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

Optimal. Leaf size=166 \[ -\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}+\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}-\frac {b n}{d^2 x \sqrt {d+e x^2}} \]

[Out]

(-a-b*ln(c*x^n))/d/x/(e*x^2+d)^(3/2)-4/3*e*x*(a+b*ln(c*x^n))/d^2/(e*x^2+d)^(3/2)+8/3*b*n*arctanh(x*e^(1/2)/(e*
x^2+d)^(1/2))*e^(1/2)/d^3-b*n/d^2/x/(e*x^2+d)^(1/2)-2/3*b*e*n*x/d^3/(e*x^2+d)^(1/2)-8/3*e*x*(a+b*ln(c*x^n))/d^
3/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {271, 192, 191, 2350, 12, 1265, 385, 217, 206} \[ -\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}-\frac {b n}{d^2 x \sqrt {d+e x^2}}+\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n)/(d^2*x*Sqrt[d + e*x^2])) - (2*b*e*n*x)/(3*d^3*Sqrt[d + e*x^2]) + (8*b*Sqrt[e]*n*ArcTanh[(Sqrt[e]*x)/Sq
rt[d + e*x^2]])/(3*d^3) - (a + b*Log[c*x^n])/(d*x*(d + e*x^2)^(3/2)) - (4*e*x*(a + b*Log[c*x^n]))/(3*d^2*(d +
e*x^2)^(3/2)) - (8*e*x*(a + b*Log[c*x^n]))/(3*d^3*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 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]

Rule 192

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

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 271

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

Rule 385

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

Rule 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

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])

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx &=-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-(b n) \int \frac {-3 d^2-12 d e x^2-8 e^2 x^4}{3 d^3 x^2 \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {(b n) \int \frac {-3 d^2-12 d e x^2-8 e^2 x^4}{x^2 \left (d+e x^2\right )^{3/2}} \, dx}{3 d^3}\\ &=-\frac {b n}{d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b n) \int \frac {6 d^2 e+8 d e^2 x^2}{\left (d+e x^2\right )^{3/2}} \, dx}{3 d^4}\\ &=-\frac {b n}{d^2 x \sqrt {d+e x^2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(8 b e n) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{3 d^3}\\ &=-\frac {b n}{d^2 x \sqrt {d+e x^2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(8 b e n) \operatorname {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{3 d^3}\\ &=-\frac {b n}{d^2 x \sqrt {d+e x^2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}+\frac {8 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 144, normalized size = 0.87 \[ \frac {-3 a d^2-12 a d e x^2-8 a e^2 x^4-b \left (3 d^2+12 d e x^2+8 e^2 x^4\right ) \log \left (c x^n\right )-3 b d^2 n-5 b d e n x^2+8 b \sqrt {e} n x \left (d+e x^2\right )^{3/2} \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )-2 b e^2 n x^4}{3 d^3 x \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*d^2 - 3*b*d^2*n - 12*a*d*e*x^2 - 5*b*d*e*n*x^2 - 8*a*e^2*x^4 - 2*b*e^2*n*x^4 - b*(3*d^2 + 12*d*e*x^2 + 8
*e^2*x^4)*Log[c*x^n] + 8*b*Sqrt[e]*n*x*(d + e*x^2)^(3/2)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(3*d^3*x*(d + e*x
^2)^(3/2))

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fricas [A]  time = 0.50, size = 399, normalized size = 2.40 \[ \left [\frac {4 \, {\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left (2 \, {\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} + {\left (5 \, b d e n + 12 \, a d e\right )} x^{2} + {\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \relax (c) + {\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}, -\frac {8 \, {\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, {\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} + {\left (5 \, b d e n + 12 \, a d e\right )} x^{2} + {\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \relax (c) + {\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(4*(b*e^2*n*x^5 + 2*b*d*e*n*x^3 + b*d^2*n*x)*sqrt(e)*log(-2*e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2
*(b*e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3*
b*d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*x
^3 + d^5*x), -1/3*(8*(b*e^2*n*x^5 + 2*b*d*e*n*x^3 + b*d^2*n*x)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (
2*(b*e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3
*b*d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*
x^3 + d^5*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {b \ln \left (c \,x^{n}\right )+a}{\left (e \,x^{2}+d \right )^{\frac {5}{2}} x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {8 \, e x}{\sqrt {e x^{2} + d} d^{3}} + \frac {4 \, e x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}} + \frac {3}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d x}\right )} + b \int \frac {\log \relax (c) + \log \left (x^{n}\right )}{{\left (e^{2} x^{6} + 2 \, d e x^{4} + d^{2} x^{2}\right )} \sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(8*e*x/(sqrt(e*x^2 + d)*d^3) + 4*e*x/((e*x^2 + d)^(3/2)*d^2) + 3/((e*x^2 + d)^(3/2)*d*x)) + b*integrate
((log(c) + log(x^n))/((e^2*x^6 + 2*d*e*x^4 + d^2*x^2)*sqrt(e*x^2 + d)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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