3.2.45 \(\int \frac {x^2 (a+b \log (c x^n))}{\sqrt {d+e x}} \, dx\) [145]

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

[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 = 0.12, 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 = {45, 2392, 12, 911, 1275, 214} \begin {gather*} \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} \end {gather*}

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 45

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 214

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

Rule 911

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 1275

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 2392

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) \text {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) \text {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 ) \text {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 = 0.12, size = 118, normalized size = 0.70 \begin {gather*} \frac {480 b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )-2 b n \left (94 d^2-17 d e x+9 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 )\right )}{225 e^3} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}\, dx\]

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 [A]
time = 0.49, size = 175, normalized size = 1.04 \begin {gather*} -\frac {4}{225} \, {\left (60 \, d^{\frac {5}{2}} e^{\left (-3\right )} \log \left (\frac {\sqrt {x e + d} - \sqrt {d}}{\sqrt {x e + d} + \sqrt {d}}\right ) + {\left (9 \, {\left (x e + d\right )}^{\frac {5}{2}} - 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {x e + d} d^{2}\right )} e^{\left (-3\right )}\right )} b n + \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} e^{\left (-3\right )} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d e^{\left (-3\right )} + 15 \, \sqrt {x e + d} d^{2} e^{\left (-3\right )}\right )} b \log \left (c x^{n}\right ) + \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} e^{\left (-3\right )} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d e^{\left (-3\right )} + 15 \, \sqrt {x e + d} d^{2} e^{\left (-3\right )}\right )} a \end {gather*}

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

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Fricas [A]
time = 0.40, size = 291, normalized size = 1.72 \begin {gather*} \left [\frac {2}{225} \, {\left (120 \, b d^{\frac {5}{2}} n \log \left (\frac {x e + 2 \, \sqrt {x e + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (188 \, b d^{2} n + 9 \, {\left (2 \, b n - 5 \, a\right )} x^{2} e^{2} - 120 \, a d^{2} - 2 \, {\left (17 \, b d n - 30 \, a d\right )} x e - 15 \, {\left (3 \, b x^{2} e^{2} - 4 \, b d x e + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{2} e^{2} - 4 \, b d n x e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-3\right )}, -\frac {2}{225} \, {\left (240 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {x e + d} \sqrt {-d}}{d}\right ) + {\left (188 \, b d^{2} n + 9 \, {\left (2 \, b n - 5 \, a\right )} x^{2} e^{2} - 120 \, a d^{2} - 2 \, {\left (17 \, b d n - 30 \, a d\right )} x e - 15 \, {\left (3 \, b x^{2} e^{2} - 4 \, b d x e + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b n x^{2} e^{2} - 4 \, b d n x e + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x e + d}\right )} e^{\left (-3\right )}\right ] \end {gather*}

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 714 vs. \(2 (170) = 340\).
time = 122.51, size = 714, normalized size = 4.22 \begin {gather*} \begin {cases} \frac {- \frac {2 a d \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^{2}} - \frac {2 a \left (- \frac {d^{3}}{\sqrt {d + e x}} - 3 d^{2} \sqrt {d + e x} + d \left (d + e x\right )^{\frac {3}{2}} - \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}} - \frac {2 b d \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^{2}} - \frac {2 b \left (- d^{3} \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 ) + 3 d^{2} \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 ) - 3 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 (- 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 ) - \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 (- d^{2} e \sqrt {d + e x} - \frac {d^{2} e \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{d}} \sqrt {d + e x}} \right )}}{\sqrt {- \frac {1}{d}}} - \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 )}{e^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {\frac {a x^{3}}{3} + b \left (- \frac {n x^{3}}{9} + \frac {x^{3} \log {\left (c x^{n} \right )}}{3}\right )}{\sqrt {d}} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise(((-2*a*d*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e**2 - 2*a*(-d**3/sqrt(d + e*
x) - 3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**2 - 2*b*d*(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**2 - 2*b*(-d**3*(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))) + 3*d**2*(-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) - 3*d*(-(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)) - (d + e*x)**(5/2)*log(c*(-d/e + (d + e*x)/e)**n)/5 - 2*n*(-d**2*e*sqrt(d + e*x) - d*
*2*e*atan(1/(sqrt(-1/d)*sqrt(d + e*x)))/sqrt(-1/d) - d*e*(d + e*x)**(3/2)/3 - e*(d + e*x)**(5/2)/5)/(5*e))/e**
2)/e, Ne(e, 0)), ((a*x**3/3 + b*(-n*x**3/9 + x**3*log(c*x**n)/3))/sqrt(d), True))

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Giac [A]
time = 5.97, size = 210, normalized size = 1.24 \begin {gather*} -\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 \left (c\right ) - 150 \, {\left (x e + d\right )}^{\frac {3}{2}} b d \log \left (c\right ) + 225 \, \sqrt {x e + d} b d^{2} \log \left (c\right ) + 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 )} \end {gather*}

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

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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