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

Optimal. Leaf size=60 \[ a x+b x \log \left (c \left (d+e \sqrt{x}\right )^n\right )-\frac{b d^2 n \log \left (d+e \sqrt{x}\right )}{e^2}+\frac{b d n \sqrt{x}}{e}-\frac{b n x}{2} \]

[Out]

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

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Rubi [A]  time = 0.040151, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2448, 266, 43} \[ a x+b x \log \left (c \left (d+e \sqrt{x}\right )^n\right )-\frac{b d^2 n \log \left (d+e \sqrt{x}\right )}{e^2}+\frac{b d n \sqrt{x}}{e}-\frac{b n x}{2} \]

Antiderivative was successfully verified.

[In]

Int[a + b*Log[c*(d + e*Sqrt[x])^n],x]

[Out]

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

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

Mathematica [A]  time = 0.0289376, size = 60, normalized size = 1. \[ a x+b x \log \left (c \left (d+e \sqrt{x}\right )^n\right )-\frac{b d^2 n \log \left (d+e \sqrt{x}\right )}{e^2}+\frac{b d n \sqrt{x}}{e}-\frac{b n x}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*Log[c*(d + e*Sqrt[x])^n],x]

[Out]

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

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Maple [A]  time = 0.084, size = 53, normalized size = 0.9 \begin{align*} ax-{\frac{bnx}{2}}-{\frac{b{d}^{2}n}{{e}^{2}}\ln \left ( d+e\sqrt{x} \right ) }+bx\ln \left ( c \left ( d+e\sqrt{x} \right ) ^{n} \right ) +{\frac{bdn}{e}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*ln(c*(d+e*x^(1/2))^n),x)

[Out]

a*x-1/2*b*n*x-b*d^2*n*ln(d+e*x^(1/2))/e^2+b*x*ln(c*(d+e*x^(1/2))^n)+b*d*n*x^(1/2)/e

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Maxima [A]  time = 1.01856, size = 77, normalized size = 1.28 \begin{align*} -\frac{1}{2} \,{\left (e n{\left (\frac{2 \, d^{2} \log \left (e \sqrt{x} + d\right )}{e^{3}} + \frac{e x - 2 \, d \sqrt{x}}{e^{2}}\right )} - 2 \, x \log \left ({\left (e \sqrt{x} + d\right )}^{n} c\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.90945, size = 158, normalized size = 2.63 \begin{align*} \frac{2 \, b e^{2} x \log \left (c\right ) + 2 \, b d e n \sqrt{x} -{\left (b e^{2} n - 2 \, a e^{2}\right )} x + 2 \,{\left (b e^{2} n x - b d^{2} n\right )} \log \left (e \sqrt{x} + d\right )}{2 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*e^2*x*log(c) + 2*b*d*e*n*sqrt(x) - (b*e^2*n - 2*a*e^2)*x + 2*(b*e^2*n*x - b*d^2*n)*log(e*sqrt(x) + d)
)/e^2

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Sympy [A]  time = 1.45738, size = 66, normalized size = 1.1 \begin{align*} a x + b \left (- \frac{e n \left (\frac{2 d^{2} \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^{2}} - \frac{2 d \sqrt{x}}{e^{2}} + \frac{x}{e}\right )}{2} + x \log{\left (c \left (d + e \sqrt{x}\right )^{n} \right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*ln(c*(d+e*x**(1/2))**n),x)

[Out]

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

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Giac [B]  time = 1.30731, size = 144, normalized size = 2.4 \begin{align*} \frac{1}{2} \,{\left ({\left (2 \,{\left (\sqrt{x} e + d\right )}^{2} \log \left (\sqrt{x} e + d\right ) - 4 \,{\left (\sqrt{x} e + d\right )} d \log \left (\sqrt{x} e + d\right ) -{\left (\sqrt{x} e + d\right )}^{2} + 4 \,{\left (\sqrt{x} e + d\right )} d\right )} n e^{\left (-1\right )} + 2 \,{\left ({\left (\sqrt{x} e + d\right )}^{2} - 2 \,{\left (\sqrt{x} e + d\right )} d\right )} e^{\left (-1\right )} \log \left (c\right )\right )} b e^{\left (-1\right )} + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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