\(\int x^3 (a+b \log (c (d+e \sqrt {x})^n)) \, dx\) [400]

Optimal result

Integrand size = 22, antiderivative size = 166 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {b d^7 n \sqrt {x}}{4 e^7}-\frac {b d^6 n x}{8 e^6}+\frac {b d^5 n x^{3/2}}{12 e^5}-\frac {b d^4 n x^2}{16 e^4}+\frac {b d^3 n x^{5/2}}{20 e^3}-\frac {b d^2 n x^3}{24 e^2}+\frac {b d n x^{7/2}}{28 e}-\frac {1}{32} b n x^4-\frac {b d^8 n \log \left (d+e \sqrt {x}\right )}{4 e^8}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \]

[Out]

-1/8*b*d^6*n*x/e^6+1/12*b*d^5*n*x^(3/2)/e^5-1/16*b*d^4*n*x^2/e^4+1/20*b*d^3*n*x^(5/2)/e^3-1/24*b*d^2*n*x^3/e^2
+1/28*b*d*n*x^(7/2)/e-1/32*b*n*x^4-1/4*b*d^8*n*ln(d+e*x^(1/2))/e^8+1/4*x^4*(a+b*ln(c*(d+e*x^(1/2))^n))+1/4*b*d
^7*n*x^(1/2)/e^7

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 166, 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} \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {b d^8 n \log \left (d+e \sqrt {x}\right )}{4 e^8}+\frac {b d^7 n \sqrt {x}}{4 e^7}-\frac {b d^6 n x}{8 e^6}+\frac {b d^5 n x^{3/2}}{12 e^5}-\frac {b d^4 n x^2}{16 e^4}+\frac {b d^3 n x^{5/2}}{20 e^3}-\frac {b d^2 n x^3}{24 e^2}+\frac {b d n x^{7/2}}{28 e}-\frac {1}{32} b n x^4 \]

[In]

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

[Out]

(b*d^7*n*Sqrt[x])/(4*e^7) - (b*d^6*n*x)/(8*e^6) + (b*d^5*n*x^(3/2))/(12*e^5) - (b*d^4*n*x^2)/(16*e^4) + (b*d^3
*n*x^(5/2))/(20*e^3) - (b*d^2*n*x^3)/(24*e^2) + (b*d*n*x^(7/2))/(28*e) - (b*n*x^4)/32 - (b*d^8*n*Log[d + e*Sqr
t[x]])/(4*e^8) + (x^4*(a + b*Log[c*(d + e*Sqrt[x])^n]))/4

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}& = 2 \text {Subst}\left (\int x^7 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \frac {x^8}{d+e x} \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {1}{4} (b e n) \text {Subst}\left (\int \left (-\frac {d^7}{e^8}+\frac {d^6 x}{e^7}-\frac {d^5 x^2}{e^6}+\frac {d^4 x^3}{e^5}-\frac {d^3 x^4}{e^4}+\frac {d^2 x^5}{e^3}-\frac {d x^6}{e^2}+\frac {x^7}{e}+\frac {d^8}{e^8 (d+e x)}\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {b d^7 n \sqrt {x}}{4 e^7}-\frac {b d^6 n x}{8 e^6}+\frac {b d^5 n x^{3/2}}{12 e^5}-\frac {b d^4 n x^2}{16 e^4}+\frac {b d^3 n x^{5/2}}{20 e^3}-\frac {b d^2 n x^3}{24 e^2}+\frac {b d n x^{7/2}}{28 e}-\frac {1}{32} b n x^4-\frac {b d^8 n \log \left (d+e \sqrt {x}\right )}{4 e^8}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.95 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}-\frac {1}{8} b e n \left (-\frac {2 d^7 \sqrt {x}}{e^8}+\frac {d^6 x}{e^7}-\frac {2 d^5 x^{3/2}}{3 e^6}+\frac {d^4 x^2}{2 e^5}-\frac {2 d^3 x^{5/2}}{5 e^4}+\frac {d^2 x^3}{3 e^3}-\frac {2 d x^{7/2}}{7 e^2}+\frac {x^4}{4 e}+\frac {2 d^8 \log \left (d+e \sqrt {x}\right )}{e^9}\right )+\frac {1}{4} b x^4 \log \left (c \left (d+e \sqrt {x}\right )^n\right ) \]

[In]

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

[Out]

(a*x^4)/4 - (b*e*n*((-2*d^7*Sqrt[x])/e^8 + (d^6*x)/e^7 - (2*d^5*x^(3/2))/(3*e^6) + (d^4*x^2)/(2*e^5) - (2*d^3*
x^(5/2))/(5*e^4) + (d^2*x^3)/(3*e^3) - (2*d*x^(7/2))/(7*e^2) + x^4/(4*e) + (2*d^8*Log[d + e*Sqrt[x]])/e^9))/8
+ (b*x^4*Log[c*(d + e*Sqrt[x])^n])/4

Maple [F]

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

[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 = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.89 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {840 \, b e^{8} x^{4} \log \left (c\right ) - 140 \, b d^{2} e^{6} n x^{3} - 210 \, b d^{4} e^{4} n x^{2} - 420 \, b d^{6} e^{2} n x - 105 \, {\left (b e^{8} n - 8 \, a e^{8}\right )} x^{4} + 840 \, {\left (b e^{8} n x^{4} - b d^{8} n\right )} \log \left (e \sqrt {x} + d\right ) + 8 \, {\left (15 \, b d e^{7} n x^{3} + 21 \, b d^{3} e^{5} n x^{2} + 35 \, b d^{5} e^{3} n x + 105 \, b d^{7} e n\right )} \sqrt {x}}{3360 \, e^{8}} \]

[In]

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

[Out]

1/3360*(840*b*e^8*x^4*log(c) - 140*b*d^2*e^6*n*x^3 - 210*b*d^4*e^4*n*x^2 - 420*b*d^6*e^2*n*x - 105*(b*e^8*n -
8*a*e^8)*x^4 + 840*(b*e^8*n*x^4 - b*d^8*n)*log(e*sqrt(x) + d) + 8*(15*b*d*e^7*n*x^3 + 21*b*d^3*e^5*n*x^2 + 35*
b*d^5*e^3*n*x + 105*b*d^7*e*n)*sqrt(x))/e^8

Sympy [A] (verification not implemented)

Time = 8.47 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a x^{4}}{4} + b \left (- \frac {e n \left (\frac {2 d^{8} \left (\begin {cases} \frac {\sqrt {x}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e \sqrt {x} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{8}} - \frac {2 d^{7} \sqrt {x}}{e^{8}} + \frac {d^{6} x}{e^{7}} - \frac {2 d^{5} x^{\frac {3}{2}}}{3 e^{6}} + \frac {d^{4} x^{2}}{2 e^{5}} - \frac {2 d^{3} x^{\frac {5}{2}}}{5 e^{4}} + \frac {d^{2} x^{3}}{3 e^{3}} - \frac {2 d x^{\frac {7}{2}}}{7 e^{2}} + \frac {x^{4}}{4 e}\right )}{8} + \frac {x^{4} \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{4}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.77 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left ({\left (e \sqrt {x} + d\right )}^{n} c\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{3360} \, b e n {\left (\frac {840 \, d^{8} \log \left (e \sqrt {x} + d\right )}{e^{9}} + \frac {105 \, e^{7} x^{4} - 120 \, d e^{6} x^{\frac {7}{2}} + 140 \, d^{2} e^{5} x^{3} - 168 \, d^{3} e^{4} x^{\frac {5}{2}} + 210 \, d^{4} e^{3} x^{2} - 280 \, d^{5} e^{2} x^{\frac {3}{2}} + 420 \, d^{6} e x - 840 \, d^{7} \sqrt {x}}{e^{8}}\right )} \]

[In]

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

[Out]

1/4*b*x^4*log((e*sqrt(x) + d)^n*c) + 1/4*a*x^4 - 1/3360*b*e*n*(840*d^8*log(e*sqrt(x) + d)/e^9 + (105*e^7*x^4 -
120*d*e^6*x^(7/2) + 140*d^2*e^5*x^3 - 168*d^3*e^4*x^(5/2) + 210*d^4*e^3*x^2 - 280*d^5*e^2*x^(3/2) + 420*d^6*e
*x - 840*d^7*sqrt(x))/e^8)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (134) = 268\).

Time = 0.32 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.10 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {840 \, b e x^{4} \log \left (c\right ) + 840 \, a e x^{4} + {\left (\frac {840 \, {\left (e \sqrt {x} + d\right )}^{8} \log \left (e \sqrt {x} + d\right )}{e^{7}} - \frac {6720 \, {\left (e \sqrt {x} + d\right )}^{7} d \log \left (e \sqrt {x} + d\right )}{e^{7}} + \frac {23520 \, {\left (e \sqrt {x} + d\right )}^{6} d^{2} \log \left (e \sqrt {x} + d\right )}{e^{7}} - \frac {47040 \, {\left (e \sqrt {x} + d\right )}^{5} d^{3} \log \left (e \sqrt {x} + d\right )}{e^{7}} + \frac {58800 \, {\left (e \sqrt {x} + d\right )}^{4} d^{4} \log \left (e \sqrt {x} + d\right )}{e^{7}} - \frac {47040 \, {\left (e \sqrt {x} + d\right )}^{3} d^{5} \log \left (e \sqrt {x} + d\right )}{e^{7}} + \frac {23520 \, {\left (e \sqrt {x} + d\right )}^{2} d^{6} \log \left (e \sqrt {x} + d\right )}{e^{7}} - \frac {6720 \, {\left (e \sqrt {x} + d\right )} d^{7} \log \left (e \sqrt {x} + d\right )}{e^{7}} - \frac {105 \, {\left (e \sqrt {x} + d\right )}^{8}}{e^{7}} + \frac {960 \, {\left (e \sqrt {x} + d\right )}^{7} d}{e^{7}} - \frac {3920 \, {\left (e \sqrt {x} + d\right )}^{6} d^{2}}{e^{7}} + \frac {9408 \, {\left (e \sqrt {x} + d\right )}^{5} d^{3}}{e^{7}} - \frac {14700 \, {\left (e \sqrt {x} + d\right )}^{4} d^{4}}{e^{7}} + \frac {15680 \, {\left (e \sqrt {x} + d\right )}^{3} d^{5}}{e^{7}} - \frac {11760 \, {\left (e \sqrt {x} + d\right )}^{2} d^{6}}{e^{7}} + \frac {6720 \, {\left (e \sqrt {x} + d\right )} d^{7}}{e^{7}}\right )} b n}{3360 \, e} \]

[In]

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

[Out]

1/3360*(840*b*e*x^4*log(c) + 840*a*e*x^4 + (840*(e*sqrt(x) + d)^8*log(e*sqrt(x) + d)/e^7 - 6720*(e*sqrt(x) + d
)^7*d*log(e*sqrt(x) + d)/e^7 + 23520*(e*sqrt(x) + d)^6*d^2*log(e*sqrt(x) + d)/e^7 - 47040*(e*sqrt(x) + d)^5*d^
3*log(e*sqrt(x) + d)/e^7 + 58800*(e*sqrt(x) + d)^4*d^4*log(e*sqrt(x) + d)/e^7 - 47040*(e*sqrt(x) + d)^3*d^5*lo
g(e*sqrt(x) + d)/e^7 + 23520*(e*sqrt(x) + d)^2*d^6*log(e*sqrt(x) + d)/e^7 - 6720*(e*sqrt(x) + d)*d^7*log(e*sqr
t(x) + d)/e^7 - 105*(e*sqrt(x) + d)^8/e^7 + 960*(e*sqrt(x) + d)^7*d/e^7 - 3920*(e*sqrt(x) + d)^6*d^2/e^7 + 940
8*(e*sqrt(x) + d)^5*d^3/e^7 - 14700*(e*sqrt(x) + d)^4*d^4/e^7 + 15680*(e*sqrt(x) + d)^3*d^5/e^7 - 11760*(e*sqr
t(x) + d)^2*d^6/e^7 + 6720*(e*sqrt(x) + d)*d^7/e^7)*b*n)/e

Mupad [B] (verification not implemented)

Time = 1.80 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.83 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,n\,x^4}{32}+\frac {b\,x^4\,\ln \left (c\,{\left (d+e\,\sqrt {x}\right )}^n\right )}{4}+\frac {b\,d\,n\,x^{7/2}}{28\,e}-\frac {b\,d^6\,n\,x}{8\,e^6}-\frac {b\,d^8\,n\,\ln \left (d+e\,\sqrt {x}\right )}{4\,e^8}-\frac {b\,d^2\,n\,x^3}{24\,e^2}-\frac {b\,d^4\,n\,x^2}{16\,e^4}+\frac {b\,d^3\,n\,x^{5/2}}{20\,e^3}+\frac {b\,d^5\,n\,x^{3/2}}{12\,e^5}+\frac {b\,d^7\,n\,\sqrt {x}}{4\,e^7} \]

[In]

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

[Out]

(a*x^4)/4 - (b*n*x^4)/32 + (b*x^4*log(c*(d + e*x^(1/2))^n))/4 + (b*d*n*x^(7/2))/(28*e) - (b*d^6*n*x)/(8*e^6) -
(b*d^8*n*log(d + e*x^(1/2)))/(4*e^8) - (b*d^2*n*x^3)/(24*e^2) - (b*d^4*n*x^2)/(16*e^4) + (b*d^3*n*x^(5/2))/(2
0*e^3) + (b*d^5*n*x^(3/2))/(12*e^5) + (b*d^7*n*x^(1/2))/(4*e^7)