\(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^{5/2}} \, dx\) [142]

Optimal result

Integrand size = 24, antiderivative size = 114 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=\frac {4 b e n}{3 g (e f-d g) \sqrt {f+g x}}-\frac {4 b e^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 g (e f-d g)^{3/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2442, 53, 65, 214} \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}}-\frac {4 b e^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 g (e f-d g)^{3/2}}+\frac {4 b e n}{3 g \sqrt {f+g x} (e f-d g)} \]

[In]

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

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 2442

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=-\frac {4 b e n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {e (f+g x)}{e f-d g}\right )}{3 g (-e f+d g) \sqrt {f+g x}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}} \]

[In]

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

[Out]

(-4*b*e*n*Hypergeometric2F1[-1/2, 1, 1/2, (e*(f + g*x))/(e*f - d*g)])/(3*g*(-(e*f) + d*g)*Sqrt[f + g*x]) - (2*
(a + b*Log[c*(d + e*x)^n]))/(3*g*(f + g*x)^(3/2))

Maple [F]

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (94) = 188\).

Time = 0.36 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.73 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=\left [-\frac {2 \, {\left ({\left (b e g^{2} n x^{2} + 2 \, b e f g n x + b e f^{2} n\right )} \sqrt {\frac {e}{e f - d g}} \log \left (\frac {e g x + 2 \, e f - d g + 2 \, {\left (e f - d g\right )} \sqrt {g x + f} \sqrt {\frac {e}{e f - d g}}}{e x + d}\right ) - {\left (2 \, b e g n x + 2 \, b e f n - a e f + a d g - {\left (b e f - b d g\right )} n \log \left (e x + d\right ) - {\left (b e f - b d g\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{3 \, {\left (e f^{3} g - d f^{2} g^{2} + {\left (e f g^{3} - d g^{4}\right )} x^{2} + 2 \, {\left (e f^{2} g^{2} - d f g^{3}\right )} x\right )}}, -\frac {2 \, {\left (2 \, {\left (b e g^{2} n x^{2} + 2 \, b e f g n x + b e f^{2} n\right )} \sqrt {-\frac {e}{e f - d g}} \arctan \left (-\frac {{\left (e f - d g\right )} \sqrt {g x + f} \sqrt {-\frac {e}{e f - d g}}}{e g x + e f}\right ) - {\left (2 \, b e g n x + 2 \, b e f n - a e f + a d g - {\left (b e f - b d g\right )} n \log \left (e x + d\right ) - {\left (b e f - b d g\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{3 \, {\left (e f^{3} g - d f^{2} g^{2} + {\left (e f g^{3} - d g^{4}\right )} x^{2} + 2 \, {\left (e f^{2} g^{2} - d f g^{3}\right )} x\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e*(d*g-e*f)>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.44 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{5/2}} \, dx=\frac {4 \, b e^{2} n \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {-e^{2} f + d e g}}\right )}{3 \, \sqrt {-e^{2} f + d e g} {\left (e f g - d g^{2}\right )}} - \frac {2 \, b n \log \left ({\left (g x + f\right )} e - e f + d g\right )}{3 \, {\left (g x + f\right )}^{\frac {3}{2}} g} + \frac {2 \, {\left (b e f n \log \left (g\right ) - b d g n \log \left (g\right ) + 2 \, {\left (g x + f\right )} b e n - b e f \log \left (c\right ) + b d g \log \left (c\right ) - a e f + a d g\right )}}{3 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} e f g - {\left (g x + f\right )}^{\frac {3}{2}} d g^{2}\right )}} \]

[In]

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

[Out]

4/3*b*e^2*n*arctan(sqrt(g*x + f)*e/sqrt(-e^2*f + d*e*g))/(sqrt(-e^2*f + d*e*g)*(e*f*g - d*g^2)) - 2/3*b*n*log(
(g*x + f)*e - e*f + d*g)/((g*x + f)^(3/2)*g) + 2/3*(b*e*f*n*log(g) - b*d*g*n*log(g) + 2*(g*x + f)*b*e*n - b*e*
f*log(c) + b*d*g*log(c) - a*e*f + a*d*g)/((g*x + f)^(3/2)*e*f*g - (g*x + f)^(3/2)*d*g^2)

Mupad [F(-1)]

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

[In]

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

[Out]

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