3.138 \(\int x^2 (d+e x)^{3/2} (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=213 \[ \frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{32 b d^4 n \sqrt{d+e x}}{315 e^3}-\frac{32 b d^3 n (d+e x)^{3/2}}{945 e^3}-\frac{32 b d^2 n (d+e x)^{5/2}}{1575 e^3}+\frac{32 b d^{9/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{315 e^3}+\frac{44 b d n (d+e x)^{7/2}}{441 e^3}-\frac{4 b n (d+e x)^{9/2}}{81 e^3} \]

[Out]

(-32*b*d^4*n*Sqrt[d + e*x])/(315*e^3) - (32*b*d^3*n*(d + e*x)^(3/2))/(945*e^3) - (32*b*d^2*n*(d + e*x)^(5/2))/
(1575*e^3) + (44*b*d*n*(d + e*x)^(7/2))/(441*e^3) - (4*b*n*(d + e*x)^(9/2))/(81*e^3) + (32*b*d^(9/2)*n*ArcTanh
[Sqrt[d + e*x]/Sqrt[d]])/(315*e^3) + (2*d^2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) - (4*d*(d + e*x)^(7/2)
*(a + b*Log[c*x^n]))/(7*e^3) + (2*(d + e*x)^(9/2)*(a + b*Log[c*x^n]))/(9*e^3)

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Rubi [A]  time = 0.197397, antiderivative size = 213, 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} \[ \frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{32 b d^4 n \sqrt{d+e x}}{315 e^3}-\frac{32 b d^3 n (d+e x)^{3/2}}{945 e^3}-\frac{32 b d^2 n (d+e x)^{5/2}}{1575 e^3}+\frac{32 b d^{9/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{315 e^3}+\frac{44 b d n (d+e x)^{7/2}}{441 e^3}-\frac{4 b n (d+e x)^{9/2}}{81 e^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

(-32*b*d^4*n*Sqrt[d + e*x])/(315*e^3) - (32*b*d^3*n*(d + e*x)^(3/2))/(945*e^3) - (32*b*d^2*n*(d + e*x)^(5/2))/
(1575*e^3) + (44*b*d*n*(d + e*x)^(7/2))/(441*e^3) - (4*b*n*(d + e*x)^(9/2))/(81*e^3) + (32*b*d^(9/2)*n*ArcTanh
[Sqrt[d + e*x]/Sqrt[d]])/(315*e^3) + (2*d^2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) - (4*d*(d + e*x)^(7/2)
*(a + b*Log[c*x^n]))/(7*e^3) + (2*(d + e*x)^(9/2)*(a + b*Log[c*x^n]))/(9*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 x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-(b n) \int \frac{2 (d+e x)^{5/2} \left (8 d^2-20 d e x+35 e^2 x^2\right )}{315 e^3 x} \, dx\\ &=\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{(2 b n) \int \frac{(d+e x)^{5/2} \left (8 d^2-20 d e x+35 e^2 x^2\right )}{x} \, dx}{315 e^3}\\ &=\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \frac{x^6 \left (63 d^2-90 d x^2+35 x^4\right )}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{315 e^4}\\ &=\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{(4 b n) \operatorname{Subst}\left (\int \left (8 d^4 e+8 d^3 e x^2+8 d^2 e x^4-55 d e x^6+35 e x^8+\frac{8 d^5}{-\frac{d}{e}+\frac{x^2}{e}}\right ) \, dx,x,\sqrt{d+e x}\right )}{315 e^4}\\ &=-\frac{32 b d^4 n \sqrt{d+e x}}{315 e^3}-\frac{32 b d^3 n (d+e x)^{3/2}}{945 e^3}-\frac{32 b d^2 n (d+e x)^{5/2}}{1575 e^3}+\frac{44 b d n (d+e x)^{7/2}}{441 e^3}-\frac{4 b n (d+e x)^{9/2}}{81 e^3}+\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac{\left (32 b d^5 n\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{315 e^4}\\ &=-\frac{32 b d^4 n \sqrt{d+e x}}{315 e^3}-\frac{32 b d^3 n (d+e x)^{3/2}}{945 e^3}-\frac{32 b d^2 n (d+e x)^{5/2}}{1575 e^3}+\frac{44 b d n (d+e x)^{7/2}}{441 e^3}-\frac{4 b n (d+e x)^{9/2}}{81 e^3}+\frac{32 b d^{9/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{315 e^3}+\frac{2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac{4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac{2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}\\ \end{align*}

Mathematica [A]  time = 0.229311, size = 153, normalized size = 0.72 \[ \frac{2 \left (\sqrt{d+e x} \left (315 a \left (8 d^2-20 d e x+35 e^2 x^2\right ) (d+e x)^2+315 b \left (8 d^2-20 d e x+35 e^2 x^2\right ) (d+e x)^2 \log \left (c x^n\right )-2 b n \left (429 d^2 e^2 x^2-677 d^3 e x+2614 d^4+2425 d e^3 x^3+1225 e^4 x^4\right )\right )+5040 b d^{9/2} n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )}{99225 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]),x]

[Out]

(2*(5040*b*d^(9/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(315*a*(d + e*x)^2*(8*d^2 - 20*d*e*x + 35*
e^2*x^2) - 2*b*n*(2614*d^4 - 677*d^3*e*x + 429*d^2*e^2*x^2 + 2425*d*e^3*x^3 + 1225*e^4*x^4) + 315*b*(d + e*x)^
2*(8*d^2 - 20*d*e*x + 35*e^2*x^2)*Log[c*x^n])))/(99225*e^3)

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Maple [F]  time = 0.561, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( ex+d \right ) ^{{\frac{3}{2}}} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(e*x+d)^(3/2)*(a+b*ln(c*x^n)),x)

[Out]

int(x^2*(e*x+d)^(3/2)*(a+b*ln(c*x^n)),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*(e*x+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47006, size = 1223, normalized size = 5.74 \begin{align*} \left [\frac{2 \,{\left (2520 \, b d^{\frac{9}{2}} n \log \left (\frac{e x + 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) -{\left (5228 \, b d^{4} n - 2520 \, a d^{4} + 1225 \,{\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} + 50 \,{\left (97 \, b d e^{3} n - 315 \, a d e^{3}\right )} x^{3} + 3 \,{\left (286 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 2 \,{\left (677 \, b d^{3} e n - 630 \, a d^{3} e\right )} x - 315 \,{\left (35 \, b e^{4} x^{4} + 50 \, b d e^{3} x^{3} + 3 \, b d^{2} e^{2} x^{2} - 4 \, b d^{3} e x + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \,{\left (35 \, b e^{4} n x^{4} + 50 \, b d e^{3} n x^{3} + 3 \, b d^{2} e^{2} n x^{2} - 4 \, b d^{3} e n x + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{99225 \, e^{3}}, -\frac{2 \,{\left (5040 \, b \sqrt{-d} d^{4} n \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) +{\left (5228 \, b d^{4} n - 2520 \, a d^{4} + 1225 \,{\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} + 50 \,{\left (97 \, b d e^{3} n - 315 \, a d e^{3}\right )} x^{3} + 3 \,{\left (286 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 2 \,{\left (677 \, b d^{3} e n - 630 \, a d^{3} e\right )} x - 315 \,{\left (35 \, b e^{4} x^{4} + 50 \, b d e^{3} x^{3} + 3 \, b d^{2} e^{2} x^{2} - 4 \, b d^{3} e x + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \,{\left (35 \, b e^{4} n x^{4} + 50 \, b d e^{3} n x^{3} + 3 \, b d^{2} e^{2} n x^{2} - 4 \, b d^{3} e n x + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt{e x + d}\right )}}{99225 \, e^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

[2/99225*(2520*b*d^(9/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (5228*b*d^4*n - 2520*a*d^4 + 1225*(2
*b*e^4*n - 9*a*e^4)*x^4 + 50*(97*b*d*e^3*n - 315*a*d*e^3)*x^3 + 3*(286*b*d^2*e^2*n - 315*a*d^2*e^2)*x^2 - 2*(6
77*b*d^3*e*n - 630*a*d^3*e)*x - 315*(35*b*e^4*x^4 + 50*b*d*e^3*x^3 + 3*b*d^2*e^2*x^2 - 4*b*d^3*e*x + 8*b*d^4)*
log(c) - 315*(35*b*e^4*n*x^4 + 50*b*d*e^3*n*x^3 + 3*b*d^2*e^2*n*x^2 - 4*b*d^3*e*n*x + 8*b*d^4*n)*log(x))*sqrt(
e*x + d))/e^3, -2/99225*(5040*b*sqrt(-d)*d^4*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (5228*b*d^4*n - 2520*a*d^4 +
1225*(2*b*e^4*n - 9*a*e^4)*x^4 + 50*(97*b*d*e^3*n - 315*a*d*e^3)*x^3 + 3*(286*b*d^2*e^2*n - 315*a*d^2*e^2)*x^
2 - 2*(677*b*d^3*e*n - 630*a*d^3*e)*x - 315*(35*b*e^4*x^4 + 50*b*d*e^3*x^3 + 3*b*d^2*e^2*x^2 - 4*b*d^3*e*x + 8
*b*d^4)*log(c) - 315*(35*b*e^4*n*x^4 + 50*b*d*e^3*n*x^3 + 3*b*d^2*e^2*n*x^2 - 4*b*d^3*e*n*x + 8*b*d^4*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*(e*x+d)**(3/2)*(a+b*ln(c*x**n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(e*x+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

integrate((e*x + d)^(3/2)*(b*log(c*x^n) + a)*x^2, x)