3.139 \(\int \sqrt {f+g x} (a+b \log (c (d+e x)^n)) \, dx\)

Optimal. Leaf size=132 \[ \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {4 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{3/2} g}-\frac {4 b n \sqrt {f+g x} (e f-d g)}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g} \]

[Out]

-4/9*b*n*(g*x+f)^(3/2)/g+4/3*b*(-d*g+e*f)^(3/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/e^(3/2)/g+2/
3*(g*x+f)^(3/2)*(a+b*ln(c*(e*x+d)^n))/g-4/3*b*(-d*g+e*f)*n*(g*x+f)^(1/2)/e/g

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Rubi [A]  time = 0.09, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2395, 50, 63, 208} \[ \frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}+\frac {4 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{3/2} g}-\frac {4 b n \sqrt {f+g x} (e f-d g)}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*(e*f - d*g)*n*Sqrt[f + g*x])/(3*e*g) - (4*b*n*(f + g*x)^(3/2))/(9*g) + (4*b*(e*f - d*g)^(3/2)*n*ArcTanh[
(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(3*e^(3/2)*g) + (2*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n]))/(3*g)

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]

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]

Rubi steps

\begin {align*} \int \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {(2 b e n) \int \frac {(f+g x)^{3/2}}{d+e x} \, dx}{3 g}\\ &=-\frac {4 b n (f+g x)^{3/2}}{9 g}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {(2 b (e f-d g) n) \int \frac {\sqrt {f+g x}}{d+e x} \, dx}{3 g}\\ &=-\frac {4 b (e f-d g) n \sqrt {f+g x}}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {\left (2 b (e f-d g)^2 n\right ) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{3 e g}\\ &=-\frac {4 b (e f-d g) n \sqrt {f+g x}}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}-\frac {\left (4 b (e f-d g)^2 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 )}{3 e g^2}\\ &=-\frac {4 b (e f-d g) n \sqrt {f+g x}}{3 e g}-\frac {4 b n (f+g x)^{3/2}}{9 g}+\frac {4 b (e f-d g)^{3/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{3 e^{3/2} g}+\frac {2 (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 118, normalized size = 0.89 \[ \frac {2 \left (\sqrt {e} \sqrt {f+g x} \left (3 a e (f+g x)+3 b e (f+g x) \log \left (c (d+e x)^n\right )-2 b n (-3 d g+4 e f+e g x)\right )+6 b n (e f-d g)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )}{9 e^{3/2} g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(6*b*(e*f - d*g)^(3/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]] + Sqrt[e]*Sqrt[f + g*x]*(3*a*e*(f
+ g*x) - 2*b*n*(4*e*f - 3*d*g + e*g*x) + 3*b*e*(f + g*x)*Log[c*(d + e*x)^n])))/(9*e^(3/2)*g)

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fricas [A]  time = 0.50, size = 311, normalized size = 2.36 \[ \left [-\frac {2 \, {\left (3 \, {\left (b e f - b d g\right )} n \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g - 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) - {\left (3 \, a e f - 2 \, {\left (4 \, b e f - 3 \, b d g\right )} n - {\left (2 \, b e g n - 3 \, a e g\right )} x + 3 \, {\left (b e g n x + b e f n\right )} \log \left (e x + d\right ) + 3 \, {\left (b e g x + b e f\right )} \log \relax (c)\right )} \sqrt {g x + f}\right )}}{9 \, e g}, \frac {2 \, {\left (6 \, {\left (b e f - b d g\right )} n \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) + {\left (3 \, a e f - 2 \, {\left (4 \, b e f - 3 \, b d g\right )} n - {\left (2 \, b e g n - 3 \, a e g\right )} x + 3 \, {\left (b e g n x + b e f n\right )} \log \left (e x + d\right ) + 3 \, {\left (b e g x + b e f\right )} \log \relax (c)\right )} \sqrt {g x + f}\right )}}{9 \, e g}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(1/2)*(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

[-2/9*(3*(b*e*f - b*d*g)*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)) - (3*a*e*f - 2*(4*b*e*f - 3*b*d*g)*n - (2*b*e*g*n - 3*a*e*g)*x + 3*(b*e*g*n*x + b*e*f*n)*log(e*x
+ d) + 3*(b*e*g*x + b*e*f)*log(c))*sqrt(g*x + f))/(e*g), 2/9*(6*(b*e*f - b*d*g)*n*sqrt(-(e*f - d*g)/e)*arctan
(-sqrt(g*x + f)*e*sqrt(-(e*f - d*g)/e)/(e*f - d*g)) + (3*a*e*f - 2*(4*b*e*f - 3*b*d*g)*n - (2*b*e*g*n - 3*a*e*
g)*x + 3*(b*e*g*n*x + b*e*f*n)*log(e*x + d) + 3*(b*e*g*x + b*e*f)*log(c))*sqrt(g*x + f))/(e*g)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {g x + f} {\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(1/2)*(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.61, size = 0, normalized size = 0.00 \[ \int \sqrt {g x +f}\, \left (b \ln \left (c \left (e x +d \right )^{n}\right )+a \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^(1/2)*(b*ln(c*(e*x+d)^n)+a),x)

[Out]

int((g*x+f)^(1/2)*(b*ln(c*(e*x+d)^n)+a),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^(1/2)*(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(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {f+g\,x}\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)^(1/2)*(a + b*log(c*(d + e*x)^n)),x)

[Out]

int((f + g*x)^(1/2)*(a + b*log(c*(d + e*x)^n)), x)

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sympy [A]  time = 4.72, size = 139, normalized size = 1.05 \[ \frac {2 \left (\frac {a \left (f + g x\right )^{\frac {3}{2}}}{3} + b \left (- \frac {2 e n \left (\frac {g \left (f + g x\right )^{\frac {3}{2}}}{3 e} + \frac {\sqrt {f + g x} \left (- d g^{2} + e f g\right )}{e^{2}} + \frac {g \left (d g - e f\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{e^{3} \sqrt {\frac {d g - e f}{e}}}\right )}{3 g} + \frac {\left (f + g x\right )^{\frac {3}{2}} \log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{3}\right )\right )}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**(1/2)*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

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

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