3.272 \(\int \frac{(d+e x^2)^{3/2} (a+b \log (c x^n))}{x^6} \, dx\)

Optimal. Leaf size=138 \[ -\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}-\frac{b e^2 n \sqrt{d+e x^2}}{5 d x}+\frac{b e^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 d}-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5} \]

[Out]

-(b*e^2*n*Sqrt[d + e*x^2])/(5*d*x) - (b*e*n*(d + e*x^2)^(3/2))/(15*d*x^3) - (b*n*(d + e*x^2)^(5/2))/(25*d*x^5)
+ (b*e^(5/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(5*d) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*d*x^5)

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Rubi [A]  time = 0.121985, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2335, 277, 217, 206} \[ -\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}-\frac{b e^2 n \sqrt{d+e x^2}}{5 d x}+\frac{b e^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 d}-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-(b*e^2*n*Sqrt[d + e*x^2])/(5*d*x) - (b*e*n*(d + e*x^2)^(3/2))/(15*d*x^3) - (b*n*(d + e*x^2)^(5/2))/(25*d*x^5)
+ (b*e^(5/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(5*d) - ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*d*x^5)

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]

Rule 277

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

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 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])

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx &=-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac{(b n) \int \frac{\left (d+e x^2\right )^{5/2}}{x^6} \, dx}{5 d}\\ &=-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac{(b e n) \int \frac{\left (d+e x^2\right )^{3/2}}{x^4} \, dx}{5 d}\\ &=-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac{\left (b e^2 n\right ) \int \frac{\sqrt{d+e x^2}}{x^2} \, dx}{5 d}\\ &=-\frac{b e^2 n \sqrt{d+e x^2}}{5 d x}-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac{\left (b e^3 n\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{5 d}\\ &=-\frac{b e^2 n \sqrt{d+e x^2}}{5 d x}-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac{\left (b e^3 n\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{5 d}\\ &=-\frac{b e^2 n \sqrt{d+e x^2}}{5 d x}-\frac{b e n \left (d+e x^2\right )^{3/2}}{15 d x^3}-\frac{b n \left (d+e x^2\right )^{5/2}}{25 d x^5}+\frac{b e^{5/2} n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{5 d}-\frac{\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\\ \end{align*}

Mathematica [A]  time = 0.182406, size = 114, normalized size = 0.83 \[ -\frac{\sqrt{d+e x^2} \left (15 a \left (d+e x^2\right )^2+b n \left (3 d^2+11 d e x^2+23 e^2 x^4\right )\right )+15 b \left (d+e x^2\right )^{5/2} \log \left (c x^n\right )-15 b e^{5/2} n x^5 \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{75 d x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-(Sqrt[d + e*x^2]*(15*a*(d + e*x^2)^2 + b*n*(3*d^2 + 11*d*e*x^2 + 23*e^2*x^4)) + 15*b*(d + e*x^2)^(5/2)*Log[c*
x^n] - 15*b*e^(5/2)*n*x^5*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(75*d*x^5)

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Maple [F]  time = 0.48, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{6}} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^6,x)

[Out]

int((e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/x^6,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56912, size = 761, normalized size = 5.51 \begin{align*} \left [\frac{15 \, b e^{\frac{5}{2}} n x^{5} \log \left (-2 \, e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{e} x - d\right ) - 2 \,{\left ({\left (23 \, b e^{2} n + 15 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 15 \, a d^{2} +{\left (11 \, b d e n + 30 \, a d e\right )} x^{2} + 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) + 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{150 \, d x^{5}}, -\frac{15 \, b \sqrt{-e} e^{2} n x^{5} \arctan \left (\frac{\sqrt{-e} x}{\sqrt{e x^{2} + d}}\right ) +{\left ({\left (23 \, b e^{2} n + 15 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 15 \, a d^{2} +{\left (11 \, b d e n + 30 \, a d e\right )} x^{2} + 15 \,{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) + 15 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{75 \, d x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**(3/2)*(a+b*ln(c*x**n))/x**6,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{\frac{3}{2}}{\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^(3/2)*(a+b*log(c*x^n))/x^6,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^(3/2)*(b*log(c*x^n) + a)/x^6, x)