3.4.1 \(\int \frac {a+b \log (c x^n)}{x (d+e x^2)^{5/2}} \, dx\) [301]

Optimal. Leaf size=251 \[ -\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}-\frac {b n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}} \]

[Out]

4/3*b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(5/2)+1/2*b*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))^2/d^(5/2)-b*n*arctan
h((e*x^2+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(5/2)-1/2*b*n*polylog(2,1-2*d^(1/2)/(d^(1
/2)-(e*x^2+d)^(1/2)))/d^(5/2)+1/3*(a+b*ln(c*x^n))*(1/d/(e*x^2+d)^(3/2)-3*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(5
/2)+3/d^2/(e*x^2+d)^(1/2))-1/3*b*n/d^2/(e*x^2+d)^(1/2)

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Rubi [A]
time = 0.29, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {272, 53, 65, 214, 2390, 6131, 6055, 2449, 2352} \begin {gather*} -\frac {b n \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{5/2}}+\frac {1}{3} \left (-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}+\frac {1}{d \left (d+e x^2\right )^{3/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}-\frac {b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}-\frac {b n}{3 d^2 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*n)/(d^2*Sqrt[d + e*x^2]) + (4*b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(3*d^(5/2)) + (b*n*ArcTanh[Sqrt[d
+ e*x^2]/Sqrt[d]]^2)/(2*d^(5/2)) + ((1/(d*(d + e*x^2)^(3/2)) + 3/(d^2*Sqrt[d + e*x^2]) - (3*ArcTanh[Sqrt[d + e
*x^2]/Sqrt[d]])/d^(5/2))*(a + b*Log[c*x^n]))/3 - (b*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d
] - Sqrt[d + e*x^2])])/d^(5/2) - (b*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(5/2))

Rule 53

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

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 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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2390

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

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^
2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6131

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

Rubi steps

\begin {align*} \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^2\right )^{5/2}} \, dx &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {1}{3 d x \left (d+e x^2\right )^{3/2}}+\frac {1}{d^2 x \sqrt {d+e x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2} x}\right ) \, dx\\ &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{x} \, dx}{d^{5/2}}-\frac {(b n) \int \frac {1}{x \sqrt {d+e x^2}} \, dx}{d^2}-\frac {(b n) \int \frac {1}{x \left (d+e x^2\right )^{3/2}} \, dx}{3 d}\\ &=\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx,x,x^2\right )}{2 d^{5/2}}-\frac {(b n) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d^2}-\frac {(b n) \text {Subst}\left (\int \frac {1}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 d}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {(b n) \text {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{-d+x^2} \, dx,x,\sqrt {d+e x^2}\right )}{d^{5/2}}-\frac {(b n) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}-\frac {(b n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{d^2 e}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {(b n) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{1-\frac {x}{\sqrt {d}}} \, dx,x,\sqrt {d+e x^2}\right )}{d^3}-\frac {(b n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^2 e}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}+\frac {(b n) \text {Subst}\left (\int \frac {\log \left (\frac {2}{1-\frac {x}{\sqrt {d}}}\right )}{1-\frac {x^2}{d}} \, dx,x,\sqrt {d+e x^2}\right )}{d^3}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}-\frac {(b n) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{d^{5/2}}\\ &=-\frac {b n}{3 d^2 \sqrt {d+e x^2}}+\frac {4 b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 d^{5/2}}+\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 d^{5/2}}+\frac {1}{3} \left (\frac {1}{d \left (d+e x^2\right )^{3/2}}+\frac {3}{d^2 \sqrt {d+e x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {b n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{d^{5/2}}-\frac {b n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x^2}}{\sqrt {d}}}\right )}{2 d^{5/2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.30, size = 273, normalized size = 1.09 \begin {gather*} \frac {b n \sqrt {1+\frac {d}{e x^2}} \left (-3 d^{5/2} \left (d+e x^2\right )^2 \, _3F_2\left (\frac {5}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {d}{e x^2}\right )+25 \sqrt {d} e^3 \sqrt {1+\frac {d}{e x^2}} x^6 \left (4 d+3 e x^2\right ) \log (x)-75 e^{5/2} x^5 \left (d+e x^2\right )^2 \sinh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right ) \log (x)\right )}{75 d^{5/2} e^2 x^4 \left (d+e x^2\right )^{5/2}}+\frac {\left (4 d+3 e x^2\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}+\frac {\log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d^{5/2}}-\frac {\left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{d^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n*Sqrt[1 + d/(e*x^2)]*(-3*d^(5/2)*(d + e*x^2)^2*HypergeometricPFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, -(d/(e*x^2))
] + 25*Sqrt[d]*e^3*Sqrt[1 + d/(e*x^2)]*x^6*(4*d + 3*e*x^2)*Log[x] - 75*e^(5/2)*x^5*(d + e*x^2)^2*ArcSinh[Sqrt[
d]/(Sqrt[e]*x)]*Log[x]))/(75*d^(5/2)*e^2*x^4*(d + e*x^2)^(5/2)) + ((4*d + 3*e*x^2)*(a - b*n*Log[x] + b*Log[c*x
^n]))/(3*d^2*(d + e*x^2)^(3/2)) + (Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]))/d^(5/2) - ((a - b*n*Log[x] + b*Log[
c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/d^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a*(3*arcsinh(sqrt(d)*e^(-1/2)/abs(x))/d^(5/2) - 3/(sqrt(x^2*e + d)*d^2) - 1/((x^2*e + d)^(3/2)*d)) + b*in
tegrate((log(c) + log(x^n))/((x^5*e^2 + 2*d*x^3*e + d^2*x)*sqrt(x^2*e + d)), x)

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

[Out]

integral((sqrt(x^2*e + d)*b*log(c*x^n) + sqrt(x^2*e + d)*a)/(x^7*e^3 + 3*d*x^5*e^2 + 3*d^2*x^3*e + d^3*x), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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