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

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

[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)

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Rubi [A]  time = 0.295893, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {266, 43, 2350, 12, 1799, 1620, 63, 208} \[ \frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt{d+e x^2}}+\frac{3 d^2 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac{11 b d^2 n \sqrt{d+e x^2}}{5 e^4}+\frac{16 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{5 e^4}+\frac{4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 e^4} \]

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 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 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 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])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1620

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 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]

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) \operatorname{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) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.208874, size = 195, normalized size = 0.93 \[ \frac{120 a d^2 e x^2+240 a d^3-30 a d e^2 x^4+15 a e^3 x^6+15 b \left (8 d^2 e x^2+16 d^3-2 d e^2 x^4+e^3 x^6\right ) \log \left (c x^n\right )-134 b d^2 e n x^2-240 b d^{5/2} n \log (x) \sqrt{d+e x^2}+240 b d^{5/2} n \sqrt{d+e x^2} \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-148 b d^3 n+11 b d e^2 n x^4-3 b e^3 n x^6}{75 e^4 \sqrt{d+e x^2}} \]

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 = 0.424, size = 0, normalized size = 0. \begin{align*} \int{{x}^{7} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.80726, size = 1062, normalized size = 5.08 \begin{align*} \left [\frac{120 \,{\left (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 (3 \,{\left (b e^{3} n - 5 \, a e^{3}\right )} x^{6} + 148 \, b d^{3} n -{\left (11 \, b d e^{2} n - 30 \, a d e^{2}\right )} x^{4} - 240 \, a d^{3} + 2 \,{\left (67 \, b d^{2} e n - 60 \, a d^{2} e\right )} x^{2} - 15 \,{\left (b e^{3} x^{6} - 2 \, b d e^{2} x^{4} + 8 \, b d^{2} e x^{2} + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \,{\left (b e^{3} n x^{6} - 2 \, b d e^{2} n x^{4} + 8 \, b d^{2} e n x^{2} + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{75 \,{\left (e^{5} x^{2} + d e^{4}\right )}}, -\frac{240 \,{\left (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 (3 \,{\left (b e^{3} n - 5 \, a e^{3}\right )} x^{6} + 148 \, b d^{3} n -{\left (11 \, b d e^{2} n - 30 \, a d e^{2}\right )} x^{4} - 240 \, a d^{3} + 2 \,{\left (67 \, b d^{2} e n - 60 \, a d^{2} e\right )} x^{2} - 15 \,{\left (b e^{3} x^{6} - 2 \, b d e^{2} x^{4} + 8 \, b d^{2} e x^{2} + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \,{\left (b e^{3} n x^{6} - 2 \, b d e^{2} n x^{4} + 8 \, b d^{2} e n x^{2} + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{75 \,{\left (e^{5} x^{2} + d e^{4}\right )}}\right ] \end{align*}

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{7}}{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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/(e*x^2 + d)^(3/2), x)