3.3.94 \(\int \frac {a+b \log (c x^n)}{x^2 (d+e x^2)^{3/2}} \, dx\) [294]

Optimal. Leaf size=110 \[ -\frac {b n \sqrt {d+e x^2}}{d^2 x}+\frac {2 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{d^2}-\frac {a+b \log \left (c x^n\right )}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \log \left (c x^n\right )\right )}{d^2 \sqrt {d+e x^2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {277, 197, 2392, 12, 462, 223, 212} \begin {gather*} -\frac {2 e x \left (a+b \log \left (c x^n\right )\right )}{d^2 \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{d x \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{d^2 x}+\frac {2 b \sqrt {e} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*n*Sqrt[d + e*x^2])/(d^2*x)) + (2*b*Sqrt[e]*n*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/d^2 - (a + b*Log[c*x^n
])/(d*x*Sqrt[d + e*x^2]) - (2*e*x*(a + b*Log[c*x^n]))/(d^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 197

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 277

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 462

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

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

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Mathematica [A]
time = 0.08, size = 103, normalized size = 0.94 \begin {gather*} \frac {-a d-b d n-2 a e x^2-b e n x^2-b \left (d+2 e x^2\right ) \log \left (c x^n\right )+2 b \sqrt {e} n x \sqrt {d+e x^2} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{d^2 x \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 128, normalized size = 1.16 \begin {gather*} \frac {{\left (b n x^{3} e + b d n x\right )} e^{\frac {1}{2}} \log \left (-2 \, x^{2} e - 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right ) - {\left ({\left (b n + 2 \, a\right )} x^{2} e + b d n + a d + {\left (2 \, b x^{2} e + b d\right )} \log \left (c\right ) + {\left (2 \, b n x^{2} e + b d n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{d^{2} x^{3} e + d^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^2/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \log {\left (c x^{n} \right )}}{x^{2} \left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^2/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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