3.427 \(\int \frac{a+b \log (c (d+\frac{e}{\sqrt{x}})^n)}{x^3} \, dx\)

Optimal. Leaf size=104 \[ -\frac{a+b \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^n\right )}{2 x^2}-\frac{b d^3 n}{2 e^3 \sqrt{x}}+\frac{b d^2 n}{4 e^2 x}+\frac{b d^4 n \log \left (d+\frac{e}{\sqrt{x}}\right )}{2 e^4}-\frac{b d n}{6 e x^{3/2}}+\frac{b n}{8 x^2} \]

[Out]

(b*n)/(8*x^2) - (b*d*n)/(6*e*x^(3/2)) + (b*d^2*n)/(4*e^2*x) - (b*d^3*n)/(2*e^3*Sqrt[x]) + (b*d^4*n*Log[d + e/S
qrt[x]])/(2*e^4) - (a + b*Log[c*(d + e/Sqrt[x])^n])/(2*x^2)

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Rubi [A]  time = 0.0757875, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2454, 2395, 43} \[ -\frac{a+b \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^n\right )}{2 x^2}-\frac{b d^3 n}{2 e^3 \sqrt{x}}+\frac{b d^2 n}{4 e^2 x}+\frac{b d^4 n \log \left (d+\frac{e}{\sqrt{x}}\right )}{2 e^4}-\frac{b d n}{6 e x^{3/2}}+\frac{b n}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n)/(8*x^2) - (b*d*n)/(6*e*x^(3/2)) + (b*d^2*n)/(4*e^2*x) - (b*d^3*n)/(2*e^3*Sqrt[x]) + (b*d^4*n*Log[d + e/S
qrt[x]])/(2*e^4) - (a + b*Log[c*(d + e/Sqrt[x])^n])/(2*x^2)

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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])

Rubi steps

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

Mathematica [A]  time = 0.0672797, size = 109, normalized size = 1.05 \[ -\frac{a}{2 x^2}-\frac{b \log \left (c \left (d+\frac{e}{\sqrt{x}}\right )^n\right )}{2 x^2}-\frac{b d^3 n}{2 e^3 \sqrt{x}}+\frac{b d^2 n}{4 e^2 x}+\frac{b d^4 n \log \left (d+\frac{e}{\sqrt{x}}\right )}{2 e^4}-\frac{b d n}{6 e x^{3/2}}+\frac{b n}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(2*x^2) + (b*n)/(8*x^2) - (b*d*n)/(6*e*x^(3/2)) + (b*d^2*n)/(4*e^2*x) - (b*d^3*n)/(2*e^3*Sqrt[x]) + (b*d^4*
n*Log[d + e/Sqrt[x]])/(2*e^4) - (b*Log[c*(d + e/Sqrt[x])^n])/(2*x^2)

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Maple [F]  time = 0.33, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( a+b\ln \left ( c \left ( d+{e{\frac{1}{\sqrt{x}}}} \right ) ^{n} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d+e/x^(1/2))^n))/x^3,x)

[Out]

int((a+b*ln(c*(d+e/x^(1/2))^n))/x^3,x)

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Maxima [A]  time = 1.04361, size = 128, normalized size = 1.23 \begin{align*} \frac{1}{24} \, b e n{\left (\frac{12 \, d^{4} \log \left (d \sqrt{x} + e\right )}{e^{5}} - \frac{6 \, d^{4} \log \left (x\right )}{e^{5}} - \frac{12 \, d^{3} x^{\frac{3}{2}} - 6 \, d^{2} e x + 4 \, d e^{2} \sqrt{x} - 3 \, e^{3}}{e^{4} x^{2}}\right )} - \frac{b \log \left (c{\left (d + \frac{e}{\sqrt{x}}\right )}^{n}\right )}{2 \, x^{2}} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^3,x, algorithm="maxima")

[Out]

1/24*b*e*n*(12*d^4*log(d*sqrt(x) + e)/e^5 - 6*d^4*log(x)/e^5 - (12*d^3*x^(3/2) - 6*d^2*e*x + 4*d*e^2*sqrt(x) -
3*e^3)/(e^4*x^2)) - 1/2*b*log(c*(d + e/sqrt(x))^n)/x^2 - 1/2*a/x^2

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Fricas [A]  time = 1.91319, size = 228, normalized size = 2.19 \begin{align*} \frac{6 \, b d^{2} e^{2} n x + 3 \, b e^{4} n - 12 \, b e^{4} \log \left (c\right ) - 12 \, a e^{4} + 12 \,{\left (b d^{4} n x^{2} - b e^{4} n\right )} \log \left (\frac{d x + e \sqrt{x}}{x}\right ) - 4 \,{\left (3 \, b d^{3} e n x + b d e^{3} n\right )} \sqrt{x}}{24 \, e^{4} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^3,x, algorithm="fricas")

[Out]

1/24*(6*b*d^2*e^2*n*x + 3*b*e^4*n - 12*b*e^4*log(c) - 12*a*e^4 + 12*(b*d^4*n*x^2 - b*e^4*n)*log((d*x + e*sqrt(
x))/x) - 4*(3*b*d^3*e*n*x + b*d*e^3*n)*sqrt(x))/(e^4*x^2)

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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*(d+e/x**(1/2))**n))/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.41518, size = 157, normalized size = 1.51 \begin{align*} \frac{{\left (12 \, b d^{4} n x^{2} \log \left (d \sqrt{x} + e\right ) - 12 \, b d^{4} n x^{2} \log \left (\sqrt{x}\right ) - 12 \, b d^{3} n x^{\frac{3}{2}} e + 6 \, b d^{2} n x e^{2} - 4 \, b d n \sqrt{x} e^{3} - 12 \, b n e^{4} \log \left (d \sqrt{x} + e\right ) + 12 \, b n e^{4} \log \left (\sqrt{x}\right ) + 3 \, b n e^{4} - 12 \, b e^{4} \log \left (c\right ) - 12 \, a e^{4}\right )} e^{\left (-4\right )}}{24 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(1/2))^n))/x^3,x, algorithm="giac")

[Out]

1/24*(12*b*d^4*n*x^2*log(d*sqrt(x) + e) - 12*b*d^4*n*x^2*log(sqrt(x)) - 12*b*d^3*n*x^(3/2)*e + 6*b*d^2*n*x*e^2
- 4*b*d*n*sqrt(x)*e^3 - 12*b*n*e^4*log(d*sqrt(x) + e) + 12*b*n*e^4*log(sqrt(x)) + 3*b*n*e^4 - 12*b*e^4*log(c)
- 12*a*e^4)*e^(-4)/x^2