3.3.95 \(\int \frac {a+b \log (c x^n)}{x^4 (d+e x^2)^{3/2}} \, dx\) [295]

Optimal. Leaf size=176 \[ -\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {8 b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}} \]

[Out]

-8/3*b*e^(3/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^3+1/3*(-a-b*ln(c*x^n))/d/x^3/(e*x^2+d)^(1/2)+4/3*e*(a+b*
ln(c*x^n))/d^2/x/(e*x^2+d)^(1/2)+8/3*e^2*x*(a+b*ln(c*x^n))/d^3/(e*x^2+d)^(1/2)-1/9*b*n*(e*x^2+d)^(1/2)/d^2/x^3
+14/9*b*e*n*(e*x^2+d)^(1/2)/d^3/x

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Rubi [A]
time = 0.12, antiderivative size = 176, 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, 197, 2392, 12, 1279, 462, 223, 212} \begin {gather*} \frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}-\frac {8 b e^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^4*(d + e*x^2)^(3/2)),x]

[Out]

-1/9*(b*n*Sqrt[d + e*x^2])/(d^2*x^3) + (14*b*e*n*Sqrt[d + e*x^2])/(9*d^3*x) - (8*b*e^(3/2)*n*ArcTanh[(Sqrt[e]*
x)/Sqrt[d + e*x^2]])/(3*d^3) - (a + b*Log[c*x^n])/(3*d*x^3*Sqrt[d + e*x^2]) + (4*e*(a + b*Log[c*x^n]))/(3*d^2*
x*Sqrt[d + e*x^2]) + (8*e^2*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 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 1279

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 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 \left (d+e x^2\right )^{3/2}} \, dx &=-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-(b n) \int \frac {-d^2+4 d e x^2+8 e^2 x^4}{3 d^3 x^4 \sqrt {d+e x^2}} \, dx\\ &=-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {(b n) \int \frac {-d^2+4 d e x^2+8 e^2 x^4}{x^4 \sqrt {d+e x^2}} \, dx}{3 d^3}\\ &=-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b n) \int \frac {-14 d^2 e-24 d e^2 x^2}{x^2 \sqrt {d+e x^2}} \, dx}{9 d^4}\\ &=-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {\left (8 b e^2 n\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{3 d^3}\\ &=-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {\left (8 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^3}\\ &=-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {8 b e^{3/2} 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 )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 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 = 0.10, size = 144, normalized size = 0.82 \begin {gather*} \frac {-3 a d^2-b d^2 n+12 a d e x^2+13 b d e n x^2+24 a e^2 x^4+14 b e^2 n x^4-3 b \left (d^2-4 d e x^2-8 e^2 x^4\right ) \log \left (c x^n\right )-24 b e^{3/2} n x^3 \sqrt {d+e x^2} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{9 d^3 x^3 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x^4*(d + e*x^2)^(3/2)),x]

[Out]

(-3*a*d^2 - b*d^2*n + 12*a*d*e*x^2 + 13*b*d*e*n*x^2 + 24*a*e^2*x^4 + 14*b*e^2*n*x^4 - 3*b*(d^2 - 4*d*e*x^2 - 8
*e^2*x^4)*Log[c*x^n] - 24*b*e^(3/2)*n*x^3*Sqrt[d + e*x^2]*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(9*d^3*x^3*Sqrt[
d + e*x^2])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x^4/(e*x^2+d)^(3/2),x)

[Out]

int((a+b*ln(c*x^n))/x^4/(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^4/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.45, size = 186, normalized size = 1.06 \begin {gather*} \frac {12 \, {\left (b n x^{5} e^{2} + b d n x^{3} e\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 (2 \, {\left (7 \, b n + 12 \, a\right )} x^{4} e^{2} - b d^{2} n + {\left (13 \, b d n + 12 \, a d\right )} x^{2} e - 3 \, a d^{2} + 3 \, {\left (8 \, b x^{4} e^{2} + 4 \, b d x^{2} e - b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (8 \, b n x^{4} e^{2} + 4 \, b d n x^{2} e - b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {x^{2} e + d}}{9 \, {\left (d^{3} x^{5} e + d^{4} x^{3}\right )}} \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)^(3/2),x, algorithm="fricas")

[Out]

1/9*(12*(b*n*x^5*e^2 + b*d*n*x^3*e)*e^(1/2)*log(-2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) - d) + (2*(7*b*n + 12*a
)*x^4*e^2 - b*d^2*n + (13*b*d*n + 12*a*d)*x^2*e - 3*a*d^2 + 3*(8*b*x^4*e^2 + 4*b*d*x^2*e - b*d^2)*log(c) + 3*(
8*b*n*x^4*e^2 + 4*b*d*n*x^2*e - b*d^2*n)*log(x))*sqrt(x^2*e + d))/(d^3*x^5*e + d^4*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} \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**4/(e*x**2+d)**(3/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**4*(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^4/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((x^2*e + d)^(3/2)*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\,{\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^4*(d + e*x^2)^(3/2)),x)

[Out]

int((a + b*log(c*x^n))/(x^4*(d + e*x^2)^(3/2)), x)

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