\(\int \frac {a+b \log (c (d+\frac {e}{\sqrt {x}})^n)}{x^3} \, dx\) [427]

Optimal result

Integrand size = 22, antiderivative size = 104 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\frac {b n}{8 x^2}-\frac {b d n}{6 e x^{3/2}}+\frac {b d^2 n}{4 e^2 x}-\frac {b d^3 n}{2 e^3 \sqrt {x}}+\frac {b d^4 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{2 e^4}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x^2} \]

[Out]

1/8*b*n/x^2-1/6*b*d*n/e/x^(3/2)+1/4*b*d^2*n/e^2/x+1/2*b*d^4*n*ln(d+e/x^(1/2))/e^4+1/2*(-a-b*ln(c*(d+e/x^(1/2))
^n))/x^2-1/2*b*d^3*n/e^3/x^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2442, 45} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x^2}+\frac {b d^4 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{2 e^4}-\frac {b d^3 n}{2 e^3 \sqrt {x}}+\frac {b d^2 n}{4 e^2 x}-\frac {b d n}{6 e x^{3/2}}+\frac {b n}{8 x^2} \]

[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 45

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])

Rule 2442

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 2504

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])

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {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) \text {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) \text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=-\frac {a}{2 x^2}-\frac {1}{4} b e n \left (-\frac {1}{2 e x^2}+\frac {2 d}{3 e^2 x^{3/2}}-\frac {d^2}{e^3 x}+\frac {2 d^3}{e^4 \sqrt {x}}-\frac {2 d^4 \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^5}\right )-\frac {b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{2 x^2} \]

[In]

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

[Out]

-1/2*a/x^2 - (b*e*n*(-1/2*1/(e*x^2) + (2*d)/(3*e^2*x^(3/2)) - d^2/(e^3*x) + (2*d^3)/(e^4*Sqrt[x]) - (2*d^4*Log
[d + e/Sqrt[x]])/e^5))/4 - (b*Log[c*(d + e/Sqrt[x])^n])/(2*x^2)

Maple [F]

\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{n}\right )}{x^{3}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\frac {6 \, b d^{2} e^{2} n x + 3 \, b e^{4} n - 12 \, b e^{4} \log \left (c\right ) - 12 \, a e^{4} + 12 \, {\left (b d^{4} n x^{2} - b e^{4} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right ) - 4 \, {\left (3 \, b d^{3} e n x + b d e^{3} n\right )} \sqrt {x}}{24 \, e^{4} x^{2}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\frac {1}{24} \, b e n {\left (\frac {12 \, d^{4} \log \left (d \sqrt {x} + e\right )}{e^{5}} - \frac {6 \, d^{4} \log \left (x\right )}{e^{5}} - \frac {12 \, d^{3} x^{\frac {3}{2}} - 6 \, d^{2} e x + 4 \, d e^{2} \sqrt {x} - 3 \, e^{3}}{e^{4} x^{2}}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (84) = 168\).

Time = 0.32 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.28 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\frac {12 \, {\left (\frac {4 \, {\left (d \sqrt {x} + e\right )} b d^{3} n}{e^{3} \sqrt {x}} - \frac {6 \, {\left (d \sqrt {x} + e\right )}^{2} b d^{2} n}{e^{3} x} + \frac {4 \, {\left (d \sqrt {x} + e\right )}^{3} b d n}{e^{3} x^{\frac {3}{2}}} - \frac {{\left (d \sqrt {x} + e\right )}^{4} b n}{e^{3} x^{2}}\right )} \log \left (\frac {d \sqrt {x} + e}{\sqrt {x}}\right ) + \frac {3 \, {\left (b n - 4 \, b \log \left (c\right ) - 4 \, a\right )} {\left (d \sqrt {x} + e\right )}^{4}}{e^{3} x^{2}} - \frac {16 \, {\left (b d n - 3 \, b d \log \left (c\right ) - 3 \, a d\right )} {\left (d \sqrt {x} + e\right )}^{3}}{e^{3} x^{\frac {3}{2}}} + \frac {36 \, {\left (b d^{2} n - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2}\right )} {\left (d \sqrt {x} + e\right )}^{2}}{e^{3} x} - \frac {48 \, {\left (b d^{3} n - b d^{3} \log \left (c\right ) - a d^{3}\right )} {\left (d \sqrt {x} + e\right )}}{e^{3} \sqrt {x}}}{24 \, e} \]

[In]

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

[Out]

1/24*(12*(4*(d*sqrt(x) + e)*b*d^3*n/(e^3*sqrt(x)) - 6*(d*sqrt(x) + e)^2*b*d^2*n/(e^3*x) + 4*(d*sqrt(x) + e)^3*
b*d*n/(e^3*x^(3/2)) - (d*sqrt(x) + e)^4*b*n/(e^3*x^2))*log((d*sqrt(x) + e)/sqrt(x)) + 3*(b*n - 4*b*log(c) - 4*
a)*(d*sqrt(x) + e)^4/(e^3*x^2) - 16*(b*d*n - 3*b*d*log(c) - 3*a*d)*(d*sqrt(x) + e)^3/(e^3*x^(3/2)) + 36*(b*d^2
*n - 2*b*d^2*log(c) - 2*a*d^2)*(d*sqrt(x) + e)^2/(e^3*x) - 48*(b*d^3*n - b*d^3*log(c) - a*d^3)*(d*sqrt(x) + e)
/(e^3*sqrt(x)))/e

Mupad [B] (verification not implemented)

Time = 1.61 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.84 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^3} \, dx=\frac {b\,n}{8\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{6\,e\,x^{3/2}}+\frac {b\,d^4\,n\,\ln \left (d+\frac {e}{\sqrt {x}}\right )}{2\,e^4}+\frac {b\,d^2\,n}{4\,e^2\,x}-\frac {b\,d^3\,n}{2\,e^3\,\sqrt {x}} \]

[In]

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

[Out]

(b*n)/(8*x^2) - a/(2*x^2) - (b*log(c*(d + e/x^(1/2))^n))/(2*x^2) - (b*d*n)/(6*e*x^(3/2)) + (b*d^4*n*log(d + e/
x^(1/2)))/(2*e^4) + (b*d^2*n)/(4*e^2*x) - (b*d^3*n)/(2*e^3*x^(1/2))