3.3.84 \(\int \frac {a+b \log (c x^n)}{x^4 \sqrt {d+e x^2}} \, dx\) [284]

Optimal. Leaf size=144 \[ \frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {2 b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x} \]

[Out]

-1/9*b*n*(e*x^2+d)^(3/2)/d^2/x^3-2/3*b*e^(3/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^2+2/3*b*e*n*(e*x^2+d)^(1
/2)/d^2/x-1/3*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d/x^3+2/3*e*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d^2/x

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Rubi [A]
time = 0.10, antiderivative size = 144, 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 = {277, 270, 2392, 12, 462, 283, 223, 212} \begin {gather*} \frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}-\frac {2 b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2}+\frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

(2*b*e*n*Sqrt[d + e*x^2])/(3*d^2*x) - (b*n*(d + e*x^2)^(3/2))/(9*d^2*x^3) - (2*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*x)
/Sqrt[d + e*x^2]])/(3*d^2) - (Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(3*d*x^3) + (2*e*Sqrt[d + e*x^2]*(a + b*Log[
c*x^n]))/(3*d^2*x)

Rule 12

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

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 270

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

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 283

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

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^4 \sqrt {d+e x^2}} \, dx &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-(b n) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{3 d^2 x^4} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {(b n) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{x^4} \, dx}{3 d^2}\\ &=-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {(2 b e n) \int \frac {\sqrt {d+e x^2}}{x^2} \, dx}{3 d^2}\\ &=\frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\left (2 b e^2 n\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{3 d^2}\\ &=\frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\left (2 b e^2 n\right ) \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{3 d^2}\\ &=\frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {2 b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 110, normalized size = 0.76 \begin {gather*} \frac {\sqrt {d+e x^2} \left (-3 a d-b d n+6 a e x^2+5 b e n x^2\right )-3 b \left (d-2 e x^2\right ) \sqrt {d+e x^2} \log \left (c x^n\right )-6 b e^{3/2} n x^3 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{9 d^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x^4*Sqrt[d + e*x^2]),x]

[Out]

(Sqrt[d + e*x^2]*(-3*a*d - b*d*n + 6*a*e*x^2 + 5*b*e*n*x^2) - 3*b*(d - 2*e*x^2)*Sqrt[d + e*x^2]*Log[c*x^n] - 6
*b*e^(3/2)*n*x^3*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(9*d^2*x^3)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \sqrt {e \,x^{2}+d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^(1/2),x)

[Out]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^(1/2),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^4/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 116, normalized size = 0.81 \begin {gather*} \frac {3 \, b n x^{3} e^{\frac {3}{2}} \log \left (-2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right ) + {\left ({\left (5 \, b n + 6 \, a\right )} x^{2} e - b d n - 3 \, a d + 3 \, {\left (2 \, b x^{2} e - b d\right )} \log \left (c\right ) + 3 \, {\left (2 \, b n x^{2} e - b d n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{9 \, d^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x^4/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

1/9*(3*b*n*x^3*e^(3/2)*log(-2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) - d) + ((5*b*n + 6*a)*x^2*e - b*d*n - 3*a*d
+ 3*(2*b*x^2*e - b*d)*log(c) + 3*(2*b*n*x^2*e - b*d*n)*log(x))*sqrt(x^2*e + d))/(d^2*x^3)

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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^{4} \sqrt {d + e x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**4/(e*x**2+d)**(1/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*sqrt(d + e*x**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^4/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/(sqrt(x^2*e + d)*x^4), 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^4\,\sqrt {e\,x^2+d}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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