3.3.86 \(\int \frac {x^7 (a+b \log (c x^n))}{(d+e x^2)^{3/2}} \, dx\) [286]

Optimal. Leaf size=209 \[ -\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4}+\frac {16 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \]

[Out]

4/15*b*d*n*(e*x^2+d)^(3/2)/e^4-1/25*b*n*(e*x^2+d)^(5/2)/e^4+16/5*b*d^(5/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/
e^4-d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^4+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^4+d^3*(a+b*ln(c*x^n))/e^4/(e*x
^2+d)^(1/2)-11/5*b*d^2*n*(e*x^2+d)^(1/2)/e^4+3*d^2*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^4

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Rubi [A]
time = 0.21, antiderivative size = 209, 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, 1813, 1634, 65, 214} \begin {gather*} \frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {16 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e^4}-\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(a + b*Log[c*x^n]))/(d + e*x^2)^(3/2),x]

[Out]

(-11*b*d^2*n*Sqrt[d + e*x^2])/(5*e^4) + (4*b*d*n*(d + e*x^2)^(3/2))/(15*e^4) - (b*n*(d + e*x^2)^(5/2))/(25*e^4
) + (16*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(5*e^4) + (d^3*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x^2])
+ (3*d^2*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^4 - (d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/e^4 + ((d + e*x^2
)^(5/2)*(a + b*Log[c*x^n]))/(5*e^4)

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 65

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 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 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1813

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/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^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-(b n) \int \frac {16 d^3+8 d^2 e x^2-2 d e^2 x^4+e^3 x^6}{5 e^4 x \sqrt {d+e x^2}} \, dx\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {(b n) \int \frac {16 d^3+8 d^2 e x^2-2 d e^2 x^4+e^3 x^6}{x \sqrt {d+e x^2}} \, dx}{5 e^4}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {(b n) \text {Subst}\left (\int \frac {16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{10 e^4}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {(b n) \text {Subst}\left (\int \left (\frac {11 d^2 e}{\sqrt {d+e x}}+\frac {16 d^3}{x \sqrt {d+e x}}-4 d e \sqrt {d+e x}+e (d+e x)^{3/2}\right ) \, dx,x,x^2\right )}{10 e^4}\\ &=-\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {\left (8 b d^3 n\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{5 e^4}\\ &=-\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {\left (16 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 )}{5 e^5}\\ &=-\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4}+\frac {16 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 195, normalized size = 0.93 \begin {gather*} \frac {240 a d^3-148 b d^3 n+120 a d^2 e x^2-134 b d^2 e n x^2-30 a d e^2 x^4+11 b d e^2 n x^4+15 a e^3 x^6-3 b e^3 n x^6-240 b d^{5/2} n \sqrt {d+e x^2} \log (x)+15 b \left (16 d^3+8 d^2 e x^2-2 d e^2 x^4+e^3 x^6\right ) \log \left (c x^n\right )+240 b d^{5/2} n \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{75 e^4 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(a + b*Log[c*x^n]))/(d + e*x^2)^(3/2),x]

[Out]

(240*a*d^3 - 148*b*d^3*n + 120*a*d^2*e*x^2 - 134*b*d^2*e*n*x^2 - 30*a*d*e^2*x^4 + 11*b*d*e^2*n*x^4 + 15*a*e^3*
x^6 - 3*b*e^3*n*x^6 - 240*b*d^(5/2)*n*Sqrt[d + e*x^2]*Log[x] + 15*b*(16*d^3 + 8*d^2*e*x^2 - 2*d*e^2*x^4 + e^3*
x^6)*Log[c*x^n] + 240*b*d^(5/2)*n*Sqrt[d + e*x^2]*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(75*e^4*Sqrt[d + e*x^2])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{7} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a+b*ln(c*x^n))/(e*x^2+d)^(3/2),x)

[Out]

int(x^7*(a+b*ln(c*x^n))/(e*x^2+d)^(3/2),x)

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Maxima [A]
time = 0.50, size = 247, normalized size = 1.18 \begin {gather*} -\frac {1}{75} \, {\left (120 \, d^{\frac {5}{2}} e^{\left (-4\right )} \log \left (\frac {\sqrt {x^{2} e + d} - \sqrt {d}}{\sqrt {x^{2} e + d} + \sqrt {d}}\right ) + {\left (3 \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} - 20 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d + 165 \, \sqrt {x^{2} e + d} d^{2}\right )} e^{\left (-4\right )}\right )} b n + \frac {1}{5} \, {\left (\frac {x^{6} e^{\left (-1\right )}}{\sqrt {x^{2} e + d}} - \frac {2 \, d x^{4} e^{\left (-2\right )}}{\sqrt {x^{2} e + d}} + \frac {8 \, d^{2} x^{2} e^{\left (-3\right )}}{\sqrt {x^{2} e + d}} + \frac {16 \, d^{3} e^{\left (-4\right )}}{\sqrt {x^{2} e + d}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{5} \, {\left (\frac {x^{6} e^{\left (-1\right )}}{\sqrt {x^{2} e + d}} - \frac {2 \, d x^{4} e^{\left (-2\right )}}{\sqrt {x^{2} e + d}} + \frac {8 \, d^{2} x^{2} e^{\left (-3\right )}}{\sqrt {x^{2} e + d}} + \frac {16 \, d^{3} e^{\left (-4\right )}}{\sqrt {x^{2} e + d}}\right )} a \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-1/75*(120*d^(5/2)*e^(-4)*log((sqrt(x^2*e + d) - sqrt(d))/(sqrt(x^2*e + d) + sqrt(d))) + (3*(x^2*e + d)^(5/2)
- 20*(x^2*e + d)^(3/2)*d + 165*sqrt(x^2*e + d)*d^2)*e^(-4))*b*n + 1/5*(x^6*e^(-1)/sqrt(x^2*e + d) - 2*d*x^4*e^
(-2)/sqrt(x^2*e + d) + 8*d^2*x^2*e^(-3)/sqrt(x^2*e + d) + 16*d^3*e^(-4)/sqrt(x^2*e + d))*b*log(c*x^n) + 1/5*(x
^6*e^(-1)/sqrt(x^2*e + d) - 2*d*x^4*e^(-2)/sqrt(x^2*e + d) + 8*d^2*x^2*e^(-3)/sqrt(x^2*e + d) + 16*d^3*e^(-4)/
sqrt(x^2*e + d))*a

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Fricas [A]
time = 0.44, size = 444, normalized size = 2.12 \begin {gather*} \left [\frac {120 \, {\left (b d^{2} n x^{2} e + b d^{3} 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 - 5 \, a\right )} x^{6} e^{3} - {\left (11 \, b d n - 30 \, a d\right )} x^{4} e^{2} + 148 \, b d^{3} n - 240 \, a d^{3} + 2 \, {\left (67 \, b d^{2} n - 60 \, a d^{2}\right )} x^{2} e - 15 \, {\left (b x^{6} e^{3} - 2 \, b d x^{4} e^{2} + 8 \, b d^{2} x^{2} e + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{6} e^{3} - 2 \, b d n x^{4} e^{2} + 8 \, b d^{2} n x^{2} e + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{75 \, {\left (x^{2} e^{5} + d e^{4}\right )}}, -\frac {240 \, {\left (b d^{2} n x^{2} e + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (3 \, {\left (b n - 5 \, a\right )} x^{6} e^{3} - {\left (11 \, b d n - 30 \, a d\right )} x^{4} e^{2} + 148 \, b d^{3} n - 240 \, a d^{3} + 2 \, {\left (67 \, b d^{2} n - 60 \, a d^{2}\right )} x^{2} e - 15 \, {\left (b x^{6} e^{3} - 2 \, b d x^{4} e^{2} + 8 \, b d^{2} x^{2} e + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{6} e^{3} - 2 \, b d n x^{4} e^{2} + 8 \, b d^{2} n x^{2} e + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{75 \, {\left (x^{2} e^{5} + d e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/75*(120*(b*d^2*n*x^2*e + b*d^3*n)*sqrt(d)*log(-(x^2*e + 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) - (3*(b*n - 5
*a)*x^6*e^3 - (11*b*d*n - 30*a*d)*x^4*e^2 + 148*b*d^3*n - 240*a*d^3 + 2*(67*b*d^2*n - 60*a*d^2)*x^2*e - 15*(b*
x^6*e^3 - 2*b*d*x^4*e^2 + 8*b*d^2*x^2*e + 16*b*d^3)*log(c) - 15*(b*n*x^6*e^3 - 2*b*d*n*x^4*e^2 + 8*b*d^2*n*x^2
*e + 16*b*d^3*n)*log(x))*sqrt(x^2*e + d))/(x^2*e^5 + d*e^4), -1/75*(240*(b*d^2*n*x^2*e + b*d^3*n)*sqrt(-d)*arc
tan(sqrt(-d)/sqrt(x^2*e + d)) + (3*(b*n - 5*a)*x^6*e^3 - (11*b*d*n - 30*a*d)*x^4*e^2 + 148*b*d^3*n - 240*a*d^3
+ 2*(67*b*d^2*n - 60*a*d^2)*x^2*e - 15*(b*x^6*e^3 - 2*b*d*x^4*e^2 + 8*b*d^2*x^2*e + 16*b*d^3)*log(c) - 15*(b*
n*x^6*e^3 - 2*b*d*n*x^4*e^2 + 8*b*d^2*n*x^2*e + 16*b*d^3*n)*log(x))*sqrt(x^2*e + d))/(x^2*e^5 + d*e^4)]

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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**7*(a+b*ln(c*x**n))/(e*x**2+d)**(3/2),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^7*(a+b*log(c*x^n))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^7/(x^2*e + d)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^7\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(a + b*log(c*x^n)))/(d + e*x^2)^(3/2),x)

[Out]

int((x^7*(a + b*log(c*x^n)))/(d + e*x^2)^(3/2), x)

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