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

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

[Out]

28/45*b*d*n*(e*x+d)^(3/2)/e^3-4/25*b*n*(e*x+d)^(5/2)/e^3+32/15*b*d^(5/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^3-
4/3*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^3+2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^3-32/15*b*d^2*n*(e*x+d)^(1/2)/e^3+
2*d^2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^3

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Rubi [A]  time = 0.17, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 12

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

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 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]

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 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])

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*}

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Mathematica [A]  time = 0.18, size = 118, normalized size = 0.70 \[ \frac {2 \sqrt {d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )+15 b \left (8 d^2-4 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )-2 b n \left (94 d^2-17 d e x+9 e^2 x^2\right )\right )+480 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{225 e^3} \]

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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fricas [A]  time = 0.45, size = 296, normalized size = 1.75 \[ \left [\frac {2 \, {\left (120 \, b d^{\frac {5}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \relax (c) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}, -\frac {2 \, {\left (240 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \relax (c) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}\right ] \]

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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giac [A]  time = 0.94, size = 210, normalized size = 1.24 \[ -\frac {32 \, b d^{3} n \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-d}}\right ) e^{\left (-3\right )}}{15 \, \sqrt {-d}} + \frac {2}{225} \, {\left (45 \, {\left (x e + d\right )}^{\frac {5}{2}} b n \log \left (x e\right ) - 150 \, {\left (x e + d\right )}^{\frac {3}{2}} b d n \log \left (x e\right ) + 225 \, \sqrt {x e + d} b d^{2} n \log \left (x e\right ) - 63 \, {\left (x e + d\right )}^{\frac {5}{2}} b n + 220 \, {\left (x e + d\right )}^{\frac {3}{2}} b d n - 465 \, \sqrt {x e + d} b d^{2} n + 45 \, {\left (x e + d\right )}^{\frac {5}{2}} b \log \relax (c) - 150 \, {\left (x e + d\right )}^{\frac {3}{2}} b d \log \relax (c) + 225 \, \sqrt {x e + d} b d^{2} \log \relax (c) + 45 \, {\left (x e + d\right )}^{\frac {5}{2}} a - 150 \, {\left (x e + d\right )}^{\frac {3}{2}} a d + 225 \, \sqrt {x e + d} a d^{2}\right )} e^{\left (-3\right )} \]

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]

-32/15*b*d^3*n*arctan(sqrt(x*e + d)/sqrt(-d))*e^(-3)/sqrt(-d) + 2/225*(45*(x*e + d)^(5/2)*b*n*log(x*e) - 150*(
x*e + d)^(3/2)*b*d*n*log(x*e) + 225*sqrt(x*e + d)*b*d^2*n*log(x*e) - 63*(x*e + d)^(5/2)*b*n + 220*(x*e + d)^(3
/2)*b*d*n - 465*sqrt(x*e + d)*b*d^2*n + 45*(x*e + d)^(5/2)*b*log(c) - 150*(x*e + d)^(3/2)*b*d*log(c) + 225*sqr
t(x*e + d)*b*d^2*log(c) + 45*(x*e + d)^(5/2)*a - 150*(x*e + d)^(3/2)*a*d + 225*sqrt(x*e + d)*a*d^2)*e^(-3)

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2}}{\sqrt {e x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 172, normalized size = 1.02 \[ -\frac {4}{225} \, b n {\left (\frac {60 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} + \frac {9 \, {\left (e x + d\right )}^{\frac {5}{2}} - 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} + \frac {2}{15} \, b {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \log \left (c x^{n}\right ) + \frac {2}{15} \, a {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \]

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]

-4/225*b*n*(60*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/e^3 + (9*(e*x + d)^(5/2) - 35*
(e*x + d)^(3/2)*d + 120*sqrt(e*x + d)*d^2)/e^3) + 2/15*b*(3*(e*x + d)^(5/2)/e^3 - 10*(e*x + d)^(3/2)*d/e^3 + 1
5*sqrt(e*x + d)*d^2/e^3)*log(c*x^n) + 2/15*a*(3*(e*x + d)^(5/2)/e^3 - 10*(e*x + d)^(3/2)*d/e^3 + 15*sqrt(e*x +
d)*d^2/e^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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