\(\int \frac {a+b \log (c (d+e \sqrt {x})^n)}{x^4} \, dx\) [407]

Optimal result

Integrand size = 22, antiderivative size = 141 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=-\frac {b e n}{15 d x^{5/2}}+\frac {b e^2 n}{12 d^2 x^2}-\frac {b e^3 n}{9 d^3 x^{3/2}}+\frac {b e^4 n}{6 d^4 x}-\frac {b e^5 n}{3 d^5 \sqrt {x}}+\frac {b e^6 n \log \left (d+e \sqrt {x}\right )}{3 d^6}-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}-\frac {b e^6 n \log (x)}{6 d^6} \]

[Out]

-1/15*b*e*n/d/x^(5/2)+1/12*b*e^2*n/d^2/x^2-1/9*b*e^3*n/d^3/x^(3/2)+1/6*b*e^4*n/d^4/x-1/6*b*e^6*n*ln(x)/d^6+1/3
*b*e^6*n*ln(d+e*x^(1/2))/d^6+1/3*(-a-b*ln(c*(d+e*x^(1/2))^n))/x^3-1/3*b*e^5*n/d^5/x^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 141, 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, 46} \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}+\frac {b e^6 n \log \left (d+e \sqrt {x}\right )}{3 d^6}-\frac {b e^6 n \log (x)}{6 d^6}-\frac {b e^5 n}{3 d^5 \sqrt {x}}+\frac {b e^4 n}{6 d^4 x}-\frac {b e^3 n}{9 d^3 x^{3/2}}+\frac {b e^2 n}{12 d^2 x^2}-\frac {b e n}{15 d x^{5/2}} \]

[In]

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

[Out]

-1/15*(b*e*n)/(d*x^(5/2)) + (b*e^2*n)/(12*d^2*x^2) - (b*e^3*n)/(9*d^3*x^(3/2)) + (b*e^4*n)/(6*d^4*x) - (b*e^5*
n)/(3*d^5*Sqrt[x]) + (b*e^6*n*Log[d + e*Sqrt[x]])/(3*d^6) - (a + b*Log[c*(d + e*Sqrt[x])^n])/(3*x^3) - (b*e^6*
n*Log[x])/(6*d^6)

Rule 46

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] && Lt
Q[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}& = 2 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^7} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}+\frac {1}{3} (b e n) \text {Subst}\left (\int \frac {1}{x^6 (d+e x)} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}+\frac {1}{3} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^6}-\frac {e}{d^2 x^5}+\frac {e^2}{d^3 x^4}-\frac {e^3}{d^4 x^3}+\frac {e^4}{d^5 x^2}-\frac {e^5}{d^6 x}+\frac {e^6}{d^6 (d+e x)}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {b e n}{15 d x^{5/2}}+\frac {b e^2 n}{12 d^2 x^2}-\frac {b e^3 n}{9 d^3 x^{3/2}}+\frac {b e^4 n}{6 d^4 x}-\frac {b e^5 n}{3 d^5 \sqrt {x}}+\frac {b e^6 n \log \left (d+e \sqrt {x}\right )}{3 d^6}-\frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}-\frac {b e^6 n \log (x)}{6 d^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{3 x^3}+\frac {1}{6} b e n \left (-\frac {2}{5 d x^{5/2}}+\frac {e}{2 d^2 x^2}-\frac {2 e^2}{3 d^3 x^{3/2}}+\frac {e^3}{d^4 x}-\frac {2 e^4}{d^5 \sqrt {x}}+\frac {2 e^5 \log \left (d+e \sqrt {x}\right )}{d^6}-\frac {e^5 \log (x)}{d^6}\right ) \]

[In]

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

[Out]

-1/3*a/x^3 - (b*Log[c*(d + e*Sqrt[x])^n])/(3*x^3) + (b*e*n*(-2/(5*d*x^(5/2)) + e/(2*d^2*x^2) - (2*e^2)/(3*d^3*
x^(3/2)) + e^3/(d^4*x) - (2*e^4)/(d^5*Sqrt[x]) + (2*e^5*Log[d + e*Sqrt[x]])/d^6 - (e^5*Log[x])/d^6))/6

Maple [F]

\[\int \frac {a +b \ln \left (c \left (d +e \sqrt {x}\right )^{n}\right )}{x^{4}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=-\frac {60 \, b e^{6} n x^{3} \log \left (\sqrt {x}\right ) - 30 \, b d^{2} e^{4} n x^{2} - 15 \, b d^{4} e^{2} n x + 60 \, b d^{6} \log \left (c\right ) + 60 \, a d^{6} - 60 \, {\left (b e^{6} n x^{3} - b d^{6} n\right )} \log \left (e \sqrt {x} + d\right ) + 4 \, {\left (15 \, b d e^{5} n x^{2} + 5 \, b d^{3} e^{3} n x + 3 \, b d^{5} e n\right )} \sqrt {x}}{180 \, d^{6} x^{3}} \]

[In]

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

[Out]

-1/180*(60*b*e^6*n*x^3*log(sqrt(x)) - 30*b*d^2*e^4*n*x^2 - 15*b*d^4*e^2*n*x + 60*b*d^6*log(c) + 60*a*d^6 - 60*
(b*e^6*n*x^3 - b*d^6*n)*log(e*sqrt(x) + d) + 4*(15*b*d*e^5*n*x^2 + 5*b*d^3*e^3*n*x + 3*b*d^5*e*n)*sqrt(x))/(d^
6*x^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.75 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=\frac {1}{180} \, b e n {\left (\frac {60 \, e^{5} \log \left (e \sqrt {x} + d\right )}{d^{6}} - \frac {30 \, e^{5} \log \left (x\right )}{d^{6}} - \frac {60 \, e^{4} x^{2} - 30 \, d e^{3} x^{\frac {3}{2}} + 20 \, d^{2} e^{2} x - 15 \, d^{3} e \sqrt {x} + 12 \, d^{4}}{d^{5} x^{\frac {5}{2}}}\right )} - \frac {b \log \left ({\left (e \sqrt {x} + d\right )}^{n} c\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \]

[In]

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

[Out]

1/180*b*e*n*(60*e^5*log(e*sqrt(x) + d)/d^6 - 30*e^5*log(x)/d^6 - (60*e^4*x^2 - 30*d*e^3*x^(3/2) + 20*d^2*e^2*x
- 15*d^3*e*sqrt(x) + 12*d^4)/(d^5*x^(5/2))) - 1/3*b*log((e*sqrt(x) + d)^n*c)/x^3 - 1/3*a/x^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (115) = 230\).

Time = 0.41 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.43 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=-\frac {\frac {60 \, b e^{7} n \log \left (e \sqrt {x} + d\right )}{{\left (e \sqrt {x} + d\right )}^{6} - 6 \, {\left (e \sqrt {x} + d\right )}^{5} d + 15 \, {\left (e \sqrt {x} + d\right )}^{4} d^{2} - 20 \, {\left (e \sqrt {x} + d\right )}^{3} d^{3} + 15 \, {\left (e \sqrt {x} + d\right )}^{2} d^{4} - 6 \, {\left (e \sqrt {x} + d\right )} d^{5} + d^{6}} - \frac {60 \, b e^{7} n \log \left (e \sqrt {x} + d\right )}{d^{6}} + \frac {60 \, b e^{7} n \log \left (e \sqrt {x}\right )}{d^{6}} + \frac {60 \, {\left (e \sqrt {x} + d\right )}^{5} b e^{7} n - 330 \, {\left (e \sqrt {x} + d\right )}^{4} b d e^{7} n + 740 \, {\left (e \sqrt {x} + d\right )}^{3} b d^{2} e^{7} n - 855 \, {\left (e \sqrt {x} + d\right )}^{2} b d^{3} e^{7} n + 522 \, {\left (e \sqrt {x} + d\right )} b d^{4} e^{7} n - 137 \, b d^{5} e^{7} n + 60 \, b d^{5} e^{7} \log \left (c\right ) + 60 \, a d^{5} e^{7}}{{\left (e \sqrt {x} + d\right )}^{6} d^{5} - 6 \, {\left (e \sqrt {x} + d\right )}^{5} d^{6} + 15 \, {\left (e \sqrt {x} + d\right )}^{4} d^{7} - 20 \, {\left (e \sqrt {x} + d\right )}^{3} d^{8} + 15 \, {\left (e \sqrt {x} + d\right )}^{2} d^{9} - 6 \, {\left (e \sqrt {x} + d\right )} d^{10} + d^{11}}}{180 \, e} \]

[In]

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

[Out]

-1/180*(60*b*e^7*n*log(e*sqrt(x) + d)/((e*sqrt(x) + d)^6 - 6*(e*sqrt(x) + d)^5*d + 15*(e*sqrt(x) + d)^4*d^2 -
20*(e*sqrt(x) + d)^3*d^3 + 15*(e*sqrt(x) + d)^2*d^4 - 6*(e*sqrt(x) + d)*d^5 + d^6) - 60*b*e^7*n*log(e*sqrt(x)
+ d)/d^6 + 60*b*e^7*n*log(e*sqrt(x))/d^6 + (60*(e*sqrt(x) + d)^5*b*e^7*n - 330*(e*sqrt(x) + d)^4*b*d*e^7*n + 7
40*(e*sqrt(x) + d)^3*b*d^2*e^7*n - 855*(e*sqrt(x) + d)^2*b*d^3*e^7*n + 522*(e*sqrt(x) + d)*b*d^4*e^7*n - 137*b
*d^5*e^7*n + 60*b*d^5*e^7*log(c) + 60*a*d^5*e^7)/((e*sqrt(x) + d)^6*d^5 - 6*(e*sqrt(x) + d)^5*d^6 + 15*(e*sqrt
(x) + d)^4*d^7 - 20*(e*sqrt(x) + d)^3*d^8 + 15*(e*sqrt(x) + d)^2*d^9 - 6*(e*sqrt(x) + d)*d^10 + d^11))/e

Mupad [B] (verification not implemented)

Time = 1.88 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x^4} \, dx=\frac {2\,b\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e\,\sqrt {x}}{d}+1\right )}{3\,d^6}-\frac {\frac {b\,e\,n}{5\,d}+\frac {b\,e^3\,n\,x}{3\,d^3}-\frac {b\,e^2\,n\,\sqrt {x}}{4\,d^2}+\frac {b\,e^5\,n\,x^2}{d^5}-\frac {b\,e^4\,n\,x^{3/2}}{2\,d^4}}{3\,x^{5/2}}-\frac {b\,\ln \left (c\,{\left (d+e\,\sqrt {x}\right )}^n\right )}{3\,x^3}-\frac {a}{3\,x^3} \]

[In]

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

[Out]

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