3.3.98 \(\int \frac {x^5 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\) [298]

Optimal. Leaf size=155 \[ \frac {b d n}{3 e^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^3}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3} \]

[Out]

-1/3*d^2*(a+b*ln(c*x^n))/e^3/(e*x^2+d)^(3/2)+8/3*b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))*d^(1/2)/e^3+1/3*b*d*n/e^
3/(e*x^2+d)^(1/2)+2*d*(a+b*ln(c*x^n))/e^3/(e*x^2+d)^(1/2)-b*n*(e*x^2+d)^(1/2)/e^3+(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 = 155, 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, 212} \begin {gather*} -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {b d n}{3 e^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^3}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*n)/(3*e^3*Sqrt[d + e*x^2]) - (b*n*Sqrt[d + e*x^2])/e^3 + (8*b*Sqrt[d]*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])
/(3*e^3) - (d^2*(a + b*Log[c*x^n]))/(3*e^3*(d + e*x^2)^(3/2)) + (2*d*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e*x^2])
+ (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 )}{\left (d+e x^2\right )^{5/2}} \, dx &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-(b n) \int \frac {8 d^2+12 d e x^2+3 e^2 x^4}{3 e^3 x \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {(b n) \int \frac {8 d^2+12 d e x^2+3 e^2 x^4}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 e^3}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {(b n) \text {Subst}\left (\int \frac {8 d^2+12 d e x+3 e^2 x^2}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^3}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {(b n) \text {Subst}\left (\int \frac {-d^2+6 d x^2+3 x^4}{x^2 \left (-\frac {d}{e}+\frac {x^2}{e}\right )} \, dx,x,\sqrt {d+e x^2}\right )}{3 e^4}\\ &=-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {(b n) \text {Subst}\left (\int \left (3 e+\frac {d e}{x^2}-\frac {8 d e}{d-x^2}\right ) \, dx,x,\sqrt {d+e x^2}\right )}{3 e^4}\\ &=\frac {b d n}{3 e^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {(8 b d n) \text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d+e x^2}\right )}{3 e^3}\\ &=\frac {b d n}{3 e^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^3}+\frac {8 b \sqrt {d} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 205, normalized size = 1.32 \begin {gather*} -\frac {8 b \sqrt {d} n \log (x)}{3 e^3}+\frac {b n \left (8 d^2+12 d e x^2+3 e^2 x^4\right ) \log (x)}{3 e^3 \left (d+e x^2\right )^{3/2}}+\sqrt {d+e x^2} \left (-\frac {d^2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3 e^3 \left (d+e x^2\right )^2}+\frac {a-b n+b \left (-n \log (x)+\log \left (c x^n\right )\right )}{e^3}+\frac {d \left (6 a+b n+6 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3 e^3 \left (d+e x^2\right )}\right )+\frac {8 b \sqrt {d} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*b*Sqrt[d]*n*Log[x])/(3*e^3) + (b*n*(8*d^2 + 12*d*e*x^2 + 3*e^2*x^4)*Log[x])/(3*e^3*(d + e*x^2)^(3/2)) + Sq
rt[d + e*x^2]*(-1/3*(d^2*(a + b*(-(n*Log[x]) + Log[c*x^n])))/(e^3*(d + e*x^2)^2) + (a - b*n + b*(-(n*Log[x]) +
Log[c*x^n]))/e^3 + (d*(6*a + b*n + 6*b*(-(n*Log[x]) + Log[c*x^n])))/(3*e^3*(d + e*x^2))) + (8*b*Sqrt[d]*n*Log
[d + Sqrt[d]*Sqrt[d + e*x^2]])/(3*e^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 194, normalized size = 1.25 \begin {gather*} -\frac {1}{3} \, {\left (4 \, \sqrt {d} e^{\left (-3\right )} \log \left (\frac {\sqrt {x^{2} e + d} - \sqrt {d}}{\sqrt {x^{2} e + d} + \sqrt {d}}\right ) + 3 \, \sqrt {x^{2} e + d} e^{\left (-3\right )} - \frac {d e^{\left (-3\right )}}{\sqrt {x^{2} e + d}}\right )} b n + \frac {1}{3} \, {\left (\frac {3 \, x^{4} e^{\left (-1\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {12 \, d x^{2} e^{\left (-2\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {8 \, d^{2} e^{\left (-3\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{3} \, {\left (\frac {3 \, x^{4} e^{\left (-1\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {12 \, d x^{2} e^{\left (-2\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {8 \, d^{2} e^{\left (-3\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}}\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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(4*sqrt(d)*e^(-3)*log((sqrt(x^2*e + d) - sqrt(d))/(sqrt(x^2*e + d) + sqrt(d))) + 3*sqrt(x^2*e + d)*e^(-3)
- d*e^(-3)/sqrt(x^2*e + d))*b*n + 1/3*(3*x^4*e^(-1)/(x^2*e + d)^(3/2) + 12*d*x^2*e^(-2)/(x^2*e + d)^(3/2) + 8
*d^2*e^(-3)/(x^2*e + d)^(3/2))*b*log(c*x^n) + 1/3*(3*x^4*e^(-1)/(x^2*e + d)^(3/2) + 12*d*x^2*e^(-2)/(x^2*e + d
)^(3/2) + 8*d^2*e^(-3)/(x^2*e + d)^(3/2))*a

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Fricas [A]
time = 0.42, size = 392, normalized size = 2.53 \begin {gather*} \left [\frac {4 \, {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (3 \, {\left (b n - a\right )} x^{4} e^{2} + 2 \, b d^{2} n + {\left (5 \, b d n - 12 \, a d\right )} x^{2} e - 8 \, a d^{2} - {\left (3 \, b x^{4} e^{2} + 12 \, b d x^{2} e + 8 \, b d^{2}\right )} \log \left (c\right ) - {\left (3 \, b n x^{4} e^{2} + 12 \, b d n x^{2} e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{3 \, {\left (x^{4} e^{5} + 2 \, d x^{2} e^{4} + d^{2} e^{3}\right )}}, -\frac {8 \, {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (3 \, {\left (b n - a\right )} x^{4} e^{2} + 2 \, b d^{2} n + {\left (5 \, b d n - 12 \, a d\right )} x^{2} e - 8 \, a d^{2} - {\left (3 \, b x^{4} e^{2} + 12 \, b d x^{2} e + 8 \, b d^{2}\right )} \log \left (c\right ) - {\left (3 \, b n x^{4} e^{2} + 12 \, b d n x^{2} e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{3 \, {\left (x^{4} e^{5} + 2 \, d x^{2} e^{4} + d^{2} e^{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)^(5/2),x, algorithm="fricas")

[Out]

[1/3*(4*(b*n*x^4*e^2 + 2*b*d*n*x^2*e + b*d^2*n)*sqrt(d)*log(-(x^2*e + 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) -
(3*(b*n - a)*x^4*e^2 + 2*b*d^2*n + (5*b*d*n - 12*a*d)*x^2*e - 8*a*d^2 - (3*b*x^4*e^2 + 12*b*d*x^2*e + 8*b*d^2)
*log(c) - (3*b*n*x^4*e^2 + 12*b*d*n*x^2*e + 8*b*d^2*n)*log(x))*sqrt(x^2*e + d))/(x^4*e^5 + 2*d*x^2*e^4 + d^2*e
^3), -1/3*(8*(b*n*x^4*e^2 + 2*b*d*n*x^2*e + b*d^2*n)*sqrt(-d)*arctan(sqrt(-d)/sqrt(x^2*e + d)) + (3*(b*n - a)*
x^4*e^2 + 2*b*d^2*n + (5*b*d*n - 12*a*d)*x^2*e - 8*a*d^2 - (3*b*x^4*e^2 + 12*b*d*x^2*e + 8*b*d^2)*log(c) - (3*
b*n*x^4*e^2 + 12*b*d*n*x^2*e + 8*b*d^2*n)*log(x))*sqrt(x^2*e + d))/(x^4*e^5 + 2*d*x^2*e^4 + d^2*e^3)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(5/2),x)

[Out]

Timed out

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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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^5/(x^2*e + d)^(5/2), 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 )}{{\left (e\,x^2+d\right )}^{5/2}} \,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)^(5/2),x)

[Out]

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

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