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

Optimal. Leaf size=109 \[ -\frac{a+b \log \left (c \left (d+e \sqrt{x}\right )^n\right )}{2 x^2}-\frac{b e^3 n}{2 d^3 \sqrt{x}}+\frac{b e^2 n}{4 d^2 x}+\frac{b e^4 n \log \left (d+e \sqrt{x}\right )}{2 d^4}-\frac{b e^4 n \log (x)}{4 d^4}-\frac{b e n}{6 d x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0742414, antiderivative size = 109, 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, 44} \[ -\frac{a+b \log \left (c \left (d+e \sqrt{x}\right )^n\right )}{2 x^2}-\frac{b e^3 n}{2 d^3 \sqrt{x}}+\frac{b e^2 n}{4 d^2 x}+\frac{b e^4 n \log \left (d+e \sqrt{x}\right )}{2 d^4}-\frac{b e^4 n \log (x)}{4 d^4}-\frac{b e n}{6 d x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0365354, size = 104, normalized size = 0.95 \[ -\frac{a}{2 x^2}-\frac{b \log \left (c \left (d+e \sqrt{x}\right )^n\right )}{2 x^2}+\frac{1}{2} b e n \left (-\frac{e^2}{d^3 \sqrt{x}}+\frac{e^3 \log \left (d+e \sqrt{x}\right )}{d^4}-\frac{e^3 \log (x)}{2 d^4}+\frac{e}{2 d^2 x}-\frac{1}{3 d x^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-a/(2*x^2) - (b*Log[c*(d + e*Sqrt[x])^n])/(2*x^2) + (b*e*n*(-1/(3*d*x^(3/2)) + e/(2*d^2*x) - e^2/(d^3*Sqrt[x])
+ (e^3*Log[d + e*Sqrt[x]])/d^4 - (e^3*Log[x])/(2*d^4)))/2

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Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( a+b\ln \left ( c \left ( d+e\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 = 1.03812, size = 113, normalized size = 1.04 \begin{align*} \frac{1}{12} \, b e n{\left (\frac{6 \, e^{3} \log \left (e \sqrt{x} + d\right )}{d^{4}} - \frac{3 \, e^{3} \log \left (x\right )}{d^{4}} - \frac{6 \, e^{2} x - 3 \, d e \sqrt{x} + 2 \, d^{2}}{d^{3} x^{\frac{3}{2}}}\right )} - \frac{b \log \left ({\left (e \sqrt{x} + d\right )}^{n} c\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]

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

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Fricas [A]  time = 2.13139, size = 240, normalized size = 2.2 \begin{align*} -\frac{6 \, b e^{4} n x^{2} \log \left (\sqrt{x}\right ) - 3 \, b d^{2} e^{2} n x + 6 \, b d^{4} \log \left (c\right ) + 6 \, a d^{4} - 6 \,{\left (b e^{4} n x^{2} - b d^{4} n\right )} \log \left (e \sqrt{x} + d\right ) + 2 \,{\left (3 \, b d e^{3} n x + b d^{3} e n\right )} \sqrt{x}}{12 \, d^{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]

-1/12*(6*b*e^4*n*x^2*log(sqrt(x)) - 3*b*d^2*e^2*n*x + 6*b*d^4*log(c) + 6*a*d^4 - 6*(b*e^4*n*x^2 - b*d^4*n)*log
(e*sqrt(x) + d) + 2*(3*b*d*e^3*n*x + b*d^3*e*n)*sqrt(x))/(d^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 [B]  time = 1.28607, size = 494, normalized size = 4.53 \begin{align*} \frac{{\left (6 \,{\left (\sqrt{x} e + d\right )}^{4} b n e^{5} \log \left (\sqrt{x} e + d\right ) - 24 \,{\left (\sqrt{x} e + d\right )}^{3} b d n e^{5} \log \left (\sqrt{x} e + d\right ) + 36 \,{\left (\sqrt{x} e + d\right )}^{2} b d^{2} n e^{5} \log \left (\sqrt{x} e + d\right ) - 24 \,{\left (\sqrt{x} e + d\right )} b d^{3} n e^{5} \log \left (\sqrt{x} e + d\right ) - 6 \,{\left (\sqrt{x} e + d\right )}^{4} b n e^{5} \log \left (\sqrt{x} e\right ) + 24 \,{\left (\sqrt{x} e + d\right )}^{3} b d n e^{5} \log \left (\sqrt{x} e\right ) - 36 \,{\left (\sqrt{x} e + d\right )}^{2} b d^{2} n e^{5} \log \left (\sqrt{x} e\right ) + 24 \,{\left (\sqrt{x} e + d\right )} b d^{3} n e^{5} \log \left (\sqrt{x} e\right ) - 6 \, b d^{4} n e^{5} \log \left (\sqrt{x} e\right ) - 6 \,{\left (\sqrt{x} e + d\right )}^{3} b d n e^{5} + 21 \,{\left (\sqrt{x} e + d\right )}^{2} b d^{2} n e^{5} - 26 \,{\left (\sqrt{x} e + d\right )} b d^{3} n e^{5} + 11 \, b d^{4} n e^{5} - 6 \, b d^{4} e^{5} \log \left (c\right ) - 6 \, a d^{4} e^{5}\right )} e^{\left (-1\right )}}{12 \,{\left ({\left (\sqrt{x} e + d\right )}^{4} d^{4} - 4 \,{\left (\sqrt{x} e + d\right )}^{3} d^{5} + 6 \,{\left (\sqrt{x} e + d\right )}^{2} d^{6} - 4 \,{\left (\sqrt{x} e + d\right )} d^{7} + d^{8}\right )}} \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]

1/12*(6*(sqrt(x)*e + d)^4*b*n*e^5*log(sqrt(x)*e + d) - 24*(sqrt(x)*e + d)^3*b*d*n*e^5*log(sqrt(x)*e + d) + 36*
(sqrt(x)*e + d)^2*b*d^2*n*e^5*log(sqrt(x)*e + d) - 24*(sqrt(x)*e + d)*b*d^3*n*e^5*log(sqrt(x)*e + d) - 6*(sqrt
(x)*e + d)^4*b*n*e^5*log(sqrt(x)*e) + 24*(sqrt(x)*e + d)^3*b*d*n*e^5*log(sqrt(x)*e) - 36*(sqrt(x)*e + d)^2*b*d
^2*n*e^5*log(sqrt(x)*e) + 24*(sqrt(x)*e + d)*b*d^3*n*e^5*log(sqrt(x)*e) - 6*b*d^4*n*e^5*log(sqrt(x)*e) - 6*(sq
rt(x)*e + d)^3*b*d*n*e^5 + 21*(sqrt(x)*e + d)^2*b*d^2*n*e^5 - 26*(sqrt(x)*e + d)*b*d^3*n*e^5 + 11*b*d^4*n*e^5
- 6*b*d^4*e^5*log(c) - 6*a*d^4*e^5)*e^(-1)/((sqrt(x)*e + d)^4*d^4 - 4*(sqrt(x)*e + d)^3*d^5 + 6*(sqrt(x)*e + d
)^2*d^6 - 4*(sqrt(x)*e + d)*d^7 + d^8)