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

Optimal. Leaf size=102 \[ \frac{1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt{x}\right )^n\right )\right )+\frac{b d^3 n \sqrt{x}}{2 e^3}-\frac{b d^2 n x}{4 e^2}-\frac{b d^4 n \log \left (d+e \sqrt{x}\right )}{2 e^4}+\frac{b d n x^{3/2}}{6 e}-\frac{1}{8} b n x^2 \]

[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

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Rubi [A]  time = 0.0721291, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2454, 2395, 43} \[ \frac{1}{2} x^2 \left (a+b \log \left (c \left (d+e \sqrt{x}\right )^n\right )\right )+\frac{b d^3 n \sqrt{x}}{2 e^3}-\frac{b d^2 n x}{4 e^2}-\frac{b d^4 n \log \left (d+e \sqrt{x}\right )}{2 e^4}+\frac{b d n x^{3/2}}{6 e}-\frac{1}{8} b n x^2 \]

Antiderivative was successfully verified.

[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 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])

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 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])

Rubi steps

\begin{align*} \int x \left (a+b \log \left (c \left (d+e \sqrt{x}\right )^n\right )\right ) \, dx &=2 \operatorname{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) \operatorname{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) \operatorname{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]  time = 0.0329987, size = 107, normalized size = 1.05 \[ \frac{a x^2}{2}+\frac{1}{2} b x^2 \log \left (c \left (d+e \sqrt{x}\right )^n\right )+\frac{b d^3 n \sqrt{x}}{2 e^3}-\frac{b d^2 n x}{4 e^2}-\frac{b d^4 n \log \left (d+e \sqrt{x}\right )}{2 e^4}+\frac{b d n x^{3/2}}{6 e}-\frac{1}{8} b n x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[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) + (a*x^2)/2 - (b*n*x^2)/8 - (b*d^4*n*L
og[d + e*Sqrt[x]])/(2*e^4) + (b*x^2*Log[c*(d + e*Sqrt[x])^n])/2

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Maple [F]  time = 0.098, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\ln \left ( c \left ( d+e\sqrt{x} \right ) ^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]  time = 1.04189, size = 113, normalized size = 1.11 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.74638, size = 224, normalized size = 2.2 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 4.7292, size = 100, normalized size = 0.98 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.24899, size = 385, normalized size = 3.77 \begin{align*} \frac{1}{24} \,{\left ({\left (12 \,{\left (\sqrt{x} e + d\right )}^{4} e^{\left (-2\right )} \log \left (\sqrt{x} e + d\right ) - 48 \,{\left (\sqrt{x} e + d\right )}^{3} d e^{\left (-2\right )} \log \left (\sqrt{x} e + d\right ) + 72 \,{\left (\sqrt{x} e + d\right )}^{2} d^{2} e^{\left (-2\right )} \log \left (\sqrt{x} e + d\right ) - 48 \,{\left (\sqrt{x} e + d\right )} d^{3} e^{\left (-2\right )} \log \left (\sqrt{x} e + d\right ) - 3 \,{\left (\sqrt{x} e + d\right )}^{4} e^{\left (-2\right )} + 16 \,{\left (\sqrt{x} e + d\right )}^{3} d e^{\left (-2\right )} - 36 \,{\left (\sqrt{x} e + d\right )}^{2} d^{2} e^{\left (-2\right )} + 48 \,{\left (\sqrt{x} e + d\right )} d^{3} e^{\left (-2\right )}\right )} b n e^{\left (-1\right )} + 12 \,{\left ({\left (\sqrt{x} e + d\right )}^{4} - 4 \,{\left (\sqrt{x} e + d\right )}^{3} d + 6 \,{\left (\sqrt{x} e + d\right )}^{2} d^{2} - 4 \,{\left (\sqrt{x} e + d\right )} d^{3}\right )} b e^{\left (-3\right )} \log \left (c\right ) + 12 \,{\left ({\left (\sqrt{x} e + d\right )}^{4} - 4 \,{\left (\sqrt{x} e + d\right )}^{3} d + 6 \,{\left (\sqrt{x} e + d\right )}^{2} d^{2} - 4 \,{\left (\sqrt{x} e + d\right )} d^{3}\right )} a e^{\left (-3\right )}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((12*(sqrt(x)*e + d)^4*e^(-2)*log(sqrt(x)*e + d) - 48*(sqrt(x)*e + d)^3*d*e^(-2)*log(sqrt(x)*e + d) + 72*
(sqrt(x)*e + d)^2*d^2*e^(-2)*log(sqrt(x)*e + d) - 48*(sqrt(x)*e + d)*d^3*e^(-2)*log(sqrt(x)*e + d) - 3*(sqrt(x
)*e + d)^4*e^(-2) + 16*(sqrt(x)*e + d)^3*d*e^(-2) - 36*(sqrt(x)*e + d)^2*d^2*e^(-2) + 48*(sqrt(x)*e + d)*d^3*e
^(-2))*b*n*e^(-1) + 12*((sqrt(x)*e + d)^4 - 4*(sqrt(x)*e + d)^3*d + 6*(sqrt(x)*e + d)^2*d^2 - 4*(sqrt(x)*e + d
)*d^3)*b*e^(-3)*log(c) + 12*((sqrt(x)*e + d)^4 - 4*(sqrt(x)*e + d)^3*d + 6*(sqrt(x)*e + d)^2*d^2 - 4*(sqrt(x)*
e + d)*d^3)*a*e^(-3))*e^(-1)