\(\int \frac {a+b \log (c x^n)}{(d+e x^2)^{5/2}} \, dx\) [306]

Optimal result

Integrand size = 22, antiderivative size = 113 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {b n x}{3 d^2 \sqrt {d+e x^2}}-\frac {2 b n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2 \sqrt {e}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \sqrt {d+e x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2360, 2351, 223, 212, 197} \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \sqrt {d+e x^2}}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac {2 b n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2 \sqrt {e}}-\frac {b n x}{3 d^2 \sqrt {d+e x^2}} \]

[In]

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

[Out]

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

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 2351

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

Rule 2360

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(q +
1)*((a + b*Log[c*x^n])/(2*d*(q + 1))), x] + (Dist[(2*q + 3)/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*(a + b*Log[
c*x^n]), x], x] + Dist[b*(n/(2*d*(q + 1))), Int[(d + e*x^2)^(q + 1), x], x]) /; FreeQ[{a, b, c, d, e, n}, x] &
& LtQ[q, -1]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {\sqrt {e} x \left (-b n \left (d+e x^2\right )+a \left (3 d+2 e x^2\right )\right )+b \sqrt {e} x \left (3 d+2 e x^2\right ) \log \left (c x^n\right )-2 b n \left (d+e x^2\right )^{3/2} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{3 d^2 \sqrt {e} \left (d+e x^2\right )^{3/2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.98 \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {{\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{3} + {\left (b d e n - 3 \, a d e\right )} x - {\left (2 \, b e^{2} x^{3} + 3 \, b d e x\right )} \log \left (c\right ) - {\left (2 \, b e^{2} n x^{3} + 3 \, b d e n x\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}, \frac {2 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left ({\left (b e^{2} n - 2 \, a e^{2}\right )} x^{3} + {\left (b d e n - 3 \, a d e\right )} x - {\left (2 \, b e^{2} x^{3} + 3 \, b d e x\right )} \log \left (c\right ) - {\left (2 \, b e^{2} n x^{3} + 3 \, b d e n x\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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