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

Optimal. Leaf size=113 \[ \frac{2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{b n x}{3 d^2 \sqrt{d+e x^2}}-\frac{2 b n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^2 \sqrt{e}} \]

[Out]

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

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Rubi [A]  time = 0.0694104, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {2323, 2314, 217, 206, 191} \[ \frac{2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{b n x}{3 d^2 \sqrt{d+e x^2}}-\frac{2 b n \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{3 d^2 \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2323

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(q + 1
)*(a + b*Log[c*x^n]))/(2*d*(q + 1)), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[c*
x^n]), x], x] + Dist[(b*n)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &&
LtQ[q, -1]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 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 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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

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

Mathematica [A]  time = 0.132886, size = 116, normalized size = 1.03 \[ \frac{\sqrt{e} x \left (a \left (3 d+2 e x^2\right )-b n \left (d+e x^2\right )\right )+b \sqrt{e} x \left (3 d+2 e x^2\right ) \log \left (c x^n\right )-2 b n \left (d+e x^2\right )^{3/2} \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )}{3 d^2 \sqrt{e} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.407, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c{x}^{n} \right ) ) \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{e x^{2} + d} d^{2}} + \frac{x}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} d}\right )} + b \int \frac{\log \left (c\right ) + \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.525, size = 756, normalized size = 6.69 \begin{align*} \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 ) -{\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{3} +{\left (b d e n - 3 \, a d e\right )} x -{\left (2 \, b e^{2} x^{3} + 3 \, b d e x\right )} \log \left (c\right ) -{\left (2 \, b e^{2} n x^{3} + 3 \, b d e n x\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}, \frac{2 \,{\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 ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{3} +{\left (b d e n - 3 \, a d e\right )} x -{\left (2 \, b e^{2} x^{3} + 3 \, b d e x\right )} \log \left (c\right ) -{\left (2 \, b e^{2} n x^{3} + 3 \, b d e n x\right )} \log \left (x\right )\right )} \sqrt{e x^{2} + d}}{3 \,{\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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