3.297 \(\int \frac {x^7 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=212 \[ \frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {16 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^4}-\frac {b d^2 n}{3 e^4 \sqrt {d+e x^2}}+\frac {8 b d n \sqrt {d+e x^2}}{3 e^4}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^4} \]

[Out]

-1/9*b*n*(e*x^2+d)^(3/2)/e^4-16/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^4+1/3*d^3*(a+b*ln(c*x^n))/e^4
/(e*x^2+d)^(3/2)+1/3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^4-1/3*b*d^2*n/e^4/(e*x^2+d)^(1/2)-3*d^2*(a+b*ln(c*x^n))
/e^4/(e*x^2+d)^(1/2)+8/3*b*d*n*(e*x^2+d)^(1/2)/e^4-3*d*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^4

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Rubi [A]  time = 0.32, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {266, 43, 2350, 12, 1799, 1619, 63, 208} \[ \frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {b d^2 n}{3 e^4 \sqrt {d+e x^2}}-\frac {16 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^4}+\frac {8 b d n \sqrt {d+e x^2}}{3 e^4}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^4} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

-(b*d^2*n)/(3*e^4*Sqrt[d + e*x^2]) + (8*b*d*n*Sqrt[d + e*x^2])/(3*e^4) - (b*n*(d + e*x^2)^(3/2))/(9*e^4) - (16
*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(3*e^4) + (d^3*(a + b*Log[c*x^n]))/(3*e^4*(d + e*x^2)^(3/2)) -
(3*d^2*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x^2]) - (3*d*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^4 + ((d + e*x^2)
^(3/2)*(a + b*Log[c*x^n]))/(3*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 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 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 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 1619

Int[((Px_)*((c_.) + (d_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[c + d*x],
(Px*(c + d*x)^(n + 1/2))/(a + b*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0] &
& GtQ[Expon[Px, x], 2]

Rule 1799

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 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^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-(b n) \int \frac {-16 d^3-24 d^2 e x^2-6 d e^2 x^4+e^3 x^6}{3 e^4 x \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {(b n) \int \frac {-16 d^3-24 d^2 e x^2-6 d e^2 x^4+e^3 x^6}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 e^4}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {(b n) \operatorname {Subst}\left (\int \frac {-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 e^4}\\ &=\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {(b n) \operatorname {Subst}\left (\int \left (-\frac {d^2 e}{(d+e x)^{3/2}}-\frac {7 d e}{\sqrt {d+e x}}-\frac {16 d^2}{x \sqrt {d+e x}}+\frac {e^2 x}{\sqrt {d+e x}}\right ) \, dx,x,x^2\right )}{6 e^4}\\ &=-\frac {b d^2 n}{3 e^4 \sqrt {d+e x^2}}+\frac {7 b d n \sqrt {d+e x^2}}{3 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {\left (8 b d^2 n\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{3 e^4}-\frac {(b n) \operatorname {Subst}\left (\int \frac {x}{\sqrt {d+e x}} \, dx,x,x^2\right )}{6 e^2}\\ &=-\frac {b d^2 n}{3 e^4 \sqrt {d+e x^2}}+\frac {7 b d n \sqrt {d+e x^2}}{3 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {\left (16 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^5}-\frac {(b n) \operatorname {Subst}\left (\int \left (-\frac {d}{e \sqrt {d+e x}}+\frac {\sqrt {d+e x}}{e}\right ) \, dx,x,x^2\right )}{6 e^2}\\ &=-\frac {b d^2 n}{3 e^4 \sqrt {d+e x^2}}+\frac {8 b d n \sqrt {d+e x^2}}{3 e^4}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^4}-\frac {16 b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \left (d+e x^2\right )^{3/2}}-\frac {3 d^2 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}-\frac {3 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 240, normalized size = 1.13 \[ \frac {-48 a d^3-72 a d^2 e x^2-18 a d e^2 x^4+3 a e^3 x^6-3 b \left (16 d^3+24 d^2 e x^2+6 d e^2 x^4-e^3 x^6\right ) \log \left (c x^n\right )-48 b d^{3/2} e n x^2 \sqrt {d+e x^2} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+48 b d^{3/2} n \log (x) \left (d+e x^2\right )^{3/2}-48 b d^{5/2} n \sqrt {d+e x^2} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+20 b d^3 n+42 b d^2 e n x^2+21 b d e^2 n x^4-b e^3 n x^6}{9 e^4 \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

(-48*a*d^3 + 20*b*d^3*n - 72*a*d^2*e*x^2 + 42*b*d^2*e*n*x^2 - 18*a*d*e^2*x^4 + 21*b*d*e^2*n*x^4 + 3*a*e^3*x^6
- b*e^3*n*x^6 + 48*b*d^(3/2)*n*(d + e*x^2)^(3/2)*Log[x] - 3*b*(16*d^3 + 24*d^2*e*x^2 + 6*d*e^2*x^4 - e^3*x^6)*
Log[c*x^n] - 48*b*d^(5/2)*n*Sqrt[d + e*x^2]*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] - 48*b*d^(3/2)*e*n*x^2*Sqrt[d + e
*x^2]*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(9*e^4*(d + e*x^2)^(3/2))

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fricas [A]  time = 0.54, size = 504, normalized size = 2.38 \[ \left [\frac {24 \, {\left (b d e^{2} n x^{4} + 2 \, b d^{2} e n x^{2} + b d^{3} n\right )} \sqrt {d} \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left ({\left (b e^{3} n - 3 \, a e^{3}\right )} x^{6} - 20 \, b d^{3} n - 3 \, {\left (7 \, b d e^{2} n - 6 \, a d e^{2}\right )} x^{4} + 48 \, a d^{3} - 6 \, {\left (7 \, b d^{2} e n - 12 \, a d^{2} e\right )} x^{2} - 3 \, {\left (b e^{3} x^{6} - 6 \, b d e^{2} x^{4} - 24 \, b d^{2} e x^{2} - 16 \, b d^{3}\right )} \log \relax (c) - 3 \, {\left (b e^{3} n x^{6} - 6 \, b d e^{2} n x^{4} - 24 \, b d^{2} e n x^{2} - 16 \, b d^{3} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{9 \, {\left (e^{6} x^{4} + 2 \, d e^{5} x^{2} + d^{2} e^{4}\right )}}, \frac {48 \, {\left (b d e^{2} n x^{4} + 2 \, b d^{2} e n x^{2} + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left ({\left (b e^{3} n - 3 \, a e^{3}\right )} x^{6} - 20 \, b d^{3} n - 3 \, {\left (7 \, b d e^{2} n - 6 \, a d e^{2}\right )} x^{4} + 48 \, a d^{3} - 6 \, {\left (7 \, b d^{2} e n - 12 \, a d^{2} e\right )} x^{2} - 3 \, {\left (b e^{3} x^{6} - 6 \, b d e^{2} x^{4} - 24 \, b d^{2} e x^{2} - 16 \, b d^{3}\right )} \log \relax (c) - 3 \, {\left (b e^{3} n x^{6} - 6 \, b d e^{2} n x^{4} - 24 \, b d^{2} e n x^{2} - 16 \, b d^{3} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{9 \, {\left (e^{6} x^{4} + 2 \, d e^{5} x^{2} + d^{2} e^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/9*(24*(b*d*e^2*n*x^4 + 2*b*d^2*e*n*x^2 + b*d^3*n)*sqrt(d)*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^
2) - ((b*e^3*n - 3*a*e^3)*x^6 - 20*b*d^3*n - 3*(7*b*d*e^2*n - 6*a*d*e^2)*x^4 + 48*a*d^3 - 6*(7*b*d^2*e*n - 12*
a*d^2*e)*x^2 - 3*(b*e^3*x^6 - 6*b*d*e^2*x^4 - 24*b*d^2*e*x^2 - 16*b*d^3)*log(c) - 3*(b*e^3*n*x^6 - 6*b*d*e^2*n
*x^4 - 24*b*d^2*e*n*x^2 - 16*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/(e^6*x^4 + 2*d*e^5*x^2 + d^2*e^4), 1/9*(48*(b*d
*e^2*n*x^4 + 2*b*d^2*e*n*x^2 + b*d^3*n)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - ((b*e^3*n - 3*a*e^3)*x^6 -
20*b*d^3*n - 3*(7*b*d*e^2*n - 6*a*d*e^2)*x^4 + 48*a*d^3 - 6*(7*b*d^2*e*n - 12*a*d^2*e)*x^2 - 3*(b*e^3*x^6 - 6
*b*d*e^2*x^4 - 24*b*d^2*e*x^2 - 16*b*d^3)*log(c) - 3*(b*e^3*n*x^6 - 6*b*d*e^2*n*x^4 - 24*b*d^2*e*n*x^2 - 16*b*
d^3*n)*log(x))*sqrt(e*x^2 + d))/(e^6*x^4 + 2*d*e^5*x^2 + d^2*e^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{7}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^7/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{7}}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b*ln(c*x^n)+a)/(e*x^2+d)^(5/2),x)

[Out]

int(x^7*(b*ln(c*x^n)+a)/(e*x^2+d)^(5/2),x)

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maxima [A]  time = 1.61, size = 246, normalized size = 1.16 \[ \frac {1}{9} \, b n {\left (\frac {24 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{e^{4}} - \frac {3 \, d^{2}}{\sqrt {e x^{2} + d} e^{4}} - \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} - 24 \, \sqrt {e x^{2} + d} d}{e^{4}}\right )} + \frac {1}{3} \, {\left (\frac {x^{6}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} - \frac {6 \, d x^{4}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} - \frac {24 \, d^{2} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{3}} - \frac {16 \, d^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{4}}\right )} b \log \left (c x^{n}\right ) + \frac {1}{3} \, {\left (\frac {x^{6}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} - \frac {6 \, d x^{4}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{2}} - \frac {24 \, d^{2} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{3}} - \frac {16 \, d^{3}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e^{4}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/9*b*n*(24*d^(3/2)*log((sqrt(e*x^2 + d) - sqrt(d))/(sqrt(e*x^2 + d) + sqrt(d)))/e^4 - 3*d^2/(sqrt(e*x^2 + d)*
e^4) - ((e*x^2 + d)^(3/2) - 24*sqrt(e*x^2 + d)*d)/e^4) + 1/3*(x^6/((e*x^2 + d)^(3/2)*e) - 6*d*x^4/((e*x^2 + d)
^(3/2)*e^2) - 24*d^2*x^2/((e*x^2 + d)^(3/2)*e^3) - 16*d^3/((e*x^2 + d)^(3/2)*e^4))*b*log(c*x^n) + 1/3*(x^6/((e
*x^2 + d)^(3/2)*e) - 6*d*x^4/((e*x^2 + d)^(3/2)*e^2) - 24*d^2*x^2/((e*x^2 + d)^(3/2)*e^3) - 16*d^3/((e*x^2 + d
)^(3/2)*e^4))*a

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^7\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2),x)

[Out]

int((x^7*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(a+b*ln(c*x**n))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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