\(\int \frac {a+b \log (c (d+\frac {e}{\sqrt {x}})^n)}{x^2} \, dx\) [426]

Optimal result

Integrand size = 22, antiderivative size = 65 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {b n}{2 x}-\frac {b d n}{e \sqrt {x}}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 65, 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} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b d n}{e \sqrt {x}}+\frac {b n}{2 x} \]

[In]

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

[Out]

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

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 \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 )}{x}+(b e n) \text {Subst}\left (\int \frac {x^2}{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 )}{x}+(b e n) \text {Subst}\left (\int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = \frac {b n}{2 x}-\frac {b d n}{e \sqrt {x}}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}+\frac {b n}{2 x}-\frac {b d n}{e \sqrt {x}}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {a}{x}+\frac {b n}{2 x}-\frac {b \ln \left (c \,{\mathrm e}^{n \ln \left (d +\frac {e}{\sqrt {x}}\right )}\right )}{x}+\frac {b \,d^{2} n \ln \left (d +\frac {e}{\sqrt {x}}\right )}{e^{2}}-\frac {b d n}{e \sqrt {x}}\)	\(63\)
default	\(-\frac {a}{x}+\frac {b n}{2 x}-\frac {b \ln \left (c \,{\mathrm e}^{n \ln \left (d +\frac {e}{\sqrt {x}}\right )}\right )}{x}+\frac {b \,d^{2} n \ln \left (d +\frac {e}{\sqrt {x}}\right )}{e^{2}}-\frac {b d n}{e \sqrt {x}}\)	\(63\)

[In]

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

[Out]

-a/x+1/2*b*n/x-b/x*ln(c*exp(n*ln(d+e/x^(1/2))))+b*d^2*n*ln(d+e/x^(1/2))/e^2-b*d*n/e/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=-\frac {2 \, b d e n \sqrt {x} - b e^{2} n + 2 \, b e^{2} \log \left (c\right ) + 2 \, a e^{2} - 2 \, {\left (b d^{2} n x - b e^{2} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right )}{2 \, e^{2} x} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (58) = 116\).

Time = 137.10 (sec) , antiderivative size = 391, normalized size of antiderivative = 6.02 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\begin {cases} - \frac {a + b \log {\left (0^{n} c \right )}}{x} & \text {for}\: \left (d = 0 \vee d = - \frac {e}{\sqrt {x}}\right ) \wedge \left (d = - \frac {e}{\sqrt {x}} \vee e = 0\right ) \\- \frac {a + b \log {\left (c d^{n} \right )}}{x} & \text {for}\: e = 0 \\- \frac {2 a d e^{2} x^{3}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 a e^{3} x^{\frac {5}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {2 b d^{3} x^{4} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b d^{2} e n x^{\frac {7}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {2 b d^{2} e x^{\frac {7}{2}} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {b d e^{2} n x^{3}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b d e^{2} x^{3} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} + \frac {b e^{3} n x^{\frac {5}{2}}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} - \frac {2 b e^{3} x^{\frac {5}{2}} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{2 d e^{2} x^{4} + 2 e^{3} x^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-(a + b*log(0**n*c))/x, (Eq(d, 0) | Eq(d, -e/sqrt(x))) & (Eq(e, 0) | Eq(d, -e/sqrt(x)))), (-(a + b*
log(c*d**n))/x, Eq(e, 0)), (-2*a*d*e**2*x**3/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*a*e**3*x**(5/2)/(2*d*e**2*x
**4 + 2*e**3*x**(7/2)) + 2*b*d**3*x**4*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*b*d**2*
e*n*x**(7/2)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) + 2*b*d**2*e*x**(7/2)*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4
+ 2*e**3*x**(7/2)) - b*d*e**2*n*x**3/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*b*d*e**2*x**3*log(c*(d + e/sqrt(x))
**n)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) + b*e**3*n*x**(5/2)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)) - 2*b*e**3*x**(5/
2)*log(c*(d + e/sqrt(x))**n)/(2*d*e**2*x**4 + 2*e**3*x**(7/2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {1}{2} \, b e n {\left (\frac {2 \, d^{2} \log \left (d \sqrt {x} + e\right )}{e^{3}} - \frac {d^{2} \log \left (x\right )}{e^{3}} - \frac {2 \, d \sqrt {x} - e}{e^{2} x}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )}{x} - \frac {a}{x} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (57) = 114\).

Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.78 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (d \sqrt {x} + e\right )} b d n}{e \sqrt {x}} - \frac {{\left (d \sqrt {x} + e\right )}^{2} b n}{e x}\right )} \log \left (\frac {d \sqrt {x} + e}{\sqrt {x}}\right ) + \frac {{\left (b n - 2 \, b \log \left (c\right ) - 2 \, a\right )} {\left (d \sqrt {x} + e\right )}^{2}}{e x} - \frac {4 \, {\left (b d n - b d \log \left (c\right ) - a d\right )} {\left (d \sqrt {x} + e\right )}}{e \sqrt {x}}}{2 \, e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.65 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx=\frac {b\,n}{2\,x}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{x}-\frac {b\,d\,n}{e\,\sqrt {x}}+\frac {b\,d^2\,n\,\ln \left (d+\frac {e}{\sqrt {x}}\right )}{e^2} \]

[In]

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

[Out]

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