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

Optimal. Leaf size=142 \[ -\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{8 b d^2 n \sqrt{d+e x}}{15 e^2}-\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^2}+\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2} \]

[Out]

(8*b*d^2*n*Sqrt[d + e*x])/(15*e^2) + (8*b*d*n*(d + e*x)^(3/2))/(45*e^2) - (4*b*n*(d + e*x)^(5/2))/(25*e^2) - (
8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^2) - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2) + (2
*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2)

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Rubi [A]  time = 0.100847, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{8 b d^2 n \sqrt{d+e x}}{15 e^2}-\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^2}+\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*d^2*n*Sqrt[d + e*x])/(15*e^2) + (8*b*d*n*(d + e*x)^(3/2))/(45*e^2) - (4*b*n*(d + e*x)^(5/2))/(25*e^2) - (
8*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^2) - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2) + (2
*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*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 x \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-(b n) \int \frac{2 (d+e x)^{3/2} (-2 d+3 e x)}{15 e^2 x} \, dx\\ &=-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac{(2 b n) \int \frac{(d+e x)^{3/2} (-2 d+3 e x)}{x} \, dx}{15 e^2}\\ &=-\frac{4 b n (d+e x)^{5/2}}{25 e^2}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{(4 b d n) \int \frac{(d+e x)^{3/2}}{x} \, dx}{15 e^2}\\ &=\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{\left (4 b d^2 n\right ) \int \frac{\sqrt{d+e x}}{x} \, dx}{15 e^2}\\ &=\frac{8 b d^2 n \sqrt{d+e x}}{15 e^2}+\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{\left (4 b d^3 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{15 e^2}\\ &=\frac{8 b d^2 n \sqrt{d+e x}}{15 e^2}+\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac{\left (8 b d^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{15 e^3}\\ &=\frac{8 b d^2 n \sqrt{d+e x}}{15 e^2}+\frac{8 b d n (d+e x)^{3/2}}{45 e^2}-\frac{4 b n (d+e x)^{5/2}}{25 e^2}-\frac{8 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^2}-\frac{2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\\ \end{align*}

Mathematica [A]  time = 0.117947, size = 116, normalized size = 0.82 \[ \frac{2 \sqrt{d+e x} \left (15 a \left (-2 d^2+d e x+3 e^2 x^2\right )+15 b \left (-2 d^2+d e x+3 e^2 x^2\right ) \log \left (c x^n\right )+2 b n \left (31 d^2-8 d e x-9 e^2 x^2\right )\right )-120 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{225 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-120*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(2*b*n*(31*d^2 - 8*d*e*x - 9*e^2*x^2) + 15*
a*(-2*d^2 + d*e*x + 3*e^2*x^2) + 15*b*(-2*d^2 + d*e*x + 3*e^2*x^2)*Log[c*x^n]))/(225*e^2)

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Maple [F]  time = 0.616, size = 0, normalized size = 0. \begin{align*} \int x \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \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 = 1.42823, size = 722, normalized size = 5.08 \begin{align*} \left [\frac{2 \,{\left (30 \, b d^{\frac{5}{2}} n \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) +{\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} -{\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \,{\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \left (c\right ) + 15 \,{\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, e^{2}}, \frac{2 \,{\left (60 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} -{\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \,{\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \left (c\right ) + 15 \,{\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, 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]

[2/225*(30*b*d^(5/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (62*b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n -
5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e)*x + 15*(3*b*e^2*x^2 + b*d*e*x - 2*b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b
*d*e*n*x - 2*b*d^2*n)*log(x))*sqrt(e*x + d))/e^2, 2/225*(60*b*sqrt(-d)*d^2*n*arctan(sqrt(e*x + d)*sqrt(-d)/d)
+ (62*b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n - 5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e)*x + 15*(3*b*e^2*x^2 + b*d*e*x
- 2*b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b*d*e*n*x - 2*b*d^2*n)*log(x))*sqrt(e*x + d))/e^2]

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Sympy [A]  time = 7.3727, size = 224, normalized size = 1.58 \begin{align*} \frac{2 \left (- \frac{a d \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{a \left (d + e x\right )^{\frac{5}{2}}}{5} - b d \left (\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 (\frac{d^{2} e \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{\sqrt{- d}} + d e \sqrt{d + e x} + \frac{e \left (d + e x\right )^{\frac{3}{2}}}{3}\right )}{3 e}\right ) + b \left (\frac{\left (d + e x\right )^{\frac{5}{2}} \log{\left (c \left (- \frac{d}{e} + \frac{d + e x}{e}\right )^{n} \right )}}{5} - \frac{2 n \left (\frac{d^{3} e \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{\sqrt{- d}} + d^{2} e \sqrt{d + e x} + \frac{d e \left (d + e x\right )^{\frac{3}{2}}}{3} + \frac{e \left (d + e x\right )^{\frac{5}{2}}}{5}\right )}{5 e}\right )\right )}{e^{2}} \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]

2*(-a*d*(d + e*x)**(3/2)/3 + a*(d + e*x)**(5/2)/5 - b*d*((d + e*x)**(3/2)*log(c*(-d/e + (d + e*x)/e)**n)/3 - 2
*n*(d**2*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d*e*sqrt(d + e*x) + e*(d + e*x)**(3/2)/3)/(3*e)) + b*((d +
e*x)**(5/2)*log(c*(-d/e + (d + e*x)/e)**n)/5 - 2*n*(d**3*e*atan(sqrt(d + e*x)/sqrt(-d))/sqrt(-d) + d**2*e*sqrt
(d + e*x) + d*e*(d + e*x)**(3/2)/3 + e*(d + e*x)**(5/2)/5)/(5*e)))/e**2

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