\(\int (f+g x)^{3/2} (a+b \log (c (d+e x)^n)) \, dx\) [138]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {4 b (e f-d g)^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g} \]

[Out]

-4/15*b*(-d*g+e*f)*n*(g*x+f)^(3/2)/e/g-4/25*b*n*(g*x+f)^(5/2)/g+4/5*b*(-d*g+e*f)^(5/2)*n*arctanh(e^(1/2)*(g*x+
f)^(1/2)/(-d*g+e*f)^(1/2))/e^(5/2)/g+2/5*(g*x+f)^(5/2)*(a+b*ln(c*(e*x+d)^n))/g-4/5*b*(-d*g+e*f)^2*n*(g*x+f)^(1
/2)/e^2/g

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 163, 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, 52, 65, 214} \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}+\frac {4 b n (e f-d g)^{5/2} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}-\frac {4 b n \sqrt {f+g x} (e f-d g)^2}{5 e^2 g}-\frac {4 b n (f+g x)^{3/2} (e f-d g)}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g} \]

[In]

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

[Out]

(-4*b*(e*f - d*g)^2*n*Sqrt[f + g*x])/(5*e^2*g) - (4*b*(e*f - d*g)*n*(f + g*x)^(3/2))/(15*e*g) - (4*b*n*(f + g*
x)^(5/2))/(25*g) + (4*b*(e*f - d*g)^(5/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(5*e^(5/2)*g) +
(2*(f + g*x)^(5/2)*(a + b*Log[c*(d + e*x)^n]))/(5*g)

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {(2 b e n) \int \frac {(f+g x)^{5/2}}{d+e x} \, dx}{5 g} \\ & = -\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {(2 b (e f-d g) n) \int \frac {(f+g x)^{3/2}}{d+e x} \, dx}{5 g} \\ & = -\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (2 b (e f-d g)^2 n\right ) \int \frac {\sqrt {f+g x}}{d+e x} \, dx}{5 e g} \\ & = -\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (2 b (e f-d g)^3 n\right ) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{5 e^2 g} \\ & = -\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac {\left (4 b (e f-d g)^3 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 )}{5 e^2 g^2} \\ & = -\frac {4 b (e f-d g)^2 n \sqrt {f+g x}}{5 e^2 g}-\frac {4 b (e f-d g) n (f+g x)^{3/2}}{15 e g}-\frac {4 b n (f+g x)^{5/2}}{25 g}+\frac {4 b (e f-d g)^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{5 e^{5/2} g}+\frac {2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.84 \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {2 \left (-\frac {2}{5} b n (f+g x)^{5/2}-\frac {2 b (e f-d g) n \left (\sqrt {e} \sqrt {f+g x} (4 e f-3 d g+e g x)-3 (e f-d g)^{3/2} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )}{3 e^{5/2}}+(f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{5 g} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (g x +f \right )^{\frac {3}{2}} \left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 538, normalized size of antiderivative = 3.30 \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\left [\frac {2 \, {\left (15 \, {\left (b e^{2} f^{2} - 2 \, b d e f g + b d^{2} g^{2}\right )} n \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g + 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) + {\left (15 \, a e^{2} f^{2} - 3 \, {\left (2 \, b e^{2} g^{2} n - 5 \, a e^{2} g^{2}\right )} x^{2} - 2 \, {\left (23 \, b e^{2} f^{2} - 35 \, b d e f g + 15 \, b d^{2} g^{2}\right )} n + 2 \, {\left (15 \, a e^{2} f g - {\left (11 \, b e^{2} f g - 5 \, b d e g^{2}\right )} n\right )} x + 15 \, {\left (b e^{2} g^{2} n x^{2} + 2 \, b e^{2} f g n x + b e^{2} f^{2} n\right )} \log \left (e x + d\right ) + 15 \, {\left (b e^{2} g^{2} x^{2} + 2 \, b e^{2} f g x + b e^{2} f^{2}\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{75 \, e^{2} g}, \frac {2 \, {\left (30 \, {\left (b e^{2} f^{2} - 2 \, b d e f g + b d^{2} g^{2}\right )} n \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) + {\left (15 \, a e^{2} f^{2} - 3 \, {\left (2 \, b e^{2} g^{2} n - 5 \, a e^{2} g^{2}\right )} x^{2} - 2 \, {\left (23 \, b e^{2} f^{2} - 35 \, b d e f g + 15 \, b d^{2} g^{2}\right )} n + 2 \, {\left (15 \, a e^{2} f g - {\left (11 \, b e^{2} f g - 5 \, b d e g^{2}\right )} n\right )} x + 15 \, {\left (b e^{2} g^{2} n x^{2} + 2 \, b e^{2} f g n x + b e^{2} f^{2} n\right )} \log \left (e x + d\right ) + 15 \, {\left (b e^{2} g^{2} x^{2} + 2 \, b e^{2} f g x + b e^{2} f^{2}\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{75 \, e^{2} g}\right ] \]

[In]

integrate((g*x+f)^(3/2)*(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

[2/75*(15*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*n*sqrt((e*f - d*g)/e)*log((e*g*x + 2*e*f - d*g + 2*sqrt(g*x +
f)*e*sqrt((e*f - d*g)/e))/(e*x + d)) + (15*a*e^2*f^2 - 3*(2*b*e^2*g^2*n - 5*a*e^2*g^2)*x^2 - 2*(23*b*e^2*f^2 -
35*b*d*e*f*g + 15*b*d^2*g^2)*n + 2*(15*a*e^2*f*g - (11*b*e^2*f*g - 5*b*d*e*g^2)*n)*x + 15*(b*e^2*g^2*n*x^2 +
2*b*e^2*f*g*n*x + b*e^2*f^2*n)*log(e*x + d) + 15*(b*e^2*g^2*x^2 + 2*b*e^2*f*g*x + b*e^2*f^2)*log(c))*sqrt(g*x
+ f))/(e^2*g), 2/75*(30*(b*e^2*f^2 - 2*b*d*e*f*g + b*d^2*g^2)*n*sqrt(-(e*f - d*g)/e)*arctan(-sqrt(g*x + f)*e*s
qrt(-(e*f - d*g)/e)/(e*f - d*g)) + (15*a*e^2*f^2 - 3*(2*b*e^2*g^2*n - 5*a*e^2*g^2)*x^2 - 2*(23*b*e^2*f^2 - 35*
b*d*e*f*g + 15*b*d^2*g^2)*n + 2*(15*a*e^2*f*g - (11*b*e^2*f*g - 5*b*d*e*g^2)*n)*x + 15*(b*e^2*g^2*n*x^2 + 2*b*
e^2*f*g*n*x + b*e^2*f^2*n)*log(e*x + d) + 15*(b*e^2*g^2*x^2 + 2*b*e^2*f*g*x + b*e^2*f^2)*log(c))*sqrt(g*x + f)
)/(e^2*g)]

Sympy [F]

\[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int \left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right ) \left (f + g x\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((g*x+f)**(3/2)*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))*(f + g*x)**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((g*x+f)^(3/2)*(a+b*log(c*(e*x+d)^n)),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]

\[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int { {\left (g x + f\right )}^{\frac {3}{2}} {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \,d x } \]

[In]

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

[Out]

integrate((g*x + f)^(3/2)*(b*log((e*x + d)^n*c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\int {\left (f+g\,x\right )}^{3/2}\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \]

[In]

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

[Out]

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