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

Optimal. Leaf size=65 \[ -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b d n}{e \sqrt {x}}+\frac {b n}{2 x} \]

[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]  time = 0.05, antiderivative size = 65, normalized size of antiderivative = 1.00, 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} \[ -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b d n}{e \sqrt {x}}+\frac {b n}{2 x} \]

Antiderivative was successfully verified.

[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 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])

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 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])

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x^2} \, dx &=-\left (2 \operatorname {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) \operatorname {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) \operatorname {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]  time = 0.03, size = 68, normalized size = 1.05 \[ -\frac {a}{x}-\frac {b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x}+\frac {b d^2 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{e^2}-\frac {b d n}{e \sqrt {x}}+\frac {b n}{2 x} \]

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 70, normalized size = 1.08 \[ -\frac {2 \, b d e n \sqrt {x} - b e^{2} n + 2 \, b e^{2} \log \relax (c) + 2 \, a e^{2} - 2 \, {\left (b d^{2} n x - b e^{2} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right )}{2 \, e^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [B]  time = 0.22, size = 162, normalized size = 2.49 \[ \frac {1}{2} \, {\left (\frac {4 \, {\left (d \sqrt {x} + e\right )} b d n \log \left (\frac {d \sqrt {x} + e}{\sqrt {x}}\right )}{\sqrt {x}} - \frac {2 \, {\left (d \sqrt {x} + e\right )}^{2} b n \log \left (\frac {d \sqrt {x} + e}{\sqrt {x}}\right )}{x} - \frac {4 \, {\left (d \sqrt {x} + e\right )} b d n}{\sqrt {x}} + \frac {4 \, {\left (d \sqrt {x} + e\right )} b d \log \relax (c)}{\sqrt {x}} + \frac {{\left (d \sqrt {x} + e\right )}^{2} b n}{x} - \frac {2 \, {\left (d \sqrt {x} + e\right )}^{2} b \log \relax (c)}{x} + \frac {4 \, {\left (d \sqrt {x} + e\right )} a d}{\sqrt {x}} - \frac {2 \, {\left (d \sqrt {x} + e\right )}^{2} a}{x}\right )} e^{\left (-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 63, normalized size = 0.97 \[ \frac {b \,d^{2} n \ln \left (d +\frac {e}{\sqrt {x}}\right )}{e^{2}}-\frac {b d n}{e \sqrt {x}}+\frac {b n}{2 x}-\frac {b \ln \left (c \,{\mathrm e}^{n \ln \left (d +\frac {e}{\sqrt {x}}\right )}\right )}{x}-\frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.81, size = 75, normalized size = 1.15 \[ \frac {1}{2} \, b e n {\left (\frac {2 \, d^{2} \log \left (d \sqrt {x} + e\right )}{e^{3}} - \frac {d^{2} \log \relax (x)}{e^{3}} - \frac {2 \, d \sqrt {x} - e}{e^{2} x}\right )} - \frac {b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )}{x} - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 0.43, size = 60, normalized size = 0.92 \[ \frac {b\,n}{2\,x}-\frac {a}{x}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{x}-\frac {b\,d\,n}{e\,\sqrt {x}}+\frac {b\,d^2\,n\,\ln \left (d+\frac {e}{\sqrt {x}}\right )}{e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________