3.3.60 \(\int \frac {\sqrt {d+e x^2} (a+b \log (c x^n))}{x^4} \, dx\) [260]

Optimal. Leaf size=112 \[ -\frac {b e n \sqrt {d+e x^2}}{3 d x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d x^3}+\frac {b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2373, 283, 223, 212} \begin {gather*} -\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d}-\frac {b e n \sqrt {d+e x^2}}{3 d x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*e*n*Sqrt[d + e*x^2])/(d*x) - (b*n*(d + e*x^2)^(3/2))/(9*d*x^3) + (b*e^(3/2)*n*ArcTanh[(Sqrt[e]*x)/Sqrt
[d + e*x^2]])/(3*d) - ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*d*x^3)

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 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 2373

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])/(d*f*(m + 1))), x] - Dist[b*(n/(d*(m + 1))), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 99, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {d+e x^2} \left (3 a \left (d+e x^2\right )+b n \left (d+4 e x^2\right )\right )+3 b \left (d+e x^2\right )^{3/2} \log \left (c x^n\right )-3 b e^{3/2} n x^3 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{9 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/9*(Sqrt[d + e*x^2]*(3*a*(d + e*x^2) + b*n*(d + 4*e*x^2)) + 3*b*(d + e*x^2)^(3/2)*Log[c*x^n] - 3*b*e^(3/2)*n
*x^3*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(d*x^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 121, normalized size = 1.08 \begin {gather*} \frac {{\left (3 \, \operatorname {arsinh}\left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\frac {3}{2}} + \frac {3 \, \sqrt {x^{2} e + d} x e^{2}}{d} - \frac {2 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} e}{d x} - \frac {{\left (x^{2} e + d\right )}^{\frac {5}{2}}}{d x^{3}}\right )} b n}{9 \, d} - \frac {{\left (x^{2} e + d\right )}^{\frac {3}{2}} b \log \left (c x^{n}\right )}{3 \, d x^{3}} - \frac {{\left (x^{2} e + d\right )}^{\frac {3}{2}} a}{3 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 112, normalized size = 1.00 \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 ) - 2 \, {\left ({\left (4 \, b n + 3 \, a\right )} x^{2} e + b d n + 3 \, a d + 3 \, {\left (b x^{2} e + b d\right )} \log \left (c\right ) + 3 \, {\left (b n x^{2} e + b d n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{18 \, d x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(3*b*n*x^3*e^(3/2)*log(-2*x^2*e - 2*sqrt(x^2*e + d)*x*e^(1/2) - d) - 2*((4*b*n + 3*a)*x^2*e + b*d*n + 3*a
*d + 3*(b*x^2*e + b*d)*log(c) + 3*(b*n*x^2*e + b*d*n)*log(x))*sqrt(x^2*e + d))/(d*x^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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