3.273 \(\int \frac{(d+e x^2)^{3/2} (a+b \log (c x^n))}{x^8} \, dx\)

Optimal. Leaf size=196 \[ \frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 b e^3 n \sqrt{d+e x^2}}{35 d^2 x}+\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}-\frac{2 b e^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{35 d^2}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7} \]

[Out]

(2*b*e^3*n*Sqrt[d + e*x^2])/(35*d^2*x) + (2*b*e^2*n*(d + e*x^2)^(3/2))/(105*d^2*x^3) + (2*b*e*n*(d + e*x^2)^(5
/2))/(175*d^2*x^5) - (b*n*(d + e*x^2)^(7/2))/(49*d^2*x^7) - (2*b*e^(7/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]
])/(35*d^2) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(7*d*x^7) + (2*e*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(3
5*d^2*x^5)

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Rubi [A]  time = 0.171883, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {271, 264, 2350, 12, 451, 277, 217, 206} \[ \frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 b e^3 n \sqrt{d+e x^2}}{35 d^2 x}+\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}-\frac{2 b e^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{35 d^2}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^8,x]

[Out]

(2*b*e^3*n*Sqrt[d + e*x^2])/(35*d^2*x) + (2*b*e^2*n*(d + e*x^2)^(3/2))/(105*d^2*x^3) + (2*b*e*n*(d + e*x^2)^(5
/2))/(175*d^2*x^5) - (b*n*(d + e*x^2)^(7/2))/(49*d^2*x^7) - (2*b*e^(7/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]
])/(35*d^2) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(7*d*x^7) + (2*e*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(3
5*d^2*x^5)

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^8} \, dx &=-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-(b n) \int \frac{\left (d+e x^2\right )^{5/2} \left (-5 d+2 e x^2\right )}{35 d^2 x^8} \, dx\\ &=-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{(b n) \int \frac{\left (d+e x^2\right )^{5/2} \left (-5 d+2 e x^2\right )}{x^8} \, dx}{35 d^2}\\ &=-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{(2 b e n) \int \frac{\left (d+e x^2\right )^{5/2}}{x^6} \, dx}{35 d^2}\\ &=\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (2 b e^2 n\right ) \int \frac{\left (d+e x^2\right )^{3/2}}{x^4} \, dx}{35 d^2}\\ &=\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (2 b e^3 n\right ) \int \frac{\sqrt{d+e x^2}}{x^2} \, dx}{35 d^2}\\ &=\frac{2 b e^3 n \sqrt{d+e x^2}}{35 d^2 x}+\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (2 b e^4 n\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{35 d^2}\\ &=\frac{2 b e^3 n \sqrt{d+e x^2}}{35 d^2 x}+\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}-\frac{\left (2 b e^4 n\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{35 d^2}\\ &=\frac{2 b e^3 n \sqrt{d+e x^2}}{35 d^2 x}+\frac{2 b e^2 n \left (d+e x^2\right )^{3/2}}{105 d^2 x^3}+\frac{2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5}-\frac{b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}-\frac{2 b e^{7/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{35 d^2}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{7 d x^7}+\frac{2 e \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{35 d^2 x^5}\\ \end{align*}

Mathematica [A]  time = 0.222432, size = 145, normalized size = 0.74 \[ -\frac{\sqrt{d+e x^2} \left (105 a \left (5 d-2 e x^2\right ) \left (d+e x^2\right )^2+b n \left (183 d^2 e x^2+75 d^3+71 d e^2 x^4-247 e^3 x^6\right )\right )+105 b \left (5 d-2 e x^2\right ) \left (d+e x^2\right )^{5/2} \log \left (c x^n\right )+210 b e^{7/2} n x^7 \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{3675 d^2 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^8,x]

[Out]

-(Sqrt[d + e*x^2]*(105*a*(5*d - 2*e*x^2)*(d + e*x^2)^2 + b*n*(75*d^3 + 183*d^2*e*x^2 + 71*d*e^2*x^4 - 247*e^3*
x^6)) + 105*b*(5*d - 2*e*x^2)*(d + e*x^2)^(5/2)*Log[c*x^n] + 210*b*e^(7/2)*n*x^7*Log[e*x + Sqrt[e]*Sqrt[d + e*
x^2]])/(3675*d^2*x^7)

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Maple [F]  time = 0.499, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{8}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^8,x)

[Out]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^8,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((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^8,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84193, size = 1013, normalized size = 5.17 \begin{align*} \left [\frac{105 \, b e^{\frac{7}{2}} n x^{7} \log \left (-2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) +{\left ({\left (247 \, b e^{3} n + 210 \, a e^{3}\right )} x^{6} - 75 \, b d^{3} n -{\left (71 \, b d e^{2} n + 105 \, a d e^{2}\right )} x^{4} - 525 \, a d^{3} - 3 \,{\left (61 \, b d^{2} e n + 280 \, a d^{2} e\right )} x^{2} + 105 \,{\left (2 \, b e^{3} x^{6} - b d e^{2} x^{4} - 8 \, b d^{2} e x^{2} - 5 \, b d^{3}\right )} \log \left (c\right ) + 105 \,{\left (2 \, b e^{3} n x^{6} - b d e^{2} n x^{4} - 8 \, b d^{2} e n x^{2} - 5 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3675 \, d^{2} x^{7}}, \frac{210 \, b \sqrt{-e} e^{3} n x^{7} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) +{\left ({\left (247 \, b e^{3} n + 210 \, a e^{3}\right )} x^{6} - 75 \, b d^{3} n -{\left (71 \, b d e^{2} n + 105 \, a d e^{2}\right )} x^{4} - 525 \, a d^{3} - 3 \,{\left (61 \, b d^{2} e n + 280 \, a d^{2} e\right )} x^{2} + 105 \,{\left (2 \, b e^{3} x^{6} - b d e^{2} x^{4} - 8 \, b d^{2} e x^{2} - 5 \, b d^{3}\right )} \log \left (c\right ) + 105 \,{\left (2 \, b e^{3} n x^{6} - b d e^{2} n x^{4} - 8 \, b d^{2} e n x^{2} - 5 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3675 \, d^{2} x^{7}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^8,x, algorithm="fricas")

[Out]

[1/3675*(105*b*e^(7/2)*n*x^7*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + ((247*b*e^3*n + 210*a*e^3)*x^6
- 75*b*d^3*n - (71*b*d*e^2*n + 105*a*d*e^2)*x^4 - 525*a*d^3 - 3*(61*b*d^2*e*n + 280*a*d^2*e)*x^2 + 105*(2*b*e^
3*x^6 - b*d*e^2*x^4 - 8*b*d^2*e*x^2 - 5*b*d^3)*log(c) + 105*(2*b*e^3*n*x^6 - b*d*e^2*n*x^4 - 8*b*d^2*e*n*x^2 -
5*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^7), 1/3675*(210*b*sqrt(-e)*e^3*n*x^7*arctan(sqrt(-e)*x/sqrt(e*x^2
+ d)) + ((247*b*e^3*n + 210*a*e^3)*x^6 - 75*b*d^3*n - (71*b*d*e^2*n + 105*a*d*e^2)*x^4 - 525*a*d^3 - 3*(61*b*d
^2*e*n + 280*a*d^2*e)*x^2 + 105*(2*b*e^3*x^6 - b*d*e^2*x^4 - 8*b*d^2*e*x^2 - 5*b*d^3)*log(c) + 105*(2*b*e^3*n*
x^6 - b*d*e^2*n*x^4 - 8*b*d^2*e*n*x^2 - 5*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^7)]

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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((e*x**2+d)**(3/2)*(a+b*ln(c*x**n))/x**8,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^8,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*log(c*x^n) + a)/x^8, x)