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

Optimal. Leaf size=119 \[ -\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac{8 b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{3 e^2}+\frac{8 b d n \sqrt{d+e x}}{3 e^2}-\frac{4 b n (d+e x)^{3/2}}{9 e^2} \]

[Out]

(8*b*d*n*Sqrt[d + e*x])/(3*e^2) - (4*b*n*(d + e*x)^(3/2))/(9*e^2) - (8*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[
d]])/(3*e^2) - (2*d*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^2 + (2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2)

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Rubi [A]  time = 0.0912596, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {43, 2350, 12, 80, 50, 63, 208} \[ -\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac{8 b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{3 e^2}+\frac{8 b d n \sqrt{d+e x}}{3 e^2}-\frac{4 b n (d+e x)^{3/2}}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d*n*Sqrt[d + e*x])/(3*e^2) - (4*b*n*(d + e*x)^(3/2))/(9*e^2) - (8*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[
d]])/(3*e^2) - (2*d*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^2 + (2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2)

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])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{x \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d+e x}} \, dx &=-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-(b n) \int \frac{2 (-2 d+e x) \sqrt{d+e x}}{3 e^2 x} \, dx\\ &=-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac{(2 b n) \int \frac{(-2 d+e x) \sqrt{d+e x}}{x} \, dx}{3 e^2}\\ &=-\frac{4 b n (d+e x)^{3/2}}{9 e^2}-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{(4 b d n) \int \frac{\sqrt{d+e x}}{x} \, dx}{3 e^2}\\ &=\frac{8 b d n \sqrt{d+e x}}{3 e^2}-\frac{4 b n (d+e x)^{3/2}}{9 e^2}-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{\left (4 b d^2 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{3 e^2}\\ &=\frac{8 b d n \sqrt{d+e x}}{3 e^2}-\frac{4 b n (d+e x)^{3/2}}{9 e^2}-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{\left (8 b d^2 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{3 e^3}\\ &=\frac{8 b d n \sqrt{d+e x}}{3 e^2}-\frac{4 b n (d+e x)^{3/2}}{9 e^2}-\frac{8 b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{3 e^2}-\frac{2 d \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac{2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0996774, size = 80, normalized size = 0.67 \[ -\frac{2 \left (\sqrt{d+e x} \left (6 a d-3 a e x+b (6 d-3 e x) \log \left (c x^n\right )-10 b d n+2 b e n x\right )+12 b d^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )}{9 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(12*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(6*a*d - 10*b*d*n - 3*a*e*x + 2*b*e*n*x + b
*(6*d - 3*e*x)*Log[c*x^n])))/(9*e^2)

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Maple [F]  time = 0.551, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ){\frac{1}{\sqrt{ex+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34229, size = 494, normalized size = 4.15 \begin{align*} \left [\frac{2 \,{\left (6 \, b d^{\frac{3}{2}} n \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) +{\left (10 \, b d n - 6 \, a d -{\left (2 \, b e n - 3 \, a e\right )} x + 3 \,{\left (b e x - 2 \, b d\right )} \log \left (c\right ) + 3 \,{\left (b e n x - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{9 \, e^{2}}, \frac{2 \,{\left (12 \, b \sqrt{-d} d n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (10 \, b d n - 6 \, a d -{\left (2 \, b e n - 3 \, a e\right )} x + 3 \,{\left (b e x - 2 \, b d\right )} \log \left (c\right ) + 3 \,{\left (b e n x - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{9 \, e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/9*(6*b*d^(3/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (10*b*d*n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3
*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*d*n)*log(x))*sqrt(e*x + d))/e^2, 2/9*(12*b*sqrt(-d)*d*n*arctan(sqrt
(e*x + d)*sqrt(-d)/d) + (10*b*d*n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*
d*n)*log(x))*sqrt(e*x + d))/e^2]

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Sympy [A]  time = 123.661, size = 473, normalized size = 3.97 \begin{align*} \begin{cases} - \frac{\frac{2 a d \left (- \frac{d}{\sqrt{d + e x}} - \sqrt{d + e x}\right )}{e} + \frac{2 a \left (\frac{d^{2}}{\sqrt{d + e x}} + 2 d \sqrt{d + e x} - \frac{\left (d + e x\right )^{\frac{3}{2}}}{3}\right )}{e} + \frac{2 b d \left (- d \left (\frac{\log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )}}{\sqrt{d + e x}} - \frac{2 n \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{d \sqrt{- \frac{1}{d}}}\right ) - \sqrt{d + e x} \log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )} - \frac{2 n \left (- e \sqrt{d + e x} - \frac{e \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{\sqrt{- \frac{1}{d}}}\right )}{e}\right )}{e} + \frac{2 b \left (d^{2} \left (\frac{\log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )}}{\sqrt{d + e x}} - \frac{2 n \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{d \sqrt{- \frac{1}{d}}}\right ) - 2 d \left (- \sqrt{d + e x} \log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )} - \frac{2 n \left (- e \sqrt{d + e x} - \frac{e \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{\sqrt{- \frac{1}{d}}}\right )}{e}\right ) - \frac{\left (d + e x\right )^{\frac{3}{2}} \log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )}}{3} - \frac{2 n \left (- d e \sqrt{d + e x} - \frac{d e \operatorname{atan}{\left (\frac{1}{\sqrt{- \frac{1}{d}} \sqrt{d + e x}} \right )}}{\sqrt{- \frac{1}{d}}} - \frac{e \left (d + e x\right )^{\frac{3}{2}}}{3}\right )}{3 e}\right )}{e}}{e} & \text{for}\: e \neq 0 \\\frac{\frac{a x^{2}}{2} + b \left (- \frac{n x^{2}}{4} + \frac{x^{2} \log{\left (c x^{n} \right )}}{2}\right )}{\sqrt{d}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-(2*a*d*(-d/sqrt(d + e*x) - sqrt(d + e*x))/e + 2*a*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e
*x)**(3/2)/3)/e + 2*b*d*(-d*(log(c*(-d/e + (d + e*x)/e)**n)/sqrt(d + e*x) - 2*n*atan(1/(sqrt(-1/d)*sqrt(d + e*
x)))/(d*sqrt(-1/d))) - sqrt(d + e*x)*log(c*(-d/e + (d + e*x)/e)**n) - 2*n*(-e*sqrt(d + e*x) - e*atan(1/(sqrt(-
1/d)*sqrt(d + e*x)))/sqrt(-1/d))/e)/e + 2*b*(d**2*(log(c*(-d/e + (d + e*x)/e)**n)/sqrt(d + e*x) - 2*n*atan(1/(
sqrt(-1/d)*sqrt(d + e*x)))/(d*sqrt(-1/d))) - 2*d*(-sqrt(d + e*x)*log(c*(-d/e + (d + e*x)/e)**n) - 2*n*(-e*sqrt
(d + e*x) - e*atan(1/(sqrt(-1/d)*sqrt(d + e*x)))/sqrt(-1/d))/e) - (d + e*x)**(3/2)*log(c*(-d/e + (d + e*x)/e)*
*n)/3 - 2*n*(-d*e*sqrt(d + e*x) - d*e*atan(1/(sqrt(-1/d)*sqrt(d + e*x)))/sqrt(-1/d) - e*(d + e*x)**(3/2)/3)/(3
*e))/e)/e, Ne(e, 0)), ((a*x**2/2 + b*(-n*x**2/4 + x**2*log(c*x**n)/2))/sqrt(d), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} x}{\sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*x^n) + a)*x/sqrt(e*x + d), x)