e^2 + 8*sqrt(e*x^2 + d)*d^2/e^3)*b*log(c*x^n) + 1/15*(3*sqrt(e*x^2 + d)*x^4/e - 4*sqrt(e*x^2 + d)*d*x^2/
e^2 + 8*sqrt(e*x^2 + d)*d^2/e^3)*a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*log(c*x^n)))/(d + e*x^2)^(1/2),x)

[Out]

int((x^5*(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 \[ \int \frac {x^{5} \left (a + b \log {\left (c x^{n} \right )}\right )}{\sqrt {d + e x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*ln(c*x**n))/(e*x**2+d)**(1/2),x)

[Out]

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

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