3.3.53 \(\int x \sqrt {d+e x^2} (a+b \log (c x^n)) \, dx\) [253]

Optimal. Leaf size=102 \[ -\frac {b d n \sqrt {d+e x^2}}{3 e}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac {b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e} \]

[Out]

-1/9*b*n*(e*x^2+d)^(3/2)/e+1/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e+1/3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^
n))/e-1/3*b*d*n*(e*x^2+d)^(1/2)/e

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Rubi [A]
time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2376, 272, 52, 65, 214} \begin {gather*} \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}+\frac {b d^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e}-\frac {b d n \sqrt {d+e x^2}}{3 e}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*d*n*Sqrt[d + e*x^2])/e - (b*n*(d + e*x^2)^(3/2))/(9*e) + (b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]
)/(3*e) + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2376

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 136, normalized size = 1.33 \begin {gather*} \frac {3 a d \sqrt {d+e x^2}-4 b d n \sqrt {d+e x^2}+3 a e x^2 \sqrt {d+e x^2}-b e n x^2 \sqrt {d+e x^2}-3 b d^{3/2} n \log (x)+3 b \left (d+e x^2\right )^{3/2} \log \left (c x^n\right )+3 b d^{3/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{9 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*d*Sqrt[d + e*x^2] - 4*b*d*n*Sqrt[d + e*x^2] + 3*a*e*x^2*Sqrt[d + e*x^2] - b*e*n*x^2*Sqrt[d + e*x^2] - 3*b
*d^(3/2)*n*Log[x] + 3*b*(d + e*x^2)^(3/2)*Log[c*x^n] + 3*b*d^(3/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(9*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 85, normalized size = 0.83 \begin {gather*} \frac {1}{3} \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} b e^{\left (-1\right )} \log \left (c x^{n}\right ) + \frac {1}{9} \, {\left (3 \, d^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {\sqrt {d} e^{\left (-\frac {1}{2}\right )}}{{\left | x \right |}}\right ) - {\left (x^{2} e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x^{2} e + d} d\right )} b n e^{\left (-1\right )} + \frac {1}{3} \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} a e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(x^2*e + d)^(3/2)*b*e^(-1)*log(c*x^n) + 1/9*(3*d^(3/2)*arcsinh(sqrt(d)*e^(-1/2)/abs(x)) - (x^2*e + d)^(3/2
) - 3*sqrt(x^2*e + d)*d)*b*n*e^(-1) + 1/3*(x^2*e + d)^(3/2)*a*e^(-1)

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Fricas [A]
time = 0.40, size = 209, normalized size = 2.05 \begin {gather*} \left [\frac {1}{18} \, {\left (3 \, b d^{\frac {3}{2}} n \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left ({\left (b n - 3 \, a\right )} x^{2} e + 4 \, b d n - 3 \, a d - 3 \, {\left (b x^{2} e + b d\right )} \log \left (c\right ) - 3 \, {\left (b n x^{2} e + b d n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-1\right )}, -\frac {1}{9} \, {\left (3 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left ({\left (b n - 3 \, a\right )} x^{2} e + 4 \, b d n - 3 \, a d - 3 \, {\left (b x^{2} e + b d\right )} \log \left (c\right ) - 3 \, {\left (b n x^{2} e + b d n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-1\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*b*d^(3/2)*n*log(-(x^2*e + 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) - 2*((b*n - 3*a)*x^2*e + 4*b*d*n - 3*
a*d - 3*(b*x^2*e + b*d)*log(c) - 3*(b*n*x^2*e + b*d*n)*log(x))*sqrt(x^2*e + d))*e^(-1), -1/9*(3*b*sqrt(-d)*d*n
*arctan(sqrt(-d)/sqrt(x^2*e + d)) + ((b*n - 3*a)*x^2*e + 4*b*d*n - 3*a*d - 3*(b*x^2*e + b*d)*log(c) - 3*(b*n*x
^2*e + b*d*n)*log(x))*sqrt(x^2*e + d))*e^(-1)]

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Sympy [A]
time = 12.32, size = 155, normalized size = 1.52 \begin {gather*} a \left (\begin {cases} \frac {\sqrt {d} x^{2}}{2} & \text {for}\: e = 0 \\\frac {\left (d + e x^{2}\right )^{\frac {3}{2}}}{3 e} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {\sqrt {d} x^{2}}{4} & \text {for}\: e = 0 \\\frac {4 d^{\frac {3}{2}} \sqrt {1 + \frac {e x^{2}}{d}}}{9 e} + \frac {d^{\frac {3}{2}} \log {\left (\frac {e x^{2}}{d} \right )}}{6 e} - \frac {d^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{3 e} + \frac {\sqrt {d} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}}{9} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {\sqrt {d} x^{2}}{2} & \text {for}\: e = 0 \\\frac {\left (d + e x^{2}\right )^{\frac {3}{2}}}{3 e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((sqrt(d)*x**2/2, Eq(e, 0)), ((d + e*x**2)**(3/2)/(3*e), True)) - b*n*Piecewise((sqrt(d)*x**2/4, Eq
(e, 0)), (4*d**(3/2)*sqrt(1 + e*x**2/d)/(9*e) + d**(3/2)*log(e*x**2/d)/(6*e) - d**(3/2)*log(sqrt(1 + e*x**2/d)
+ 1)/(3*e) + sqrt(d)*x**2*sqrt(1 + e*x**2/d)/9, True)) + b*Piecewise((sqrt(d)*x**2/2, Eq(e, 0)), ((d + e*x**2
)**(3/2)/(3*e), True))*log(c*x**n)

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Giac [A]
time = 5.38, size = 145, normalized size = 1.42 \begin {gather*} \frac {1}{3} \, \sqrt {x^{2} e + d} b x^{2} \log \left (c\right ) + \frac {1}{3} \, \sqrt {x^{2} e + d} b d e^{\left (-1\right )} \log \left (c\right ) + \frac {1}{3} \, \sqrt {x^{2} e + d} a x^{2} + \frac {1}{3} \, \sqrt {x^{2} e + d} a d e^{\left (-1\right )} + \frac {1}{9} \, {\left (3 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} e^{\left (-1\right )} \log \left (x\right ) - {\left (\frac {3 \, d^{2} \arctan \left (\frac {\sqrt {x^{2} e + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + {\left (x^{2} e + d\right )}^{\frac {3}{2}} + 3 \, \sqrt {x^{2} e + d} d\right )} e^{\left (-1\right )}\right )} b n \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(x^2*e + d)*b*x^2*log(c) + 1/3*sqrt(x^2*e + d)*b*d*e^(-1)*log(c) + 1/3*sqrt(x^2*e + d)*a*x^2 + 1/3*sqr
t(x^2*e + d)*a*d*e^(-1) + 1/9*(3*(x^2*e + d)^(3/2)*e^(-1)*log(x) - (3*d^2*arctan(sqrt(x^2*e + d)/sqrt(-d))/sqr
t(-d) + (x^2*e + d)^(3/2) + 3*sqrt(x^2*e + d)*d)*e^(-1))*b*n

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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