3.1.48 \(\int x \log (c (a+b \sqrt {x})^p) \, dx\) [48]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {2504, 2442, 45} \begin {gather*} -\frac {a^4 p \log \left (a+b \sqrt {x}\right )}{2 b^4}+\frac {a^3 p \sqrt {x}}{2 b^3}-\frac {a^2 p x}{4 b^2}+\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} \end {gather*}

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 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*} \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx &=2 \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.03, size = 88, normalized size = 0.95 \begin {gather*} \frac {b p \sqrt {x} \left (12 a^3-6 a^2 b \sqrt {x}+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} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int x \ln \left (c \left (a +b \sqrt {x}\right )^{p}\right )\, dx\]

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 = 0.28, size = 76, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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 = 0.43, size = 80, normalized size = 0.86 \begin {gather*} -\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 {gather*}

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 = 1.46, size = 92, normalized size = 0.99 \begin {gather*} - \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (73) = 146\).
time = 3.39, size = 171, normalized size = 1.84 \begin {gather*} \frac {12 \, b x^{2} \log \left (c\right ) + {\left (\frac {12 \, {\left (b \sqrt {x} + a\right )}^{4} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {48 \, {\left (b \sqrt {x} + a\right )}^{3} a \log \left (b \sqrt {x} + a\right )}{b^{3}} + \frac {72 \, {\left (b \sqrt {x} + a\right )}^{2} a^{2} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {48 \, {\left (b \sqrt {x} + a\right )} a^{3} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {3 \, {\left (b \sqrt {x} + a\right )}^{4}}{b^{3}} + \frac {16 \, {\left (b \sqrt {x} + a\right )}^{3} a}{b^{3}} - \frac {36 \, {\left (b \sqrt {x} + a\right )}^{2} a^{2}}{b^{3}} + \frac {48 \, {\left (b \sqrt {x} + a\right )} a^{3}}{b^{3}}\right )} p}{24 \, b} \end {gather*}

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

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Mupad [B]
time = 0.27, size = 73, normalized size = 0.78 \begin {gather*} \frac {x^2\,\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{2}-\frac {p\,x^2}{8}-\frac {a^4\,p\,\ln \left (a+b\,\sqrt {x}\right )}{2\,b^4}+\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(c*(a + b*x^(1/2))^p),x)

[Out]

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

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