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

Optimal. Leaf size=166 \[ -\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{2 b e n x}{3 d^3 \sqrt{d+e x^2}}-\frac{b n}{d^2 x \sqrt{d+e x^2}}+\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^3} \]

[Out]

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

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Rubi [A]  time = 0.168664, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {271, 192, 191, 2350, 12, 1265, 385, 217, 206} \[ -\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{2 b e n x}{3 d^3 \sqrt{d+e x^2}}-\frac{b n}{d^2 x \sqrt{d+e x^2}}+\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 192

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 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 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rule 385

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

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^2 \left (d+e x^2\right )^{5/2}} \, dx &=-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}-(b n) \int \frac{-3 d^2-12 d e x^2-8 e^2 x^4}{3 d^3 x^2 \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}-\frac{(b n) \int \frac{-3 d^2-12 d e x^2-8 e^2 x^4}{x^2 \left (d+e x^2\right )^{3/2}} \, dx}{3 d^3}\\ &=-\frac{b n}{d^2 x \sqrt{d+e x^2}}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}+\frac{(b n) \int \frac{6 d^2 e+8 d e^2 x^2}{\left (d+e x^2\right )^{3/2}} \, dx}{3 d^4}\\ &=-\frac{b n}{d^2 x \sqrt{d+e x^2}}-\frac{2 b e n x}{3 d^3 \sqrt{d+e x^2}}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}+\frac{(8 b e n) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{3 d^3}\\ &=-\frac{b n}{d^2 x \sqrt{d+e x^2}}-\frac{2 b e n x}{3 d^3 \sqrt{d+e x^2}}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}+\frac{(8 b e n) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{3 d^3}\\ &=-\frac{b n}{d^2 x \sqrt{d+e x^2}}-\frac{2 b e n x}{3 d^3 \sqrt{d+e x^2}}+\frac{8 b \sqrt{e} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^3}-\frac{a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac{4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac{8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt{d+e x^2}}\\ \end{align*}

Mathematica [A]  time = 0.182835, size = 144, normalized size = 0.87 \[ \frac{-3 a d^2-12 a d e x^2-8 a e^2 x^4-b \left (3 d^2+12 d e x^2+8 e^2 x^4\right ) \log \left (c x^n\right )-3 b d^2 n-5 b d e n x^2+8 b \sqrt{e} n x \left (d+e x^2\right )^{3/2} \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )-2 b e^2 n x^4}{3 d^3 x \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*d^2 - 3*b*d^2*n - 12*a*d*e*x^2 - 5*b*d*e*n*x^2 - 8*a*e^2*x^4 - 2*b*e^2*n*x^4 - b*(3*d^2 + 12*d*e*x^2 + 8
*e^2*x^4)*Log[c*x^n] + 8*b*Sqrt[e]*n*x*(d + e*x^2)^(3/2)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(3*d^3*x*(d + e*x
^2)^(3/2))

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Maple [F]  time = 0.412, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{2}} \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78833, size = 914, normalized size = 5.51 \begin{align*} \left [\frac{4 \,{\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt{e} \log \left (-2 \, e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) -{\left (2 \,{\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} +{\left (5 \, b d e n + 12 \, a d e\right )} x^{2} +{\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) +{\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}, -\frac{8 \,{\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt{-e} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) +{\left (2 \,{\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} +{\left (5 \, b d e n + 12 \, a d e\right )} x^{2} +{\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) +{\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(4*(b*e^2*n*x^5 + 2*b*d*e*n*x^3 + b*d^2*n*x)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2
*(b*e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3*
b*d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*x
^3 + d^5*x), -1/3*(8*(b*e^2*n*x^5 + 2*b*d*e*n*x^3 + b*d^2*n*x)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (
2*(b*e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3
*b*d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*
x^3 + d^5*x)]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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