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

Optimal. Leaf size=163 \[ \frac{2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{4 b n \sqrt{f+g x} (e f-d g)^2}{5 e^2 g}+\frac{4 b n (e f-d g)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{5 e^{5/2} g}-\frac{4 b n (f+g x)^{3/2} (e f-d g)}{15 e g}-\frac{4 b n (f+g x)^{5/2}}{25 g} \]

[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)

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Rubi [A]  time = 0.163095, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2395, 50, 63, 208} \[ \frac{2 (f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{4 b n \sqrt{f+g x} (e f-d g)^2}{5 e^2 g}+\frac{4 b n (e f-d g)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{5 e^{5/2} g}-\frac{4 b n (f+g x)^{3/2} (e f-d g)}{15 e g}-\frac{4 b n (f+g x)^{5/2}}{25 g} \]

Antiderivative was successfully verified.

[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 2395

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]

Rule 50

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 63

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 208

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

Rubi steps

\begin{align*} \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{(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 ) \operatorname{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]  time = 0.221988, size = 137, normalized size = 0.84 \[ \frac{2 \left ((f+g x)^{5/2} \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{2 b n (e f-d g) \left (\sqrt{e} \sqrt{f+g x} (-3 d g+4 e f+e g x)-3 (e f-d g)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )\right )}{3 e^{5/2}}-\frac{2}{5} b n (f+g x)^{5/2}\right )}{5 g} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 1.185, size = 0, normalized size = 0. \begin{align*} \int \left ( gx+f \right ) ^{{\frac{3}{2}}} \left ( a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98157, size = 1207, normalized size = 7.4 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)