3.284 \(\int \frac{a+b \log (c x^n)}{x^4 \sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{2 e \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}-\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^2}+\frac{2 b e n \sqrt{d+e x^2}}{3 d^2 x}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13297, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{2 e \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac{\sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}-\frac{2 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^2}+\frac{2 b e n \sqrt{d+e x^2}}{3 d^2 x}-\frac{b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

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

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

Mathematica [A]  time = 0.142161, size = 110, normalized size = 0.76 \[ \frac{\sqrt{d+e x^2} \left (-3 a d+6 a e x^2-b d n+5 b e n x^2\right )-3 b \left (d-2 e x^2\right ) \sqrt{d+e x^2} \log \left (c x^n\right )-6 b e^{3/2} n x^3 \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{9 d^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

(Sqrt[d + e*x^2]*(-3*a*d - b*d*n + 6*a*e*x^2 + 5*b*e*n*x^2) - 3*b*(d - 2*e*x^2)*Sqrt[d + e*x^2]*Log[c*x^n] - 6
*b*e^(3/2)*n*x^3*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(9*d^2*x^3)

________________________________________________________________________________________

Maple [F]  time = 0.434, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{4}}{\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^(1/2),x)

________________________________________________________________________________________

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((a+b*log(c*x^n))/x^4/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.50646, size = 540, normalized size = 3.75 \begin{align*} \left [\frac{3 \, b e^{\frac{3}{2}} n x^{3} \log \left (-2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) -{\left (b d n -{\left (5 \, b e n + 6 \, a e\right )} x^{2} + 3 \, a d - 3 \,{\left (2 \, b e x^{2} - b d\right )} \log \left (c\right ) - 3 \,{\left (2 \, b e n x^{2} - b d n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{9 \, d^{2} x^{3}}, \frac{6 \, b \sqrt{-e} e n x^{3} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) -{\left (b d n -{\left (5 \, b e n + 6 \, a e\right )} x^{2} + 3 \, a d - 3 \,{\left (2 \, b e x^{2} - b d\right )} \log \left (c\right ) - 3 \,{\left (2 \, b e n x^{2} - b d n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{9 \, d^{2} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/9*(3*b*e^(3/2)*n*x^3*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (b*d*n - (5*b*e*n + 6*a*e)*x^2 + 3*a
*d - 3*(2*b*e*x^2 - b*d)*log(c) - 3*(2*b*e*n*x^2 - b*d*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^3), 1/9*(6*b*sqrt(-e
)*e*n*x^3*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (b*d*n - (5*b*e*n + 6*a*e)*x^2 + 3*a*d - 3*(2*b*e*x^2 - b*d)*lo
g(c) - 3*(2*b*e*n*x^2 - b*d*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \log{\left (c x^{n} \right )}}{x^{4} \sqrt{d + e x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**4/(e*x**2+d)**(1/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*sqrt(d + e*x**2)), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{\sqrt{e x^{2} + d} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/(sqrt(e*x^2 + d)*x^4), x)