\(\int x^3 (a+b \log (c (d+\frac {e}{\sqrt {x}})^n)) \, dx\) [421]

Optimal result

Integrand size = 22, antiderivative size = 171 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {b e^7 n \sqrt {x}}{4 d^7}-\frac {b e^6 n x}{8 d^6}+\frac {b e^5 n x^{3/2}}{12 d^5}-\frac {b e^4 n x^2}{16 d^4}+\frac {b e^3 n x^{5/2}}{20 d^3}-\frac {b e^2 n x^3}{24 d^2}+\frac {b e n x^{7/2}}{28 d}-\frac {b e^8 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{4 d^8}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^8 n \log (x)}{8 d^8} \]

[Out]

-1/8*b*e^6*n*x/d^6+1/12*b*e^5*n*x^(3/2)/d^5-1/16*b*e^4*n*x^2/d^4+1/20*b*e^3*n*x^(5/2)/d^3-1/24*b*e^2*n*x^3/d^2
+1/28*b*e*n*x^(7/2)/d-1/8*b*e^8*n*ln(x)/d^8-1/4*b*e^8*n*ln(d+e/x^(1/2))/d^8+1/4*x^4*(a+b*ln(c*(d+e/x^(1/2))^n)
)+1/4*b*e^7*n*x^(1/2)/d^7

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 171, 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} \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^8 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{4 d^8}-\frac {b e^8 n \log (x)}{8 d^8}+\frac {b e^7 n \sqrt {x}}{4 d^7}-\frac {b e^6 n x}{8 d^6}+\frac {b e^5 n x^{3/2}}{12 d^5}-\frac {b e^4 n x^2}{16 d^4}+\frac {b e^3 n x^{5/2}}{20 d^3}-\frac {b e^2 n x^3}{24 d^2}+\frac {b e n x^{7/2}}{28 d} \]

[In]

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

[Out]

(b*e^7*n*Sqrt[x])/(4*d^7) - (b*e^6*n*x)/(8*d^6) + (b*e^5*n*x^(3/2))/(12*d^5) - (b*e^4*n*x^2)/(16*d^4) + (b*e^3
*n*x^(5/2))/(20*d^3) - (b*e^2*n*x^3)/(24*d^2) + (b*e*n*x^(7/2))/(28*d) - (b*e^8*n*Log[d + e/Sqrt[x]])/(4*d^8)
+ (x^4*(a + b*Log[c*(d + e/Sqrt[x])^n]))/4 - (b*e^8*n*Log[x])/(8*d^8)

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}& = -\left (2 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^9} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \frac {1}{x^8 (d+e x)} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \left (\frac {1}{d x^8}-\frac {e}{d^2 x^7}+\frac {e^2}{d^3 x^6}-\frac {e^3}{d^4 x^5}+\frac {e^4}{d^5 x^4}-\frac {e^5}{d^6 x^3}+\frac {e^6}{d^7 x^2}-\frac {e^7}{d^8 x}+\frac {e^8}{d^8 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = \frac {b e^7 n \sqrt {x}}{4 d^7}-\frac {b e^6 n x}{8 d^6}+\frac {b e^5 n x^{3/2}}{12 d^5}-\frac {b e^4 n x^2}{16 d^4}+\frac {b e^3 n x^{5/2}}{20 d^3}-\frac {b e^2 n x^3}{24 d^2}+\frac {b e n x^{7/2}}{28 d}-\frac {b e^8 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{4 d^8}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^8 n \log (x)}{8 d^8} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.91 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}+\frac {1}{4} b x^4 \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )+\frac {1}{8} b e n \left (\frac {2 e^6 \sqrt {x}}{d^7}-\frac {e^5 x}{d^6}+\frac {2 e^4 x^{3/2}}{3 d^5}-\frac {e^3 x^2}{2 d^4}+\frac {2 e^2 x^{5/2}}{5 d^3}-\frac {e x^3}{3 d^2}+\frac {2 x^{7/2}}{7 d}-\frac {2 e^7 \log \left (d+\frac {e}{\sqrt {x}}\right )}{d^8}-\frac {e^7 \log (x)}{d^8}\right ) \]

[In]

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

[Out]

(a*x^4)/4 + (b*x^4*Log[c*(d + e/Sqrt[x])^n])/4 + (b*e*n*((2*e^6*Sqrt[x])/d^7 - (e^5*x)/d^6 + (2*e^4*x^(3/2))/(
3*d^5) - (e^3*x^2)/(2*d^4) + (2*e^2*x^(5/2))/(5*d^3) - (e*x^3)/(3*d^2) + (2*x^(7/2))/(7*d) - (2*e^7*Log[d + e/
Sqrt[x]])/d^8 - (e^7*Log[x])/d^8))/8

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {420 \, b d^{8} x^{4} \log \left (c\right ) - 70 \, b d^{6} e^{2} n x^{3} + 420 \, a d^{8} x^{4} - 105 \, b d^{4} e^{4} n x^{2} - 210 \, b d^{2} e^{6} n x - 420 \, b d^{8} n \log \left (\sqrt {x}\right ) + 420 \, {\left (b d^{8} - b e^{8}\right )} n \log \left (d \sqrt {x} + e\right ) + 420 \, {\left (b d^{8} n x^{4} - b d^{8} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right ) + 4 \, {\left (15 \, b d^{7} e n x^{3} + 21 \, b d^{5} e^{3} n x^{2} + 35 \, b d^{3} e^{5} n x + 105 \, b d e^{7} n\right )} \sqrt {x}}{1680 \, d^{8}} \]

[In]

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

[Out]

1/1680*(420*b*d^8*x^4*log(c) - 70*b*d^6*e^2*n*x^3 + 420*a*d^8*x^4 - 105*b*d^4*e^4*n*x^2 - 210*b*d^2*e^6*n*x -
420*b*d^8*n*log(sqrt(x)) + 420*(b*d^8 - b*e^8)*n*log(d*sqrt(x) + e) + 420*(b*d^8*n*x^4 - b*d^8*n)*log((d*x + e
*sqrt(x))/x) + 4*(15*b*d^7*e*n*x^3 + 21*b*d^5*e^3*n*x^2 + 35*b*d^3*e^5*n*x + 105*b*d*e^7*n)*sqrt(x))/d^8

Sympy [A] (verification not implemented)

Time = 58.18 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.84 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {a x^{4}}{4} + b \left (\frac {e n \left (\frac {2 x^{\frac {7}{2}}}{7 d} - \frac {e x^{3}}{3 d^{2}} + \frac {2 e^{2} x^{\frac {5}{2}}}{5 d^{3}} - \frac {e^{3} x^{2}}{2 d^{4}} + \frac {2 e^{4} x^{\frac {3}{2}}}{3 d^{5}} - \frac {e^{5} x}{d^{6}} - \frac {2 e^{7} \left (\begin {cases} \frac {\sqrt {x}}{e} & \text {for}\: d = 0 \\\frac {\log {\left (d \sqrt {x} + e \right )}}{d} & \text {otherwise} \end {cases}\right )}{d^{7}} + \frac {2 e^{6} \sqrt {x}}{d^{7}}\right )}{8} + \frac {x^{4} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{4}\right ) \]

[In]

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

[Out]

a*x**4/4 + b*(e*n*(2*x**(7/2)/(7*d) - e*x**3/(3*d**2) + 2*e**2*x**(5/2)/(5*d**3) - e**3*x**2/(2*d**4) + 2*e**4
*x**(3/2)/(3*d**5) - e**5*x/d**6 - 2*e**7*Piecewise((sqrt(x)/e, Eq(d, 0)), (log(d*sqrt(x) + e)/d, True))/d**7
+ 2*e**6*sqrt(x)/d**7)/8 + x**4*log(c*(d + e/sqrt(x))**n)/4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{1680} \, b e n {\left (\frac {420 \, e^{7} \log \left (d \sqrt {x} + e\right )}{d^{8}} - \frac {60 \, d^{6} x^{\frac {7}{2}} - 70 \, d^{5} e x^{3} + 84 \, d^{4} e^{2} x^{\frac {5}{2}} - 105 \, d^{3} e^{3} x^{2} + 140 \, d^{2} e^{4} x^{\frac {3}{2}} - 210 \, d e^{5} x + 420 \, e^{6} \sqrt {x}}{d^{7}}\right )} \]

[In]

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

[Out]

1/4*b*x^4*log(c*(d + e/sqrt(x))^n) + 1/4*a*x^4 - 1/1680*b*e*n*(420*e^7*log(d*sqrt(x) + e)/d^8 - (60*d^6*x^(7/2
) - 70*d^5*e*x^3 + 84*d^4*e^2*x^(5/2) - 105*d^3*e^3*x^2 + 140*d^2*e^4*x^(3/2) - 210*d*e^5*x + 420*e^6*sqrt(x))
/d^7)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.56 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left (c\right ) + \frac {1}{4} \, a x^{4} - \frac {{\left (e^{9} {\left (\frac {420 \, \log \left (\frac {{\left | d \sqrt {x} + e \right |}}{\sqrt {{\left | x \right |}}}\right )}{d^{8}} - \frac {420 \, \log \left ({\left | -d + \frac {d \sqrt {x} + e}{\sqrt {x}} \right |}\right )}{d^{8}} + \frac {1089 \, d^{7} - \frac {4683 \, {\left (d \sqrt {x} + e\right )} d^{6}}{\sqrt {x}} + \frac {9639 \, {\left (d \sqrt {x} + e\right )}^{2} d^{5}}{x} - \frac {11165 \, {\left (d \sqrt {x} + e\right )}^{3} d^{4}}{x^{\frac {3}{2}}} + \frac {7490 \, {\left (d \sqrt {x} + e\right )}^{4} d^{3}}{x^{2}} - \frac {2730 \, {\left (d \sqrt {x} + e\right )}^{5} d^{2}}{x^{\frac {5}{2}}} + \frac {420 \, {\left (d \sqrt {x} + e\right )}^{6} d}{x^{3}}}{{\left (d - \frac {d \sqrt {x} + e}{\sqrt {x}}\right )}^{7} d^{8}}\right )} - \frac {420 \, e^{9} \log \left (-{\left (e - \frac {d}{\frac {d}{e} - \frac {d \sqrt {x} + e}{e \sqrt {x}}}\right )} {\left (\frac {d}{e} - \frac {d \sqrt {x} + e}{e \sqrt {x}}\right )}\right )}{{\left (d - \frac {d \sqrt {x} + e}{\sqrt {x}}\right )}^{8}}\right )} b n}{1680 \, e} \]

[In]

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

[Out]

1/4*b*x^4*log(c) + 1/4*a*x^4 - 1/1680*(e^9*(420*log(abs(d*sqrt(x) + e)/sqrt(abs(x)))/d^8 - 420*log(abs(-d + (d
*sqrt(x) + e)/sqrt(x)))/d^8 + (1089*d^7 - 4683*(d*sqrt(x) + e)*d^6/sqrt(x) + 9639*(d*sqrt(x) + e)^2*d^5/x - 11
165*(d*sqrt(x) + e)^3*d^4/x^(3/2) + 7490*(d*sqrt(x) + e)^4*d^3/x^2 - 2730*(d*sqrt(x) + e)^5*d^2/x^(5/2) + 420*
(d*sqrt(x) + e)^6*d/x^3)/((d - (d*sqrt(x) + e)/sqrt(x))^7*d^8)) - 420*e^9*log(-(e - d/(d/e - (d*sqrt(x) + e)/(
e*sqrt(x))))*(d/e - (d*sqrt(x) + e)/(e*sqrt(x))))/(d - (d*sqrt(x) + e)/sqrt(x))^8)*b*n/e

Mupad [B] (verification not implemented)

Time = 2.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.82 \[ \int x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx=\frac {\frac {b\,d\,e^7\,n\,\sqrt {x}}{4}-\frac {b\,d^2\,e^6\,n\,x}{8}+\frac {b\,d^7\,e\,n\,x^{7/2}}{28}-\frac {b\,d^4\,e^4\,n\,x^2}{16}-\frac {b\,d^6\,e^2\,n\,x^3}{24}+\frac {b\,d^3\,e^5\,n\,x^{3/2}}{12}+\frac {b\,d^5\,e^3\,n\,x^{5/2}}{20}+\frac {b\,e^8\,n\,\mathrm {atan}\left (\frac {d\,1{}\mathrm {i}+\frac {e\,2{}\mathrm {i}}{\sqrt {x}}}{d}\right )\,1{}\mathrm {i}}{2}}{d^8}+\frac {a\,x^4}{4}+\frac {b\,x^4\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )}{4} \]

[In]

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

[Out]

((b*e^8*n*atan((d*1i + (e*2i)/x^(1/2))/d)*1i)/2 - (b*d^2*e^6*n*x)/8 + (b*d*e^7*n*x^(1/2))/4 + (b*d^7*e*n*x^(7/
2))/28 - (b*d^4*e^4*n*x^2)/16 - (b*d^6*e^2*n*x^3)/24 + (b*d^3*e^5*n*x^(3/2))/12 + (b*d^5*e^3*n*x^(5/2))/20)/d^
8 + (a*x^4)/4 + (b*x^4*log(c*(d + e/x^(1/2))^n))/4