3.305 \(\int \frac {x^2 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d e^{3/2}}+\frac {b n x}{3 d e \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2335, 288, 217, 206} \[ \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d e^{3/2}}+\frac {b n x}{3 d e \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 2335

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

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 101, normalized size = 1.13 \[ \frac {\sqrt {e} x \left (a e x^2+b n \left (d+e x^2\right )\right )+b e^{3/2} x^3 \log \left (c x^n\right )-b n \left (d+e x^2\right )^{3/2} \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{3 d e^{3/2} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 0.47, size = 277, normalized size = 3.11 \[ \left [\frac {{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + 2 \, {\left (b e^{2} n x^{3} \log \relax (x) + b e^{2} x^{3} \log \relax (c) + b d e n x + {\left (b e^{2} n + a e^{2}\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (d e^{4} x^{4} + 2 \, d^{2} e^{3} x^{2} + d^{3} e^{2}\right )}}, \frac {{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (b e^{2} n x^{3} \log \relax (x) + b e^{2} x^{3} \log \relax (c) + b d e n x + {\left (b e^{2} n + a e^{2}\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d e^{4} x^{4} + 2 \, d^{2} e^{3} x^{2} + d^{3} e^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/6*((b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + 2*(b*e
^2*n*x^3*log(x) + b*e^2*x^3*log(c) + b*d*e*n*x + (b*e^2*n + a*e^2)*x^3)*sqrt(e*x^2 + d))/(d*e^4*x^4 + 2*d^2*e^
3*x^2 + d^3*e^2), 1/3*((b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (
b*e^2*n*x^3*log(x) + b*e^2*x^3*log(c) + b*d*e*n*x + (b*e^2*n + a*e^2)*x^3)*sqrt(e*x^2 + d))/(d*e^4*x^4 + 2*d^2
*e^3*x^2 + d^3*e^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{2}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^2/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) x^{2}}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*ln(c*x^n)+a)/(e*x^2+d)^(5/2),x)

[Out]

int(x^2*(b*ln(c*x^n)+a)/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} e} - \frac {x}{\sqrt {e x^{2} + d} d e}\right )} + b \int \frac {x^{2} \log \relax (c) + x^{2} \log \left (x^{n}\right )}{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(x/((e*x^2 + d)^(3/2)*e) - x/(sqrt(e*x^2 + d)*d*e)) + b*integrate((x^2*log(c) + x^2*log(x^n))/((e^2*x^4
+ 2*d*e*x^2 + d^2)*sqrt(e*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2),x)

[Out]

int((x^2*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2), 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**2+d)**(5/2),x)

[Out]

Timed out

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