\(\int (a+b \log (c (d+\frac {e}{\sqrt {x}})^n)) \, dx\) [424]

Optimal result

Integrand size = 18, antiderivative size = 53 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {b e n \sqrt {x}}{d}+a x+b x \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-\frac {b e^2 n \log \left (e+d \sqrt {x}\right )}{d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2498, 269, 196, 45} \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=a x+b x \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-\frac {b e^2 n \log \left (d \sqrt {x}+e\right )}{d^2}+\frac {b e n \sqrt {x}}{d} \]

[In]

Int[a + b*Log[c*(d + e/Sqrt[x])^n],x]

[Out]

(b*e*n*Sqrt[x])/d + a*x + b*x*Log[c*(d + e/Sqrt[x])^n] - (b*e^2*n*Log[e + d*Sqrt[x]])/d^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 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=a x+b x \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )+\frac {1}{2} b e n \left (\frac {2 \sqrt {x}}{d}-\frac {2 e \log \left (d+\frac {e}{\sqrt {x}}\right )}{d^2}-\frac {e \log (x)}{d^2}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.62

method	result	size
default	\(a x +b \left (x \ln \left (c \left (\frac {e +d \sqrt {x}}{\sqrt {x}}\right )^{n}\right )+\frac {e n \left (\frac {2 \sqrt {x}}{d}-\frac {e \ln \left (e +d \sqrt {x}\right )}{d^{2}}+\frac {e \ln \left (d \sqrt {x}-e \right )}{d^{2}}-\frac {e \ln \left (d^{2} x -e^{2}\right )}{d^{2}}\right )}{2}\right )\)	\(86\)
parts	\(a x +b \left (x \ln \left (c \left (\frac {e +d \sqrt {x}}{\sqrt {x}}\right )^{n}\right )+\frac {e n \left (\frac {2 \sqrt {x}}{d}-\frac {e \ln \left (e +d \sqrt {x}\right )}{d^{2}}+\frac {e \ln \left (d \sqrt {x}-e \right )}{d^{2}}-\frac {e \ln \left (d^{2} x -e^{2}\right )}{d^{2}}\right )}{2}\right )\)	\(86\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.70 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {b d^{2} x \log \left (c\right ) - b d^{2} n \log \left (\sqrt {x}\right ) + b d e n \sqrt {x} + a d^{2} x + {\left (b d^{2} - b e^{2}\right )} n \log \left (d \sqrt {x} + e\right ) + {\left (b d^{2} n x - b d^{2} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right )}{d^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=a x + b \left (\frac {e n \left (- \frac {2 e \left (\begin {cases} \frac {\sqrt {x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt {x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d} + \frac {2 \sqrt {x}}{d}\right )}{2} + x \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}\right ) \]

[In]

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

[Out]

a*x + b*(e*n*(-2*e*Piecewise((sqrt(x)/e, Eq(d, 0)), (log(d*sqrt(x) + e)/d, True))/d + 2*sqrt(x)/d)/2 + x*log(c
*(d + e/sqrt(x))**n))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=-{\left (e n {\left (\frac {e \log \left (d \sqrt {x} + e\right )}{d^{2}} - \frac {\sqrt {x}}{d}\right )} - x \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right )\right )} b + a x \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=-{\left ({\left (e {\left (\frac {e \log \left ({\left | d \sqrt {x} + e \right |}\right )}{d^{2}} - \frac {\sqrt {x}}{d}\right )} - x \log \left (d + \frac {e}{\sqrt {x}}\right )\right )} n - x \log \left (c\right )\right )} b + a x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=a\,x+b\,x\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )-\frac {b\,e\,n\,\left (e\,\ln \left (e+d\,\sqrt {x}\right )-d\,\sqrt {x}\right )}{d^2} \]

[In]

int(a + b*log(c*(d + e/x^(1/2))^n),x)

[Out]

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