Integrand size = 16, antiderivative size = 93 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {a^3 p \sqrt {x}}{2 b^3}-\frac {a^2 p x}{4 b^2}+\frac {a p x^{3/2}}{6 b}-\frac {p x^2}{8}-\frac {a^4 p \log \left (a+b \sqrt {x}\right )}{2 b^4}+\frac {1}{2} x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \] Output:
1/2*a^3*p*x^(1/2)/b^3-1/4*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)
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.95 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {b p \sqrt {x} \left (12 a^3-6 a^2 b \sqrt {x}+4 a b^2 x-3 b^3 x^{3/2}\right )-12 a^4 p \log \left (a+b \sqrt {x}\right )+12 b^4 x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right )}{24 b^4} \] Input:
Integrate[x*Log[c*(a + b*Sqrt[x])^p],x]
Output:
(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)
Time = 0.46 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2904, 2842, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 2 \int x^{3/2} \log \left (c \left (a+b \sqrt {x}\right )^p\right )d\sqrt {x}\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle 2 \left (\frac {1}{4} x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-\frac {1}{4} b p \int \frac {x^2}{a+b \sqrt {x}}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{4} x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-\frac {1}{4} b p \int \left (\frac {a^4}{b^4 \left (a+b \sqrt {x}\right )}-\frac {a^3}{b^4}+\frac {\sqrt {x} a^2}{b^3}-\frac {x a}{b^2}+\frac {x^{3/2}}{b}\right )d\sqrt {x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{4} x^2 \log \left (c \left (a+b \sqrt {x}\right )^p\right )-\frac {1}{4} b p \left (\frac {a^4 \log \left (a+b \sqrt {x}\right )}{b^5}-\frac {a^3 \sqrt {x}}{b^4}+\frac {a^2 x}{2 b^3}-\frac {a x^{3/2}}{3 b^2}+\frac {x^2}{4 b}\right )\right )\) |
Input:
Int[x*Log[c*(a + b*Sqrt[x])^p],x]
Output:
2*(-1/4*(b*p*(-((a^3*Sqrt[x])/b^4) + (a^2*x)/(2*b^3) - (a*x^(3/2))/(3*b^2) + x^2/(4*b) + (a^4*Log[a + b*Sqrt[x]])/b^5)) + (x^2*Log[c*(a + b*Sqrt[x]) ^p])/4)
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[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])
Time = 0.65 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.83
method | result | size |
parts | \(\frac {x^{2} \ln \left (c \left (a +b \sqrt {x}\right )^{p}\right )}{2}-\frac {p b \left (-\frac {2 \left (-\frac {b^{3} x^{2}}{4}+\frac {a \,x^{\frac {3}{2}} b^{2}}{3}-\frac {a^{2} b x}{2}+a^{3} \sqrt {x}\right )}{b^{4}}+\frac {2 a^{4} \ln \left (a +b \sqrt {x}\right )}{b^{5}}\right )}{4}\) | \(77\) |
Input:
int(x*ln(c*(a+b*x^(1/2))^p),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*ln(c*(a+b*x^(1/2))^p)-1/4*p*b*(-2/b^4*(-1/4*b^3*x^2+1/3*a*x^(3/2)* b^2-1/2*a^2*b*x+a^3*x^(1/2))+2*a^4/b^5*ln(a+b*x^(1/2)))
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=-\frac {3 \, b^{4} p x^{2} - 12 \, b^{4} x^{2} \log \left (c\right ) + 6 \, a^{2} b^{2} p x - 12 \, {\left (b^{4} p x^{2} - a^{4} p\right )} \log \left (b \sqrt {x} + a\right ) - 4 \, {\left (a b^{3} p x + 3 \, a^{3} b p\right )} \sqrt {x}}{24 \, b^{4}} \] Input:
integrate(x*log(c*(a+b*x^(1/2))^p),x, algorithm="fricas")
Output:
-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
Time = 1.00 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=- \frac {b p \left (\frac {2 a^{4} \left (\begin {cases} \frac {\sqrt {x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{4}} - \frac {2 a^{3} \sqrt {x}}{b^{4}} + \frac {a^{2} x}{b^{3}} - \frac {2 a x^{\frac {3}{2}}}{3 b^{2}} + \frac {x^{2}}{2 b}\right )}{4} + \frac {x^{2} \log {\left (c \left (a + b \sqrt {x}\right )^{p} \right )}}{2} \] Input:
integrate(x*ln(c*(a+b*x**(1/2))**p),x)
Output:
-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
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=-\frac {1}{24} \, b p {\left (\frac {12 \, a^{4} \log \left (b \sqrt {x} + a\right )}{b^{5}} + \frac {3 \, b^{3} x^{2} - 4 \, a b^{2} x^{\frac {3}{2}} + 6 \, a^{2} b x - 12 \, a^{3} \sqrt {x}}{b^{4}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right ) \] Input:
integrate(x*log(c*(a+b*x^(1/2))^p),x, algorithm="maxima")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (73) = 146\).
Time = 0.12 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.84 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {12 \, b x^{2} \log \left (c\right ) + {\left (\frac {12 \, {\left (b \sqrt {x} + a\right )}^{4} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {48 \, {\left (b \sqrt {x} + a\right )}^{3} a \log \left (b \sqrt {x} + a\right )}{b^{3}} + \frac {72 \, {\left (b \sqrt {x} + a\right )}^{2} a^{2} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {48 \, {\left (b \sqrt {x} + a\right )} a^{3} \log \left (b \sqrt {x} + a\right )}{b^{3}} - \frac {3 \, {\left (b \sqrt {x} + a\right )}^{4}}{b^{3}} + \frac {16 \, {\left (b \sqrt {x} + a\right )}^{3} a}{b^{3}} - \frac {36 \, {\left (b \sqrt {x} + a\right )}^{2} a^{2}}{b^{3}} + \frac {48 \, {\left (b \sqrt {x} + a\right )} a^{3}}{b^{3}}\right )} p}{24 \, b} \] Input:
integrate(x*log(c*(a+b*x^(1/2))^p),x, algorithm="giac")
Output:
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
Time = 15.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.78 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {x^2\,\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{2}-\frac {p\,x^2}{8}-\frac {a^4\,p\,\ln \left (a+b\,\sqrt {x}\right )}{2\,b^4}+\frac {a^3\,p\,\sqrt {x}}{2\,b^3}-\frac {a^2\,p\,x}{4\,b^2}+\frac {a\,p\,x^{3/2}}{6\,b} \] Input:
int(x*log(c*(a + b*x^(1/2))^p),x)
Output:
(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 )
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.85 \[ \int x \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \, dx=\frac {12 \sqrt {x}\, a^{3} b p +4 \sqrt {x}\, a \,b^{3} p x -12 \,\mathrm {log}\left (\left (\sqrt {x}\, b +a \right )^{p} c \right ) a^{4}+12 \,\mathrm {log}\left (\left (\sqrt {x}\, b +a \right )^{p} c \right ) b^{4} x^{2}-6 a^{2} b^{2} p x -3 b^{4} p \,x^{2}}{24 b^{4}} \] Input:
int(x*log(c*(a+b*x^(1/2))^p),x)
Output:
(12*sqrt(x)*a**3*b*p + 4*sqrt(x)*a*b**3*p*x - 12*log((sqrt(x)*b + a)**p*c) *a**4 + 12*log((sqrt(x)*b + a)**p*c)*b**4*x**2 - 6*a**2*b**2*p*x - 3*b**4* p*x**2)/(24*b**4)