\(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^{9/2}} \, dx\) [144]

Optimal result

Integrand size = 24, antiderivative size = 176 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {4 b e^{7/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}} \]

[Out]

4/35*b*e*n/g/(-d*g+e*f)/(g*x+f)^(5/2)+4/21*b*e^2*n/g/(-d*g+e*f)^2/(g*x+f)^(3/2)-4/7*b*e^(7/2)*n*arctanh(e^(1/2
)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/g/(-d*g+e*f)^(7/2)-2/7*(a+b*ln(c*(e*x+d)^n))/g/(g*x+f)^(7/2)+4/7*b*e^3*n/g/(
-d*g+e*f)^3/(g*x+f)^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2442, 53, 65, 214} \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}-\frac {4 b e^{7/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}+\frac {4 b e^3 n}{7 g \sqrt {f+g x} (e f-d g)^3}+\frac {4 b e^2 n}{21 g (f+g x)^{3/2} (e f-d g)^2}+\frac {4 b e n}{35 g (f+g x)^{5/2} (e f-d g)} \]

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(9/2),x]

[Out]

(4*b*e*n)/(35*g*(e*f - d*g)*(f + g*x)^(5/2)) + (4*b*e^2*n)/(21*g*(e*f - d*g)^2*(f + g*x)^(3/2)) + (4*b*e^3*n)/
(7*g*(e*f - d*g)^3*Sqrt[f + g*x]) - (4*b*e^(7/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(7*g*(e*f
- d*g)^(7/2)) - (2*(a + b*Log[c*(d + e*x)^n]))/(7*g*(f + g*x)^(7/2))

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 )}{7 g (f+g x)^{7/2}}+\frac {(2 b e n) \int \frac {1}{(d+e x) (f+g x)^{7/2}} \, dx}{7 g} \\ & = \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^2 n\right ) \int \frac {1}{(d+e x) (f+g x)^{5/2}} \, dx}{7 g (e f-d g)} \\ & = \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^3 n\right ) \int \frac {1}{(d+e x) (f+g x)^{3/2}} \, dx}{7 g (e f-d g)^2} \\ & = \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^4 n\right ) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{7 g (e f-d g)^3} \\ & = \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (4 b e^4 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 )}{7 g^2 (e f-d g)^3} \\ & = \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {4 b e^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}} \\ \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 = 78, normalized size of antiderivative = 0.44 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\frac {2 \left (\frac {2 b e n (f+g x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {e (f+g x)}{e f-d g}\right )}{e f-d g}-5 \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{35 g (f+g x)^{7/2}} \]

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(9/2),x]

[Out]

(2*((2*b*e*n*(f + g*x)*Hypergeometric2F1[-5/2, 1, -3/2, (e*(f + g*x))/(e*f - d*g)])/(e*f - d*g) - 5*(a + b*Log
[c*(d + e*x)^n])))/(35*g*(f + g*x)^(7/2))

Maple [F]

\[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\left (g x +f \right )^{\frac {9}{2}}}d x\]

[In]

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(9/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(9/2),x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 621 vs. \(2 (148) = 296\).

Time = 0.43 (sec) , antiderivative size = 1252, normalized size of antiderivative = 7.11 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(9/2),x, algorithm="fricas")

[Out]

[-2/105*(15*(b*e^3*g^4*n*x^4 + 4*b*e^3*f*g^3*n*x^3 + 6*b*e^3*f^2*g^2*n*x^2 + 4*b*e^3*f^3*g*n*x + b*e^3*f^4*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)) - (
30*b*e^3*g^3*n*x^3 - 15*a*e^3*f^3 + 45*a*d*e^2*f^2*g - 45*a*d^2*e*f*g^2 + 15*a*d^3*g^3 + 10*(10*b*e^3*f*g^2 -
b*d*e^2*g^3)*n*x^2 + 2*(58*b*e^3*f^2*g - 16*b*d*e^2*f*g^2 + 3*b*d^2*e*g^3)*n*x - 15*(b*e^3*f^3 - 3*b*d*e^2*f^2
*g + 3*b*d^2*e*f*g^2 - b*d^3*g^3)*n*log(e*x + d) + 2*(23*b*e^3*f^3 - 11*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2)*n - 1
5*(b*e^3*f^3 - 3*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2 - b*d^3*g^3)*log(c))*sqrt(g*x + f))/(e^3*f^7*g - 3*d*e^2*f^6*
g^2 + 3*d^2*e*f^5*g^3 - d^3*f^4*g^4 + (e^3*f^3*g^5 - 3*d*e^2*f^2*g^6 + 3*d^2*e*f*g^7 - d^3*g^8)*x^4 + 4*(e^3*f
^4*g^4 - 3*d*e^2*f^3*g^5 + 3*d^2*e*f^2*g^6 - d^3*f*g^7)*x^3 + 6*(e^3*f^5*g^3 - 3*d*e^2*f^4*g^4 + 3*d^2*e*f^3*g
^5 - d^3*f^2*g^6)*x^2 + 4*(e^3*f^6*g^2 - 3*d*e^2*f^5*g^3 + 3*d^2*e*f^4*g^4 - d^3*f^3*g^5)*x), -2/105*(30*(b*e^
3*g^4*n*x^4 + 4*b*e^3*f*g^3*n*x^3 + 6*b*e^3*f^2*g^2*n*x^2 + 4*b*e^3*f^3*g*n*x + b*e^3*f^4*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)) - (30*b*e^3*g^3*n*x^3 - 15*a*e^3*f^3
+ 45*a*d*e^2*f^2*g - 45*a*d^2*e*f*g^2 + 15*a*d^3*g^3 + 10*(10*b*e^3*f*g^2 - b*d*e^2*g^3)*n*x^2 + 2*(58*b*e^3*
f^2*g - 16*b*d*e^2*f*g^2 + 3*b*d^2*e*g^3)*n*x - 15*(b*e^3*f^3 - 3*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2 - b*d^3*g^3)
*n*log(e*x + d) + 2*(23*b*e^3*f^3 - 11*b*d*e^2*f^2*g + 3*b*d^2*e*f*g^2)*n - 15*(b*e^3*f^3 - 3*b*d*e^2*f^2*g +
3*b*d^2*e*f*g^2 - b*d^3*g^3)*log(c))*sqrt(g*x + f))/(e^3*f^7*g - 3*d*e^2*f^6*g^2 + 3*d^2*e*f^5*g^3 - d^3*f^4*g
^4 + (e^3*f^3*g^5 - 3*d*e^2*f^2*g^6 + 3*d^2*e*f*g^7 - d^3*g^8)*x^4 + 4*(e^3*f^4*g^4 - 3*d*e^2*f^3*g^5 + 3*d^2*
e*f^2*g^6 - d^3*f*g^7)*x^3 + 6*(e^3*f^5*g^3 - 3*d*e^2*f^4*g^4 + 3*d^2*e*f^3*g^5 - d^3*f^2*g^6)*x^2 + 4*(e^3*f^
6*g^2 - 3*d*e^2*f^5*g^3 + 3*d^2*e*f^4*g^4 - d^3*f^3*g^5)*x)]

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*ln(c*(e*x+d)**n))/(g*x+f)**(9/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(9/2),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 [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x + f\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x+f)^(9/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/(g*x + f)^(9/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{(f+g x)^{9/2}} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{{\left (f+g\,x\right )}^{9/2}} \,d x \]

[In]

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^(9/2),x)

[Out]

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^(9/2), x)