\(\int x (a+b \log (c (d+e \sqrt {x})^n)) \, dx\) [402]

Optimal result

Integrand size = 20, antiderivative size = 102 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {b d^3 n \sqrt {x}}{2 e^3}-\frac {b d^2 n x}{4 e^2}+\frac {b d n x^{3/2}}{6 e}-\frac {1}{8} b n x^2-\frac {b d^4 n \log \left (d+e \sqrt {x}\right )}{2 e^4}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \]

[Out]

-1/4*b*d^2*n*x/e^2+1/6*b*d*n*x^(3/2)/e-1/8*b*n*x^2-1/2*b*d^4*n*ln(d+e*x^(1/2))/e^4+1/2*x^2*(a+b*ln(c*(d+e*x^(1
/2))^n))+1/2*b*d^3*n*x^(1/2)/e^3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2504, 2442, 45} \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {b d^4 n \log \left (d+e \sqrt {x}\right )}{2 e^4}+\frac {b d^3 n \sqrt {x}}{2 e^3}-\frac {b d^2 n x}{4 e^2}+\frac {b d n x^{3/2}}{6 e}-\frac {1}{8} b n x^2 \]

[In]

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

[Out]

(b*d^3*n*Sqrt[x])/(2*e^3) - (b*d^2*n*x)/(4*e^2) + (b*d*n*x^(3/2))/(6*e) - (b*n*x^2)/8 - (b*d^4*n*Log[d + e*Sqr
t[x]])/(2*e^4) + (x^2*(a + b*Log[c*(d + e*Sqrt[x])^n]))/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 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^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {x^4}{d+e x} \, dx,x,\sqrt {x}\right ) \\ & = \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )-\frac {1}{2} (b e n) \text {Subst}\left (\int \left (-\frac {d^3}{e^4}+\frac {d^2 x}{e^3}-\frac {d x^2}{e^2}+\frac {x^3}{e}+\frac {d^4}{e^4 (d+e x)}\right ) \, dx,x,\sqrt {x}\right ) \\ & = \frac {b d^3 n \sqrt {x}}{2 e^3}-\frac {b d^2 n x}{4 e^2}+\frac {b d n x^{3/2}}{6 e}-\frac {1}{8} b n x^2-\frac {b d^4 n \log \left (d+e \sqrt {x}\right )}{2 e^4}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a x^2}{2}-\frac {1}{4} b e n \left (-\frac {2 d^3 \sqrt {x}}{e^4}+\frac {d^2 x}{e^3}-\frac {2 d x^{3/2}}{3 e^2}+\frac {x^2}{2 e}+\frac {2 d^4 \log \left (d+e \sqrt {x}\right )}{e^5}\right )+\frac {1}{2} b x^2 \log \left (c \left (d+e \sqrt {x}\right )^n\right ) \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {12 \, b e^{4} x^{2} \log \left (c\right ) - 6 \, b d^{2} e^{2} n x - 3 \, {\left (b e^{4} n - 4 \, a e^{4}\right )} x^{2} + 12 \, {\left (b e^{4} n x^{2} - b d^{4} n\right )} \log \left (e \sqrt {x} + d\right ) + 4 \, {\left (b d e^{3} n x + 3 \, b d^{3} e n\right )} \sqrt {x}}{24 \, e^{4}} \]

[In]

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

[Out]

1/24*(12*b*e^4*x^2*log(c) - 6*b*d^2*e^2*n*x - 3*(b*e^4*n - 4*a*e^4)*x^2 + 12*(b*e^4*n*x^2 - b*d^4*n)*log(e*sqr
t(x) + d) + 4*(b*d*e^3*n*x + 3*b*d^3*e*n)*sqrt(x))/e^4

Sympy [A] (verification not implemented)

Time = 1.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.98 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a x^{2}}{2} + b \left (- \frac {e n \left (\frac {2 d^{4} \left (\begin {cases} \frac {\sqrt {x}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e \sqrt {x} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{4}} - \frac {2 d^{3} \sqrt {x}}{e^{4}} + \frac {d^{2} x}{e^{3}} - \frac {2 d x^{\frac {3}{2}}}{3 e^{2}} + \frac {x^{2}}{2 e}\right )}{4} + \frac {x^{2} \log {\left (c \left (d + e \sqrt {x}\right )^{n} \right )}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=-\frac {1}{24} \, b e n {\left (\frac {12 \, d^{4} \log \left (e \sqrt {x} + d\right )}{e^{5}} + \frac {3 \, e^{3} x^{2} - 4 \, d e^{2} x^{\frac {3}{2}} + 6 \, d^{2} e x - 12 \, d^{3} \sqrt {x}}{e^{4}}\right )} + \frac {1}{2} \, b x^{2} \log \left ({\left (e \sqrt {x} + d\right )}^{n} c\right ) + \frac {1}{2} \, a x^{2} \]

[In]

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

[Out]

-1/24*b*e*n*(12*d^4*log(e*sqrt(x) + d)/e^5 + (3*e^3*x^2 - 4*d*e^2*x^(3/2) + 6*d^2*e*x - 12*d^3*sqrt(x))/e^4) +
1/2*b*x^2*log((e*sqrt(x) + d)^n*c) + 1/2*a*x^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (82) = 164\).

Time = 0.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.76 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {12 \, b e x^{2} \log \left (c\right ) + 12 \, a e x^{2} + {\left (\frac {12 \, {\left (e \sqrt {x} + d\right )}^{4} \log \left (e \sqrt {x} + d\right )}{e^{3}} - \frac {48 \, {\left (e \sqrt {x} + d\right )}^{3} d \log \left (e \sqrt {x} + d\right )}{e^{3}} + \frac {72 \, {\left (e \sqrt {x} + d\right )}^{2} d^{2} \log \left (e \sqrt {x} + d\right )}{e^{3}} - \frac {48 \, {\left (e \sqrt {x} + d\right )} d^{3} \log \left (e \sqrt {x} + d\right )}{e^{3}} - \frac {3 \, {\left (e \sqrt {x} + d\right )}^{4}}{e^{3}} + \frac {16 \, {\left (e \sqrt {x} + d\right )}^{3} d}{e^{3}} - \frac {36 \, {\left (e \sqrt {x} + d\right )}^{2} d^{2}}{e^{3}} + \frac {48 \, {\left (e \sqrt {x} + d\right )} d^{3}}{e^{3}}\right )} b n}{24 \, e} \]

[In]

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

[Out]

1/24*(12*b*e*x^2*log(c) + 12*a*e*x^2 + (12*(e*sqrt(x) + d)^4*log(e*sqrt(x) + d)/e^3 - 48*(e*sqrt(x) + d)^3*d*l
og(e*sqrt(x) + d)/e^3 + 72*(e*sqrt(x) + d)^2*d^2*log(e*sqrt(x) + d)/e^3 - 48*(e*sqrt(x) + d)*d^3*log(e*sqrt(x)
+ d)/e^3 - 3*(e*sqrt(x) + d)^4/e^3 + 16*(e*sqrt(x) + d)^3*d/e^3 - 36*(e*sqrt(x) + d)^2*d^2/e^3 + 48*(e*sqrt(x
) + d)*d^3/e^3)*b*n)/e

Mupad [B] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int x \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right ) \, dx=\frac {a\,x^2}{2}-\frac {b\,n\,x^2}{8}+\frac {b\,x^2\,\ln \left (c\,{\left (d+e\,\sqrt {x}\right )}^n\right )}{2}-\frac {b\,d^2\,n\,x}{4\,e^2}+\frac {b\,d\,n\,x^{3/2}}{6\,e}-\frac {b\,d^4\,n\,\ln \left (d+e\,\sqrt {x}\right )}{2\,e^4}+\frac {b\,d^3\,n\,\sqrt {x}}{2\,e^3} \]

[In]

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

[Out]

(a*x^2)/2 - (b*n*x^2)/8 + (b*x^2*log(c*(d + e*x^(1/2))^n))/2 - (b*d^2*n*x)/(4*e^2) + (b*d*n*x^(3/2))/(6*e) - (
b*d^4*n*log(d + e*x^(1/2)))/(2*e^4) + (b*d^3*n*x^(1/2))/(2*e^3)