3.3.99 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\) [299]

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

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {272, 45, 2392, 12, 457, 79, 65, 214} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*n)/(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 12

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

Rule 45

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 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 79

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] || L
tQ[p, n]))))

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 457

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 2392

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])

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) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.17, size = 137, normalized size = 1.27 \begin {gather*} \frac {\frac {2 b n \log (x)}{\sqrt {d}}-\frac {b n \left (2 d+3 e x^2\right ) \log (x)}{\left (d+e x^2\right )^{3/2}}+\frac {d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-\left (d+e x^2\right ) \left (3 a+b n-3 b n \log (x)+3 b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}}-\frac {2 b n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{\sqrt {d}}}{3 e^2} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

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 [A]
time = 0.51, size = 138, normalized size = 1.28 \begin {gather*} \frac {1}{3} \, {\left (\frac {e^{\left (-2\right )} \log \left (\frac {\sqrt {x^{2} e + d} - \sqrt {d}}{\sqrt {x^{2} e + d} + \sqrt {d}}\right )}{\sqrt {d}} - \frac {e^{\left (-2\right )}}{\sqrt {x^{2} e + d}}\right )} b n - \frac {1}{3} \, {\left (\frac {3 \, x^{2} e^{\left (-1\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {2 \, d e^{\left (-2\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}}\right )} b \log \left (c x^{n}\right ) - \frac {1}{3} \, {\left (\frac {3 \, x^{2} e^{\left (-1\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}} + \frac {2 \, d e^{\left (-2\right )}}{{\left (x^{2} e + d\right )}^{\frac {3}{2}}}\right )} a \end {gather*}

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]

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

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Fricas [A]
time = 0.49, size = 328, normalized size = 3.04 \begin {gather*} \left [\frac {{\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {x^{2} e - 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (b d^{2} n + {\left (b d n + 3 \, a d\right )} x^{2} e + 2 \, a d^{2} + {\left (3 \, b d x^{2} e + 2 \, b d^{2}\right )} \log \left (c\right ) + {\left (3 \, b d n x^{2} e + 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{3 \, {\left (d x^{4} e^{4} + 2 \, d^{2} x^{2} e^{3} + d^{3} e^{2}\right )}}, \frac {2 \, {\left (b n x^{4} e^{2} + 2 \, b d n x^{2} e + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) - {\left (b d^{2} n + {\left (b d n + 3 \, a d\right )} x^{2} e + 2 \, a d^{2} + {\left (3 \, b d x^{2} e + 2 \, b d^{2}\right )} \log \left (c\right ) + {\left (3 \, b d n x^{2} e + 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{3 \, {\left (d x^{4} e^{4} + 2 \, d^{2} x^{2} e^{3} + d^{3} e^{2}\right )}}\right ] \end {gather*}

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

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Sympy [A]
time = 30.87, size = 333, normalized size = 3.08 \begin {gather*} 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 {gather*}

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 = 0.00, 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^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/(x^2*e + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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