3.3.65 \(\int x (d+e x^2)^{3/2} (a+b \log (c x^n)) \, dx\) [265]

Optimal. Leaf size=125 \[ -\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, 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 )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}+\frac {b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e}-\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*d^2*n*Sqrt[d + e*x^2])/e - (b*d*n*(d + e*x^2)^(3/2))/(15*e) - (b*n*(d + e*x^2)^(5/2))/(25*e) + (b*d^(5
/2)*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(5*e) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*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 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {(b n) \int \frac {\left (d+e x^2\right )^{5/2}}{x} \, dx}{5 e}\\ &=\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {(b n) \text {Subst}\left (\int \frac {(d+e x)^{5/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {(b d n) \text {Subst}\left (\int \frac {(d+e x)^{3/2}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {\left (b d^2 n\right ) \text {Subst}\left (\int \frac {\sqrt {d+e x}}{x} \, dx,x,x^2\right )}{10 e}\\ &=-\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {\left (b d^3 n\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{10 e}\\ &=-\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {\left (b d^3 n\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{5 e^2}\\ &=-\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {b d^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 181, normalized size = 1.45 \begin {gather*} -\frac {b d^{5/2} n \log (x)}{5 e}+\frac {b n \left (d+e x^2\right )^{5/2} \log (x)}{5 e}+\sqrt {d+e x^2} \left (\frac {1}{25} e x^4 \left (5 a-b n+5 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d^2 \left (15 a-23 b n+15 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{75 e}+\frac {1}{75} d x^2 \left (30 a-11 b n+30 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+\frac {b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(b*d^(5/2)*n*Log[x])/e + (b*n*(d + e*x^2)^(5/2)*Log[x])/(5*e) + Sqrt[d + e*x^2]*((e*x^4*(5*a - b*n + 5*b*
(-(n*Log[x]) + Log[c*x^n])))/25 + (d^2*(15*a - 23*b*n + 15*b*(-(n*Log[x]) + Log[c*x^n])))/(75*e) + (d*x^2*(30*
a - 11*b*n + 30*b*(-(n*Log[x]) + Log[c*x^n])))/75) + (b*d^(5/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(5*e)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 100, normalized size = 0.80 \begin {gather*} \frac {1}{5} \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} b e^{\left (-1\right )} \log \left (c x^{n}\right ) + \frac {1}{5} \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} a e^{\left (-1\right )} + \frac {1}{75} \, {\left (15 \, d^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {\sqrt {d} e^{\left (-\frac {1}{2}\right )}}{{\left | x \right |}}\right ) - 3 \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} - 5 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d - 15 \, \sqrt {x^{2} e + d} d^{2}\right )} b n e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(x^2*e + d)^(5/2)*b*e^(-1)*log(c*x^n) + 1/5*(x^2*e + d)^(5/2)*a*e^(-1) + 1/75*(15*d^(5/2)*arcsinh(sqrt(d)*
e^(-1/2)/abs(x)) - 3*(x^2*e + d)^(5/2) - 5*(x^2*e + d)^(3/2)*d - 15*sqrt(x^2*e + d)*d^2)*b*n*e^(-1)

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Fricas [A]
time = 0.44, size = 299, normalized size = 2.39 \begin {gather*} \left [\frac {1}{150} \, {\left (15 \, b d^{\frac {5}{2}} n \log \left (-\frac {x^{2} e + 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} + 23 \, b d^{2} n + {\left (11 \, b d n - 30 \, a d\right )} x^{2} e - 15 \, a d^{2} - 15 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}\right )} e^{\left (-1\right )}, -\frac {1}{75} \, {\left (15 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (3 \, {\left (b n - 5 \, a\right )} x^{4} e^{2} + 23 \, b d^{2} n + {\left (11 \, b d n - 30 \, a d\right )} x^{2} e - 15 \, a d^{2} - 15 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} 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*(e*x^2+d)^(3/2)*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

[1/150*(15*b*d^(5/2)*n*log(-(x^2*e + 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) - 2*(3*(b*n - 5*a)*x^4*e^2 + 23*b*d
^2*n + (11*b*d*n - 30*a*d)*x^2*e - 15*a*d^2 - 15*(b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*log(c) - 15*(b*n*x^4*e^2 +
2*b*d*n*x^2*e + b*d^2*n)*log(x))*sqrt(x^2*e + d))*e^(-1), -1/75*(15*b*sqrt(-d)*d^2*n*arctan(sqrt(-d)/sqrt(x^2*
e + d)) + (3*(b*n - 5*a)*x^4*e^2 + 23*b*d^2*n + (11*b*d*n - 30*a*d)*x^2*e - 15*a*d^2 - 15*(b*x^4*e^2 + 2*b*d*x
^2*e + b*d^2)*log(c) - 15*(b*n*x^4*e^2 + 2*b*d*n*x^2*e + b*d^2*n)*log(x))*sqrt(x^2*e + d))*e^(-1)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a + b \log {\left (c x^{n} \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (e\,x^2+d\right )}^{3/2}\,\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)^(3/2)*(a + b*log(c*x^n)),x)

[Out]

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

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