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

Optimal. Leaf size=53 \[ -\frac{a^2 p \log \left (a+b \sqrt{x}\right )}{b^2}+x \log \left (c \left (a+b \sqrt{x}\right )^p\right )+\frac{a p \sqrt{x}}{b}-\frac{p x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0283768, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2448, 266, 43} \[ -\frac{a^2 p \log \left (a+b \sqrt{x}\right )}{b^2}+x \log \left (c \left (a+b \sqrt{x}\right )^p\right )+\frac{a p \sqrt{x}}{b}-\frac{p x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*Sqrt[x])^p],x]

[Out]

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

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 \log \left (c \left (a+b \sqrt{x}\right )^p\right ) \, dx &=x \log \left (c \left (a+b \sqrt{x}\right )^p\right )-\frac{1}{2} (b p) \int \frac{\sqrt{x}}{a+b \sqrt{x}} \, dx\\ &=x \log \left (c \left (a+b \sqrt{x}\right )^p\right )-(b p) \operatorname{Subst}\left (\int \frac{x^2}{a+b x} \, dx,x,\sqrt{x}\right )\\ &=x \log \left (c \left (a+b \sqrt{x}\right )^p\right )-(b p) \operatorname{Subst}\left (\int \left (-\frac{a}{b^2}+\frac{x}{b}+\frac{a^2}{b^2 (a+b x)}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a p \sqrt{x}}{b}-\frac{p x}{2}-\frac{a^2 p \log \left (a+b \sqrt{x}\right )}{b^2}+x \log \left (c \left (a+b \sqrt{x}\right )^p\right )\\ \end{align*}

Mathematica [A]  time = 0.0235122, size = 53, normalized size = 1. \[ -\frac{a^2 p \log \left (a+b \sqrt{x}\right )}{b^2}+x \log \left (c \left (a+b \sqrt{x}\right )^p\right )+\frac{a p \sqrt{x}}{b}-\frac{p x}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*Sqrt[x])^p],x]

[Out]

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

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Maple [A]  time = 0.065, size = 46, normalized size = 0.9 \begin{align*} -{\frac{px}{2}}-{\frac{{a}^{2}p}{{b}^{2}}\ln \left ( a+b\sqrt{x} \right ) }+x\ln \left ( c \left ( a+b\sqrt{x} \right ) ^{p} \right ) +{\frac{ap}{b}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05733, size = 68, normalized size = 1.28 \begin{align*} -\frac{1}{2} \, b p{\left (\frac{2 \, a^{2} \log \left (b \sqrt{x} + a\right )}{b^{3}} + \frac{b x - 2 \, a \sqrt{x}}{b^{2}}\right )} + x \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(log(c*(a+b*x^(1/2))^p),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.42336, size = 130, normalized size = 2.45 \begin{align*} -\frac{b^{2} p x - 2 \, b^{2} x \log \left (c\right ) - 2 \, a b p \sqrt{x} - 2 \,{\left (b^{2} p x - a^{2} p\right )} \log \left (b \sqrt{x} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*p*x - 2*b^2*x*log(c) - 2*a*b*p*sqrt(x) - 2*(b^2*p*x - a^2*p)*log(b*sqrt(x) + a))/b^2

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Sympy [A]  time = 2.10022, size = 61, normalized size = 1.15 \begin{align*} - \frac{b p \left (\frac{2 a^{2} \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^{2}} - \frac{2 a \sqrt{x}}{b^{2}} + \frac{x}{b}\right )}{2} + x \log{\left (c \left (a + b \sqrt{x}\right )^{p} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.30654, size = 131, normalized size = 2.47 \begin{align*} \frac{\frac{{\left (2 \,{\left (b \sqrt{x} + a\right )}^{2} \log \left (b \sqrt{x} + a\right ) - 4 \,{\left (b \sqrt{x} + a\right )} a \log \left (b \sqrt{x} + a\right ) -{\left (b \sqrt{x} + a\right )}^{2} + 4 \,{\left (b \sqrt{x} + a\right )} a\right )} p}{b} + \frac{2 \,{\left ({\left (b \sqrt{x} + a\right )}^{2} - 2 \,{\left (b \sqrt{x} + a\right )} a\right )} \log \left (c\right )}{b}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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