\(\int (f+g x) (a+b \log (c (d+e x)^n))^{5/2} \, dx\) [118]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F(-2)]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 24, antiderivative size = 413 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx=-\frac {15 b^{5/2} e^{-\frac {a}{b n}} (e f-d g) n^{5/2} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e^2}-\frac {15 b^{5/2} e^{-\frac {2 a}{b n}} g n^{5/2} \sqrt {\frac {\pi }{2}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{64 e^2}+\frac {15 b^2 (e f-d g) n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}-\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2} \]

[Out]

-5/2*b*(-d*g+e*f)*n*(e*x+d)*(a+b*ln(c*(e*x+d)^n))^(3/2)/e^2-5/8*b*g*n*(e*x+d)^2*(a+b*ln(c*(e*x+d)^n))^(3/2)/e^
2+(-d*g+e*f)*(e*x+d)*(a+b*ln(c*(e*x+d)^n))^(5/2)/e^2+1/2*g*(e*x+d)^2*(a+b*ln(c*(e*x+d)^n))^(5/2)/e^2-15/128*b^
(5/2)*g*n^(5/2)*(e*x+d)^2*erfi(2^(1/2)*(a+b*ln(c*(e*x+d)^n))^(1/2)/b^(1/2)/n^(1/2))*2^(1/2)*Pi^(1/2)/e^2/exp(2
*a/b/n)/((c*(e*x+d)^n)^(2/n))-15/8*b^(5/2)*(-d*g+e*f)*n^(5/2)*(e*x+d)*erfi((a+b*ln(c*(e*x+d)^n))^(1/2)/b^(1/2)
/n^(1/2))*Pi^(1/2)/e^2/exp(a/b/n)/((c*(e*x+d)^n)^(1/n))+15/4*b^2*(-d*g+e*f)*n^2*(e*x+d)*(a+b*ln(c*(e*x+d)^n))^
(1/2)/e^2+15/32*b^2*g*n^2*(e*x+d)^2*(a+b*ln(c*(e*x+d)^n))^(1/2)/e^2

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2448, 2436, 2333, 2337, 2211, 2235, 2437, 2342, 2347} \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx=-\frac {15 \sqrt {\pi } b^{5/2} n^{5/2} e^{-\frac {a}{b n}} (d+e x) (e f-d g) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e^2}-\frac {15 \sqrt {\frac {\pi }{2}} b^{5/2} g n^{5/2} e^{-\frac {2 a}{b n}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{64 e^2}+\frac {15 b^2 n^2 (d+e x) (e f-d g) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}+\frac {(d+e x) (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}-\frac {5 b n (d+e x) (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2} \]

[In]

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

[Out]

(-15*b^(5/2)*(e*f - d*g)*n^(5/2)*Sqrt[Pi]*(d + e*x)*Erfi[Sqrt[a + b*Log[c*(d + e*x)^n]]/(Sqrt[b]*Sqrt[n])])/(8
*e^2*E^(a/(b*n))*(c*(d + e*x)^n)^n^(-1)) - (15*b^(5/2)*g*n^(5/2)*Sqrt[Pi/2]*(d + e*x)^2*Erfi[(Sqrt[2]*Sqrt[a +
 b*Log[c*(d + e*x)^n]])/(Sqrt[b]*Sqrt[n])])/(64*e^2*E^((2*a)/(b*n))*(c*(d + e*x)^n)^(2/n)) + (15*b^2*(e*f - d*
g)*n^2*(d + e*x)*Sqrt[a + b*Log[c*(d + e*x)^n]])/(4*e^2) + (15*b^2*g*n^2*(d + e*x)^2*Sqrt[a + b*Log[c*(d + e*x
)^n]])/(32*e^2) - (5*b*(e*f - d*g)*n*(d + e*x)*(a + b*Log[c*(d + e*x)^n])^(3/2))/(2*e^2) - (5*b*g*n*(d + e*x)^
2*(a + b*Log[c*(d + e*x)^n])^(3/2))/(8*e^2) + ((e*f - d*g)*(d + e*x)*(a + b*Log[c*(d + e*x)^n])^(5/2))/e^2 + (
g*(d + e*x)^2*(a + b*Log[c*(d + e*x)^n])^(5/2))/(2*e^2)

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2437

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

Rule 2448

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}+\frac {g (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e}\right ) \, dx \\ & = \frac {g \int (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx}{e}+\frac {(e f-d g) \int \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx}{e} \\ & = \frac {g \text {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right )^{5/2} \, dx,x,d+e x\right )}{e^2}+\frac {(e f-d g) \text {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^{5/2} \, dx,x,d+e x\right )}{e^2} \\ & = \frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}-\frac {(5 b g n) \text {Subst}\left (\int x \left (a+b \log \left (c x^n\right )\right )^{3/2} \, dx,x,d+e x\right )}{4 e^2}-\frac {(5 b (e f-d g) n) \text {Subst}\left (\int \left (a+b \log \left (c x^n\right )\right )^{3/2} \, dx,x,d+e x\right )}{2 e^2} \\ & = -\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}+\frac {\left (15 b^2 g n^2\right ) \text {Subst}\left (\int x \sqrt {a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{16 e^2}+\frac {\left (15 b^2 (e f-d g) n^2\right ) \text {Subst}\left (\int \sqrt {a+b \log \left (c x^n\right )} \, dx,x,d+e x\right )}{4 e^2} \\ & = \frac {15 b^2 (e f-d g) n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}-\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}-\frac {\left (15 b^3 g n^3\right ) \text {Subst}\left (\int \frac {x}{\sqrt {a+b \log \left (c x^n\right )}} \, dx,x,d+e x\right )}{64 e^2}-\frac {\left (15 b^3 (e f-d g) n^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b \log \left (c x^n\right )}} \, dx,x,d+e x\right )}{8 e^2} \\ & = \frac {15 b^2 (e f-d g) n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}-\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}-\frac {\left (15 b^3 g n^2 (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {2 x}{n}}}{\sqrt {a+b x}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{64 e^2}-\frac {\left (15 b^3 (e f-d g) n^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{n}}}{\sqrt {a+b x}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{8 e^2} \\ & = \frac {15 b^2 (e f-d g) n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}-\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2}-\frac {\left (15 b^2 g n^2 (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n}\right ) \text {Subst}\left (\int e^{-\frac {2 a}{b n}+\frac {2 x^2}{b n}} \, dx,x,\sqrt {a+b \log \left (c (d+e x)^n\right )}\right )}{32 e^2}-\frac {\left (15 b^2 (e f-d g) n^2 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int e^{-\frac {a}{b n}+\frac {x^2}{b n}} \, dx,x,\sqrt {a+b \log \left (c (d+e x)^n\right )}\right )}{4 e^2} \\ & = -\frac {15 b^{5/2} e^{-\frac {a}{b n}} (e f-d g) n^{5/2} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{8 e^2}-\frac {15 b^{5/2} e^{-\frac {2 a}{b n}} g n^{5/2} \sqrt {\frac {\pi }{2}} (d+e x)^2 \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{64 e^2}+\frac {15 b^2 (e f-d g) n^2 (d+e x) \sqrt {a+b \log \left (c (d+e x)^n\right )}}{4 e^2}+\frac {15 b^2 g n^2 (d+e x)^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}{32 e^2}-\frac {5 b (e f-d g) n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{2 e^2}-\frac {5 b g n (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}{8 e^2}+\frac {(e f-d g) (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{e^2}+\frac {g (d+e x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}}{2 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.79 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2} \, dx=\frac {(d+e x) \left (128 (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}+64 g (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}-80 b (e f-d g) n \left (3 b^{3/2} e^{-\frac {a}{b n}} n^{3/2} \sqrt {\pi } \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )+2 \sqrt {a+b \log \left (c (d+e x)^n\right )} \left (2 a-3 b n+2 b \log \left (c (d+e x)^n\right )\right )\right )-5 b g n (d+e x) \left (3 b^{3/2} e^{-\frac {2 a}{b n}} n^{3/2} \sqrt {2 \pi } \left (c (d+e x)^n\right )^{-2/n} \text {erfi}\left (\frac {\sqrt {2} \sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )+4 \sqrt {a+b \log \left (c (d+e x)^n\right )} \left (4 a-3 b n+4 b \log \left (c (d+e x)^n\right )\right )\right )\right )}{128 e^2} \]

[In]

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

[Out]

((d + e*x)*(128*(e*f - d*g)*(a + b*Log[c*(d + e*x)^n])^(5/2) + 64*g*(d + e*x)*(a + b*Log[c*(d + e*x)^n])^(5/2)
 - 80*b*(e*f - d*g)*n*((3*b^(3/2)*n^(3/2)*Sqrt[Pi]*Erfi[Sqrt[a + b*Log[c*(d + e*x)^n]]/(Sqrt[b]*Sqrt[n])])/(E^
(a/(b*n))*(c*(d + e*x)^n)^n^(-1)) + 2*Sqrt[a + b*Log[c*(d + e*x)^n]]*(2*a - 3*b*n + 2*b*Log[c*(d + e*x)^n])) -
 5*b*g*n*(d + e*x)*((3*b^(3/2)*n^(3/2)*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*Log[c*(d + e*x)^n]])/(Sqrt[b]*Sqrt[
n])])/(E^((2*a)/(b*n))*(c*(d + e*x)^n)^(2/n)) + 4*Sqrt[a + b*Log[c*(d + e*x)^n]]*(4*a - 3*b*n + 4*b*Log[c*(d +
 e*x)^n]))))/(128*e^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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