3.299 \(\int \frac{x^3 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=108 \[ -\frac{a+b \log \left (c x^n\right )}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{b n}{3 e^2 \sqrt{d+e x^2}}-\frac{2 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 \sqrt{d} e^2} \]

[Out]

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

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Rubi [A]  time = 0.160831, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {266, 43, 2350, 12, 446, 78, 63, 208} \[ -\frac{a+b \log \left (c x^n\right )}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{b n}{3 e^2 \sqrt{d+e x^2}}-\frac{2 b n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{3 \sqrt{d} e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 266

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.268131, size = 137, normalized size = 1.27 \[ \frac{\frac{d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-\left (d+e x^2\right ) \left (3 a+3 b \log \left (c x^n\right )-3 b n \log (x)+b n\right )}{\left (d+e x^2\right )^{3/2}}-\frac{b n \log (x) \left (2 d+3 e x^2\right )}{\left (d+e x^2\right )^{3/2}}-\frac{2 b n \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )}{\sqrt{d}}+\frac{2 b n \log (x)}{\sqrt{d}}}{3 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

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Maple [F]  time = 0.402, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b\ln \left ( c{x}^{n} \right ) \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(x^3*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

int(x^3*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),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(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6605, size = 737, normalized size = 6.82 \begin{align*} \left [\frac{{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{d} \log \left (-\frac{e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) -{\left (b d^{2} n + 2 \, a d^{2} +{\left (b d e n + 3 \, a d e\right )} x^{2} +{\left (3 \, b d e x^{2} + 2 \, b d^{2}\right )} \log \left (c\right ) +{\left (3 \, b d e n x^{2} + 2 \, b d^{2} n\right )} \log \left (x\right )\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 )}}, \frac{2 \,{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt{-d} \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) -{\left (b d^{2} n + 2 \, a d^{2} +{\left (b d e n + 3 \, a d e\right )} x^{2} +{\left (3 \, b d e x^{2} + 2 \, b d^{2}\right )} \log \left (c\right ) +{\left (3 \, b d e n x^{2} + 2 \, b d^{2} n\right )} \log \left (x\right )\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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(d)*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (b
*d^2*n + 2*a*d^2 + (b*d*e*n + 3*a*d*e)*x^2 + (3*b*d*e*x^2 + 2*b*d^2)*log(c) + (3*b*d*e*n*x^2 + 2*b*d^2*n)*log(
x))*sqrt(e*x^2 + d))/(d*e^4*x^4 + 2*d^2*e^3*x^2 + d^3*e^2), 1/3*(2*(b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqr
t(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - (b*d^2*n + 2*a*d^2 + (b*d*e*n + 3*a*d*e)*x^2 + (3*b*d*e*x^2 + 2*b*d^2
)*log(c) + (3*b*d*e*n*x^2 + 2*b*d^2*n)*log(x))*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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Sympy [A]  time = 80.4278, size = 333, normalized size = 3.08 \begin{align*} a \left (\begin{cases} \frac{x^{4}}{4 d^{\frac{5}{2}}} & \text{for}\: e = 0 \\\frac{d}{3 e^{2} \left (d + e x^{2}\right )^{\frac{3}{2}}} - \frac{1}{e^{2} \sqrt{d + e x^{2}}} & \text{otherwise} \end{cases}\right ) - b n \left (\begin{cases} \frac{x^{4}}{16 d^{\frac{5}{2}}} & \text{for}\: e = 0 \\\frac{2 d^{4} \sqrt{1 + \frac{e x^{2}}{d}}}{6 d^{\frac{9}{2}} e^{2} + 6 d^{\frac{7}{2}} e^{3} x^{2}} + \frac{d^{4} \log{\left (\frac{e x^{2}}{d} \right )}}{6 d^{\frac{9}{2}} e^{2} + 6 d^{\frac{7}{2}} e^{3} x^{2}} - \frac{2 d^{4} \log{\left (\sqrt{1 + \frac{e x^{2}}{d}} + 1 \right )}}{6 d^{\frac{9}{2}} e^{2} + 6 d^{\frac{7}{2}} e^{3} x^{2}} + \frac{d^{3} x^{2} \log{\left (\frac{e x^{2}}{d} \right )}}{6 d^{\frac{9}{2}} e + 6 d^{\frac{7}{2}} e^{2} x^{2}} - \frac{2 d^{3} x^{2} \log{\left (\sqrt{1 + \frac{e x^{2}}{d}} + 1 \right )}}{6 d^{\frac{9}{2}} e + 6 d^{\frac{7}{2}} e^{2} x^{2}} + \frac{\operatorname{asinh}{\left (\frac{\sqrt{d}}{\sqrt{e} x} \right )}}{\sqrt{d} e^{2}} & \text{otherwise} \end{cases}\right ) + b \left (\begin{cases} \frac{x^{4}}{4 d^{\frac{5}{2}}} & \text{for}\: e = 0 \\\frac{d}{3 e^{2} \left (d + e x^{2}\right )^{\frac{3}{2}}} - \frac{1}{e^{2} \sqrt{d + e x^{2}}} & \text{otherwise} \end{cases}\right ) \log{\left (c x^{n} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d)**(5/2),x)

[Out]

a*Piecewise((x**4/(4*d**(5/2)), Eq(e, 0)), (d/(3*e**2*(d + e*x**2)**(3/2)) - 1/(e**2*sqrt(d + e*x**2)), True))
- b*n*Piecewise((x**4/(16*d**(5/2)), Eq(e, 0)), (2*d**4*sqrt(1 + e*x**2/d)/(6*d**(9/2)*e**2 + 6*d**(7/2)*e**3
*x**2) + d**4*log(e*x**2/d)/(6*d**(9/2)*e**2 + 6*d**(7/2)*e**3*x**2) - 2*d**4*log(sqrt(1 + e*x**2/d) + 1)/(6*d
**(9/2)*e**2 + 6*d**(7/2)*e**3*x**2) + d**3*x**2*log(e*x**2/d)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) - 2*d**3*
x**2*log(sqrt(1 + e*x**2/d) + 1)/(6*d**(9/2)*e + 6*d**(7/2)*e**2*x**2) + asinh(sqrt(d)/(sqrt(e)*x))/(sqrt(d)*e
**2), True)) + b*Piecewise((x**4/(4*d**(5/2)), Eq(e, 0)), (d/(3*e**2*(d + e*x**2)**(3/2)) - 1/(e**2*sqrt(d + e
*x**2)), True))*log(c*x**n)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\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(x^3*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^3/(e*x^2 + d)^(5/2), x)