3.3.76 \(\int \frac {x^5 (a+b \log (c x^n))}{\sqrt {d+e x^2}} \, dx\) [276]

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

[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 = 0.16, 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 = {272, 45, 2392, 12, 1265, 911, 1275, 214} \begin {gather*} \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} \end {gather*}

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 45

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 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 911

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 1265

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 1275

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 2392

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) \text {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) \text {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) \text {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 ) \text {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 = 0.12, size = 204, normalized size = 1.12 \begin {gather*} \frac {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} \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 (d+\sqrt {d} \sqrt {d+e x^2}\right )}{225 e^3} \end {gather*}

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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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{5} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 209, normalized size = 1.15 \begin {gather*} -\frac {1}{225} \, {\left (60 \, d^{\frac {5}{2}} e^{\left (-3\right )} \log \left (\frac {\sqrt {x^{2} e + d} - \sqrt {d}}{\sqrt {x^{2} e + d} + \sqrt {d}}\right ) + {\left (9 \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} - 35 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {x^{2} e + d} d^{2}\right )} e^{\left (-3\right )}\right )} b n + \frac {1}{15} \, {\left (3 \, \sqrt {x^{2} e + d} x^{4} e^{\left (-1\right )} - 4 \, \sqrt {x^{2} e + d} d x^{2} e^{\left (-2\right )} + 8 \, \sqrt {x^{2} e + d} d^{2} e^{\left (-3\right )}\right )} b \log \left (c x^{n}\right ) + \frac {1}{15} \, {\left (3 \, \sqrt {x^{2} e + d} x^{4} e^{\left (-1\right )} - 4 \, \sqrt {x^{2} e + d} d x^{2} e^{\left (-2\right )} + 8 \, \sqrt {x^{2} e + d} d^{2} e^{\left (-3\right )}\right )} a \end {gather*}

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

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Fricas [A]
time = 0.41, size = 309, normalized size = 1.70 \begin {gather*} \left [\frac {1}{225} \, {\left (60 \, b d^{\frac {5}{2}} n \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} + 94 \, b d^{2} n - {\left (17 \, b d n - 60 \, a d\right )} x^{2} e - 120 \, a d^{2} - 15 \, {\left (3 \, b x^{4} e^{2} - 4 \, b d x^{2} e + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{4} e^{2} - 4 \, b d n x^{2} e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-3\right )}, -\frac {1}{225} \, {\left (120 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (9 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} + 94 \, b d^{2} n - {\left (17 \, b d n - 60 \, a d\right )} x^{2} e - 120 \, a d^{2} - 15 \, {\left (3 \, b x^{4} e^{2} - 4 \, b d x^{2} e + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{4} e^{2} - 4 \, b d n x^{2} e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-3\right )}\right ] \end {gather*}

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

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

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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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^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(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^5\,\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^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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