3.3.73 \(\int \frac {(d+e x^2)^{3/2} (a+b \log (c x^n))}{x^8} \, dx\) [273]

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

[Out]

2/105*b*e^2*n*(e*x^2+d)^(3/2)/d^2/x^3+2/175*b*e*n*(e*x^2+d)^(5/2)/d^2/x^5-1/49*b*n*(e*x^2+d)^(7/2)/d^2/x^7-2/3
5*b*e^(7/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^2-1/7*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/d/x^7+2/35*e*(e*x^2+d
)^(5/2)*(a+b*ln(c*x^n))/d^2/x^5+2/35*b*e^3*n*(e*x^2+d)^(1/2)/d^2/x

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Rubi [A]
time = 0.12, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {277, 270, 2392, 12, 462, 283, 223, 212} \begin {gather*} \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^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {e} 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 {b n \left (d+e x^2\right )^{7/2}}{49 d^2 x^7}+\frac {2 b e n \left (d+e x^2\right )^{5/2}}{175 d^2 x^5} \end {gather*}

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 12

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

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 223

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 270

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 277

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 283

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 462

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 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 {\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 ) \text {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*}

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Mathematica [A]
time = 0.15, size = 145, normalized size = 0.74 \begin {gather*} -\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 (75 d^3+183 d^2 e x^2+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 (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{3675 d^2 x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3675*(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 - 2
47*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]])/(d^2*x^7)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{8}}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

1/35*a*(2*(x^2*e + d)^(5/2)*e/(d^2*x^5) - 5*(x^2*e + d)^(5/2)/(d*x^7)) + b*integrate((x^2*e*log(c) + d*log(c)
+ (x^2*e + d)*log(x^n))*sqrt(x^2*e + d)/x^8, x)

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Fricas [A]
time = 0.43, size = 204, normalized size = 1.04 \begin {gather*} \frac {105 \, b n x^{7} e^{\frac {7}{2}} \log \left (-2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right ) + {\left ({\left (247 \, b n + 210 \, a\right )} x^{6} e^{3} - {\left (71 \, b d n + 105 \, a d\right )} x^{4} e^{2} - 75 \, b d^{3} n - 525 \, a d^{3} - 3 \, {\left (61 \, b d^{2} n + 280 \, a d^{2}\right )} x^{2} e + 105 \, {\left (2 \, b x^{6} e^{3} - b d x^{4} e^{2} - 8 \, b d^{2} x^{2} e - 5 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (2 \, b n x^{6} e^{3} - b d n x^{4} e^{2} - 8 \, b d^{2} n x^{2} e - 5 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{3675 \, d^{2} x^{7}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}}{x^{8}}\, dx \end {gather*}

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]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^8} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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