3.48 \(\int x \log (c (a+b \sqrt{x})^p) \, dx\)

Optimal. Leaf size=93 \[ \frac{a^3 p \sqrt{x}}{2 b^3}-\frac{a^2 p x}{4 b^2}-\frac{a^4 p \log \left (a+b \sqrt{x}\right )}{2 b^4}+\frac{1}{2} x^2 \log \left (c \left (a+b \sqrt{x}\right )^p\right )+\frac{a p x^{3/2}}{6 b}-\frac{p x^2}{8} \]

[Out]

(a^3*p*Sqrt[x])/(2*b^3) - (a^2*p*x)/(4*b^2) + (a*p*x^(3/2))/(6*b) - (p*x^2)/8 - (a^4*p*Log[a + b*Sqrt[x]])/(2*
b^4) + (x^2*Log[c*(a + b*Sqrt[x])^p])/2

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Rubi [A]  time = 0.0602953, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2454, 2395, 43} \[ \frac{a^3 p \sqrt{x}}{2 b^3}-\frac{a^2 p x}{4 b^2}-\frac{a^4 p \log \left (a+b \sqrt{x}\right )}{2 b^4}+\frac{1}{2} x^2 \log \left (c \left (a+b \sqrt{x}\right )^p\right )+\frac{a p x^{3/2}}{6 b}-\frac{p x^2}{8} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b*Sqrt[x])^p],x]

[Out]

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

Mathematica [A]  time = 0.0361773, size = 88, normalized size = 0.95 \[ \frac{b p \sqrt{x} \left (-6 a^2 b \sqrt{x}+12 a^3+4 a b^2 x-3 b^3 x^{3/2}\right )-12 a^4 p \log \left (a+b \sqrt{x}\right )+12 b^4 x^2 \log \left (c \left (a+b \sqrt{x}\right )^p\right )}{24 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b*Sqrt[x])^p],x]

[Out]

(b*p*Sqrt[x]*(12*a^3 - 6*a^2*b*Sqrt[x] + 4*a*b^2*x - 3*b^3*x^(3/2)) - 12*a^4*p*Log[a + b*Sqrt[x]] + 12*b^4*x^2
*Log[c*(a + b*Sqrt[x])^p])/(24*b^4)

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int x\ln \left ( c \left ( a+b\sqrt{x} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(c*(a+b*x^(1/2))^p),x)

[Out]

int(x*ln(c*(a+b*x^(1/2))^p),x)

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Maxima [A]  time = 1.06029, size = 103, normalized size = 1.11 \begin{align*} -\frac{1}{24} \, b p{\left (\frac{12 \, a^{4} \log \left (b \sqrt{x} + a\right )}{b^{5}} + \frac{3 \, b^{3} x^{2} - 4 \, a b^{2} x^{\frac{3}{2}} + 6 \, a^{2} b x - 12 \, a^{3} \sqrt{x}}{b^{4}}\right )} + \frac{1}{2} \, x^{2} \log \left ({\left (b \sqrt{x} + a\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.36408, size = 190, normalized size = 2.04 \begin{align*} -\frac{3 \, b^{4} p x^{2} - 12 \, b^{4} x^{2} \log \left (c\right ) + 6 \, a^{2} b^{2} p x - 12 \,{\left (b^{4} p x^{2} - a^{4} p\right )} \log \left (b \sqrt{x} + a\right ) - 4 \,{\left (a b^{3} p x + 3 \, a^{3} b p\right )} \sqrt{x}}{24 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 8.12539, size = 92, normalized size = 0.99 \begin{align*} - \frac{b p \left (\frac{2 a^{4} \left (\begin{cases} \frac{\sqrt{x}}{a} & \text{for}\: b = 0 \\\frac{\log{\left (a + b \sqrt{x} \right )}}{b} & \text{otherwise} \end{cases}\right )}{b^{4}} - \frac{2 a^{3} \sqrt{x}}{b^{4}} + \frac{a^{2} x}{b^{3}} - \frac{2 a x^{\frac{3}{2}}}{3 b^{2}} + \frac{x^{2}}{2 b}\right )}{4} + \frac{x^{2} \log{\left (c \left (a + b \sqrt{x}\right )^{p} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(c*(a+b*x**(1/2))**p),x)

[Out]

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

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Giac [B]  time = 1.30632, size = 298, normalized size = 3.2 \begin{align*} \frac{\frac{{\left (\frac{12 \,{\left (b \sqrt{x} + a\right )}^{4} \log \left (b \sqrt{x} + a\right )}{b^{2}} - \frac{48 \,{\left (b \sqrt{x} + a\right )}^{3} a \log \left (b \sqrt{x} + a\right )}{b^{2}} + \frac{72 \,{\left (b \sqrt{x} + a\right )}^{2} a^{2} \log \left (b \sqrt{x} + a\right )}{b^{2}} - \frac{48 \,{\left (b \sqrt{x} + a\right )} a^{3} \log \left (b \sqrt{x} + a\right )}{b^{2}} - \frac{3 \,{\left (b \sqrt{x} + a\right )}^{4}}{b^{2}} + \frac{16 \,{\left (b \sqrt{x} + a\right )}^{3} a}{b^{2}} - \frac{36 \,{\left (b \sqrt{x} + a\right )}^{2} a^{2}}{b^{2}} + \frac{48 \,{\left (b \sqrt{x} + a\right )} a^{3}}{b^{2}}\right )} p}{b} + \frac{12 \,{\left ({\left (b \sqrt{x} + a\right )}^{4} - 4 \,{\left (b \sqrt{x} + a\right )}^{3} a + 6 \,{\left (b \sqrt{x} + a\right )}^{2} a^{2} - 4 \,{\left (b \sqrt{x} + a\right )} a^{3}\right )} \log \left (c\right )}{b^{3}}}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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