3.145 \(\int \frac{x^2 (a+b \log (c x^n))}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=169 \[ \frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3} \]

[Out]

(-32*b*d^2*n*Sqrt[d + e*x])/(15*e^3) + (28*b*d*n*(d + e*x)^(3/2))/(45*e^3) - (4*b*n*(d + e*x)^(5/2))/(25*e^3)
+ (32*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^3) + (2*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^3 - (4
*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + (2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)

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Rubi [A]  time = 0.169124, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {43, 2350, 12, 897, 1261, 208} \[ \frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*b*d^2*n*Sqrt[d + e*x])/(15*e^3) + (28*b*d*n*(d + e*x)^(3/2))/(45*e^3) - (4*b*n*(d + e*x)^(5/2))/(25*e^3)
+ (32*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(15*e^3) + (2*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^3 - (4
*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + (2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)

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 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt{d+e x}} \, dx &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-(b n) \int \frac{2 \sqrt{d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{15 e^3 x} \, dx\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(2 b n) \int \frac{\sqrt{d+e x} \left (8 d^2-4 d e x+3 e^2 x^2\right )}{x} \, dx}{15 e^3}\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \frac{x^2 \left (15 d^2-10 d x^2+3 x^4\right )}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{15 e^4}\\ &=\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \left (8 d^2 e-7 d e x^2+3 e x^4+\frac{8 d^3}{-\frac{d}{e}+\frac{x^2}{e}}\right ) \, dx,x,\sqrt{d+e x}\right )}{15 e^4}\\ &=-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3}+\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{\left (32 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^4}\\ &=-\frac{32 b d^2 n \sqrt{d+e x}}{15 e^3}+\frac{28 b d n (d+e x)^{3/2}}{45 e^3}-\frac{4 b n (d+e x)^{5/2}}{25 e^3}+\frac{32 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{15 e^3}+\frac{2 d^2 \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac{4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac{2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\\ \end{align*}

Mathematica [A]  time = 0.175731, size = 118, normalized size = 0.7 \[ \frac{2 \sqrt{d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )+15 b \left (8 d^2-4 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )-2 b n \left (94 d^2-17 d e x+9 e^2 x^2\right )\right )+480 b d^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{225 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(480*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(15*a*(8*d^2 - 4*d*e*x + 3*e^2*x^2) - 2*b*n*
(94*d^2 - 17*d*e*x + 9*e^2*x^2) + 15*b*(8*d^2 - 4*d*e*x + 3*e^2*x^2)*Log[c*x^n]))/(225*e^3)

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Maple [F]  time = 0.598, size = 0, normalized size = 0. \begin{align*} \int{{x}^{2} \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^2*(a+b*ln(c*x^n))/(e*x+d)^(1/2),x)

[Out]

int(x^2*(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^2*(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 = 1.45459, size = 748, normalized size = 4.43 \begin{align*} \left [\frac{2 \,{\left (120 \, b d^{\frac{5}{2}} n \log \left (\frac{e x + 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) -{\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \,{\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \,{\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, e^{3}}, -\frac{2 \,{\left (240 \, b \sqrt{-d} d^{2} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \,{\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \,{\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \,{\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \,{\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{225 \, e^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/225*(120*b*d^(5/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (188*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n
- 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a*d*e)*x - 15*(3*b*e^2*x^2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*
x^2 - 4*b*d*e*n*x + 8*b*d^2*n)*log(x))*sqrt(e*x + d))/e^3, -2/225*(240*b*sqrt(-d)*d^2*n*arctan(sqrt(e*x + d)*s
qrt(-d)/d) + (188*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n - 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a*d*e)*x - 15*(3*b*e^
2*x^2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^2 - 4*b*d*e*n*x + 8*b*d^2*n)*log(x))*sqrt(e*x + d))/e^3]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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^{2}}{\sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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