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

Optimal. Leaf size=139 \[ \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{3 d^6}-\frac {b e^6 n \log (x)}{6 d^6}+\frac {b e^5 n \sqrt {x}}{3 d^5}-\frac {b e^4 n x}{6 d^4}+\frac {b e^3 n x^{3/2}}{9 d^3}-\frac {b e^2 n x^2}{12 d^2}+\frac {b e n x^{5/2}}{15 d} \]

[Out]

-1/6*b*e^4*n*x/d^4+1/9*b*e^3*n*x^(3/2)/d^3-1/12*b*e^2*n*x^2/d^2+1/15*b*e*n*x^(5/2)/d-1/6*b*e^6*n*ln(x)/d^6-1/3
*b*e^6*n*ln(d+e/x^(1/2))/d^6+1/3*x^3*(a+b*ln(c*(d+e/x^(1/2))^n))+1/3*b*e^5*n*x^(1/2)/d^5

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Rubi [A]  time = 0.10, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2454, 2395, 44} \[ \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )+\frac {b e^3 n x^{3/2}}{9 d^3}-\frac {b e^2 n x^2}{12 d^2}+\frac {b e^5 n \sqrt {x}}{3 d^5}-\frac {b e^4 n x}{6 d^4}-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{3 d^6}-\frac {b e^6 n \log (x)}{6 d^6}+\frac {b e n x^{5/2}}{15 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^5*n*Sqrt[x])/(3*d^5) - (b*e^4*n*x)/(6*d^4) + (b*e^3*n*x^(3/2))/(9*d^3) - (b*e^2*n*x^2)/(12*d^2) + (b*e*n*
x^(5/2))/(15*d) - (b*e^6*n*Log[d + e/Sqrt[x]])/(3*d^6) + (x^3*(a + b*Log[c*(d + e/Sqrt[x])^n]))/3 - (b*e^6*n*L
og[x])/(6*d^6)

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x^7} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \operatorname {Subst}\left (\int \frac {1}{x^6 (d+e x)} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {1}{3} (b e n) \operatorname {Subst}\left (\int \left (\frac {1}{d x^6}-\frac {e}{d^2 x^5}+\frac {e^2}{d^3 x^4}-\frac {e^3}{d^4 x^3}+\frac {e^4}{d^5 x^2}-\frac {e^5}{d^6 x}+\frac {e^6}{d^6 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=\frac {b e^5 n \sqrt {x}}{3 d^5}-\frac {b e^4 n x}{6 d^4}+\frac {b e^3 n x^{3/2}}{9 d^3}-\frac {b e^2 n x^2}{12 d^2}+\frac {b e n x^{5/2}}{15 d}-\frac {b e^6 n \log \left (d+\frac {e}{\sqrt {x}}\right )}{3 d^6}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-\frac {b e^6 n \log (x)}{6 d^6}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 130, normalized size = 0.94 \[ \frac {a x^3}{3}+\frac {1}{3} b x^3 \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )-\frac {1}{3} b e n \left (\frac {e^5 \log \left (d+\frac {e}{\sqrt {x}}\right )}{d^6}+\frac {e^5 \log (x)}{2 d^6}-\frac {e^4 \sqrt {x}}{d^5}+\frac {e^3 x}{2 d^4}-\frac {e^2 x^{3/2}}{3 d^3}+\frac {e x^2}{4 d^2}-\frac {x^{5/2}}{5 d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^3)/3 + (b*x^3*Log[c*(d + e/Sqrt[x])^n])/3 - (b*e*n*(-((e^4*Sqrt[x])/d^5) + (e^3*x)/(2*d^4) - (e^2*x^(3/2)
)/(3*d^3) + (e*x^2)/(4*d^2) - x^(5/2)/(5*d) + (e^5*Log[d + e/Sqrt[x]])/d^6 + (e^5*Log[x])/(2*d^6)))/3

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fricas [A]  time = 0.45, size = 153, normalized size = 1.10 \[ \frac {60 \, b d^{6} x^{3} \log \relax (c) - 15 \, b d^{4} e^{2} n x^{2} + 60 \, a d^{6} x^{3} - 30 \, b d^{2} e^{4} n x - 60 \, b d^{6} n \log \left (\sqrt {x}\right ) + 60 \, {\left (b d^{6} - b e^{6}\right )} n \log \left (d \sqrt {x} + e\right ) + 60 \, {\left (b d^{6} n x^{3} - b d^{6} n\right )} \log \left (\frac {d x + e \sqrt {x}}{x}\right ) + 4 \, {\left (3 \, b d^{5} e n x^{2} + 5 \, b d^{3} e^{3} n x + 15 \, b d e^{5} n\right )} \sqrt {x}}{180 \, d^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/180*(60*b*d^6*x^3*log(c) - 15*b*d^4*e^2*n*x^2 + 60*a*d^6*x^3 - 30*b*d^2*e^4*n*x - 60*b*d^6*n*log(sqrt(x)) +
60*(b*d^6 - b*e^6)*n*log(d*sqrt(x) + e) + 60*(b*d^6*n*x^3 - b*d^6*n)*log((d*x + e*sqrt(x))/x) + 4*(3*b*d^5*e*n
*x^2 + 5*b*d^3*e^3*n*x + 15*b*d*e^5*n)*sqrt(x))/d^6

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giac [B]  time = 0.39, size = 236, normalized size = 1.70 \[ \frac {1}{3} \, b x^{3} \log \relax (c) + \frac {1}{3} \, a x^{3} - \frac {1}{180} \, {\left ({\left (\frac {60 \, \log \left (\frac {{\left | d \sqrt {x} + e \right |}}{\sqrt {{\left | x \right |}}}\right )}{d^{6}} - \frac {60 \, \log \left ({\left | -d + \frac {d \sqrt {x} + e}{\sqrt {x}} \right |}\right )}{d^{6}} + \frac {137 \, d^{5} - \frac {385 \, {\left (d \sqrt {x} + e\right )} d^{4}}{\sqrt {x}} + \frac {470 \, {\left (d \sqrt {x} + e\right )}^{2} d^{3}}{x} - \frac {270 \, {\left (d \sqrt {x} + e\right )}^{3} d^{2}}{x^{\frac {3}{2}}} + \frac {60 \, {\left (d \sqrt {x} + e\right )}^{4} d}{x^{2}}}{{\left (d - \frac {d \sqrt {x} + e}{\sqrt {x}}\right )}^{5} d^{6}}\right )} e^{7} - \frac {60 \, e^{7} \log \left ({\left (d e^{\left (-1\right )} - \frac {{\left (d \sqrt {x} + e\right )} e^{\left (-1\right )}}{\sqrt {x}}\right )} {\left (\frac {d}{d e^{\left (-1\right )} - \frac {{\left (d \sqrt {x} + e\right )} e^{\left (-1\right )}}{\sqrt {x}}} - e\right )}\right )}{{\left (d - \frac {d \sqrt {x} + e}{\sqrt {x}}\right )}^{6}}\right )} b n e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3*log(c) + 1/3*a*x^3 - 1/180*((60*log(abs(d*sqrt(x) + e)/sqrt(abs(x)))/d^6 - 60*log(abs(-d + (d*sqrt(x
) + e)/sqrt(x)))/d^6 + (137*d^5 - 385*(d*sqrt(x) + e)*d^4/sqrt(x) + 470*(d*sqrt(x) + e)^2*d^3/x - 270*(d*sqrt(
x) + e)^3*d^2/x^(3/2) + 60*(d*sqrt(x) + e)^4*d/x^2)/((d - (d*sqrt(x) + e)/sqrt(x))^5*d^6))*e^7 - 60*e^7*log((d
*e^(-1) - (d*sqrt(x) + e)*e^(-1)/sqrt(x))*(d/(d*e^(-1) - (d*sqrt(x) + e)*e^(-1)/sqrt(x)) - e))/(d - (d*sqrt(x)
+ e)/sqrt(x))^6)*b*n*e^(-1)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \left (d +\frac {e}{\sqrt {x}}\right )^{n}\right )+a \right ) x^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 96, normalized size = 0.69 \[ \frac {1}{3} \, b x^{3} \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{n}\right ) + \frac {1}{3} \, a x^{3} - \frac {1}{180} \, b e n {\left (\frac {60 \, e^{5} \log \left (d \sqrt {x} + e\right )}{d^{6}} - \frac {12 \, d^{4} x^{\frac {5}{2}} - 15 \, d^{3} e x^{2} + 20 \, d^{2} e^{2} x^{\frac {3}{2}} - 30 \, d e^{3} x + 60 \, e^{4} \sqrt {x}}{d^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*x^3*log(c*(d + e/sqrt(x))^n) + 1/3*a*x^3 - 1/180*b*e*n*(60*e^5*log(d*sqrt(x) + e)/d^6 - (12*d^4*x^(5/2)
- 15*d^3*e*x^2 + 20*d^2*e^2*x^(3/2) - 30*d*e^3*x + 60*e^4*sqrt(x))/d^5)

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mupad [B]  time = 0.69, size = 106, normalized size = 0.76 \[ \frac {a\,x^3}{3}+\frac {b\,\left (60\,d^6\,x^3\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^n\right )-120\,e^6\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,\sqrt {x}}+1\right )-15\,d^4\,e^2\,n\,x^2+20\,d^3\,e^3\,n\,x^{3/2}-30\,d^2\,e^4\,n\,x+60\,d\,e^5\,n\,\sqrt {x}+12\,d^5\,e\,n\,x^{5/2}\right )}{180\,d^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x^3)/3 + (b*(60*d^6*x^3*log(c*(d + e/x^(1/2))^n) - 120*e^6*n*atanh((2*e)/(d*x^(1/2)) + 1) - 15*d^4*e^2*n*x^
2 + 20*d^3*e^3*n*x^(3/2) - 30*d^2*e^4*n*x + 60*d*e^5*n*x^(1/2) + 12*d^5*e*n*x^(5/2)))/(180*d^6)

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sympy [A]  time = 49.02, size = 134, normalized size = 0.96 \[ \frac {a x^{3}}{3} + b \left (\frac {e n \left (\frac {2 x^{\frac {5}{2}}}{5 d} - \frac {e x^{2}}{2 d^{2}} + \frac {2 e^{2} x^{\frac {3}{2}}}{3 d^{3}} - \frac {e^{3} x}{d^{4}} + \frac {2 e^{4} \sqrt {x}}{d^{5}} - \frac {2 e^{6} \left (\begin {cases} \frac {1}{d \sqrt {x}} & \text {for}\: e = 0 \\\frac {\log {\left (d + \frac {e}{\sqrt {x}} \right )}}{e} & \text {otherwise} \end {cases}\right )}{d^{6}} + \frac {2 e^{5} \log {\left (\frac {1}{\sqrt {x}} \right )}}{d^{6}}\right )}{6} + \frac {x^{3} \log {\left (c \left (d + \frac {e}{\sqrt {x}}\right )^{n} \right )}}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 + b*(e*n*(2*x**(5/2)/(5*d) - e*x**2/(2*d**2) + 2*e**2*x**(3/2)/(3*d**3) - e**3*x/d**4 + 2*e**4*sqrt(x
)/d**5 - 2*e**6*Piecewise((1/(d*sqrt(x)), Eq(e, 0)), (log(d + e/sqrt(x))/e, True))/d**6 + 2*e**5*log(1/sqrt(x)
)/d**6)/6 + x**3*log(c*(d + e/sqrt(x))**n)/3)

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