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

Optimal. Leaf size=97 \[ \frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac {4 b n \sqrt {e f-d g} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} g}-\frac {4 b n \sqrt {f+g x}}{g} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2395, 50, 63, 208} \[ \frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}+\frac {4 b n \sqrt {e f-d g} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} g}-\frac {4 b n \sqrt {f+g x}}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*n*Sqrt[f + g*x])/g + (4*b*Sqrt[e*f - d*g]*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(Sqrt[e]*g
) + (2*Sqrt[f + g*x]*(a + b*Log[c*(d + e*x)^n]))/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 \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {f+g x}} \, dx &=\frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {(2 b e n) \int \frac {\sqrt {f+g x}}{d+e x} \, dx}{g}\\ &=-\frac {4 b n \sqrt {f+g x}}{g}+\frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {(2 b (e f-d g) n) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{g}\\ &=-\frac {4 b n \sqrt {f+g x}}{g}+\frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}-\frac {(4 b (e f-d g) n) \operatorname {Subst}\left (\int \frac {1}{d-\frac {e f}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{g^2}\\ &=-\frac {4 b n \sqrt {f+g x}}{g}+\frac {4 b \sqrt {e f-d g} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e} g}+\frac {2 \sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )\right )}{g}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 83, normalized size = 0.86 \[ \frac {2 \left (\sqrt {f+g x} \left (a+b \log \left (c (d+e x)^n\right )-2 b n\right )+\frac {2 b n \sqrt {e f-d g} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{\sqrt {e}}\right )}{g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 185, normalized size = 1.91 \[ \left [\frac {2 \, {\left (b n \sqrt {\frac {e f - d g}{e}} \log \left (\frac {e g x + 2 \, e f - d g + 2 \, \sqrt {g x + f} e \sqrt {\frac {e f - d g}{e}}}{e x + d}\right ) + {\left (b n \log \left (e x + d\right ) - 2 \, b n + b \log \relax (c) + a\right )} \sqrt {g x + f}\right )}}{g}, \frac {2 \, {\left (2 \, b n \sqrt {-\frac {e f - d g}{e}} \arctan \left (-\frac {\sqrt {g x + f} e \sqrt {-\frac {e f - d g}{e}}}{e f - d g}\right ) + {\left (b n \log \left (e x + d\right ) - 2 \, b n + b \log \relax (c) + a\right )} \sqrt {g x + f}\right )}}{g}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 110, normalized size = 1.13 \[ \frac {2 \, {\left ({\left (2 \, {\left (\frac {{\left (d g - f e\right )} \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right ) e^{\left (-1\right )}}{\sqrt {d g e - f e^{2}}} - \sqrt {g x + f} e^{\left (-1\right )}\right )} e + \sqrt {g x + f} \log \left (x e + d\right )\right )} b n + \sqrt {g x + f} b \log \relax (c) + \sqrt {g x + f} a\right )}}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*((d*g - f*e)*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))*e^(-1)/sqrt(d*g*e - f*e^2) - sqrt(g*x + f)*e^(-
1))*e + sqrt(g*x + f)*log(x*e + d))*b*n + sqrt(g*x + f)*b*log(c) + sqrt(g*x + f)*a)/g

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maple [A]  time = 0.08, size = 148, normalized size = 1.53 \[ \frac {4 b d n \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}}-\frac {4 b e f n \arctan \left (\frac {\sqrt {g x +f}\, e}{\sqrt {\left (d g -e f \right ) e}}\right )}{\sqrt {\left (d g -e f \right ) e}\, g}-\frac {4 \sqrt {g x +f}\, b n}{g}+\frac {2 \sqrt {g x +f}\, b \ln \left (c \left (\frac {d g -e f +\left (g x +f \right ) e}{g}\right )^{n}\right )}{g}+\frac {2 \sqrt {g x +f}\, a}{g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+b*log(c*(e*x+d)^n))/(g*x+f)^(1/2),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(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 \[ \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{\sqrt {f+g\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 38.95, size = 326, normalized size = 3.36 \[ \begin {cases} \frac {- \frac {2 a f}{\sqrt {f + g x}} - 2 a \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right ) - 2 b f \left (\frac {2 e n \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {e}{d g - e f}} \sqrt {f + g x}} \right )}}{\sqrt {\frac {e}{d g - e f}} \left (d g - e f\right )} + \frac {\log {\left (c \left (d + e x\right )^{n} \right )}}{\sqrt {f + g x}}\right ) - 2 b \left (- \frac {2 e n \left (- \frac {g \sqrt {f + g x}}{e} - \frac {g \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {e}{d g - e f}} \sqrt {f + g x}} \right )}}{e \sqrt {\frac {e}{d g - e f}}}\right )}{g} - f \left (\frac {2 e n \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {e}{d g - e f}} \sqrt {f + g x}} \right )}}{\sqrt {\frac {e}{d g - e f}} \left (d g - e f\right )} + \frac {\log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{\sqrt {f + g x}}\right ) - \sqrt {f + g x} \log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}\right )}{g} & \text {for}\: g \neq 0 \\\frac {a x + b \left (- e n \left (- \frac {d \left (\begin {cases} \frac {x}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x \right )}}{e} & \text {otherwise} \end {cases}\right )}{e} + \frac {x}{e}\right ) + x \log {\left (c \left (d + e x\right )^{n} \right )}\right )}{\sqrt {f}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*a*f/sqrt(f + g*x) - 2*a*(-f/sqrt(f + g*x) - sqrt(f + g*x)) - 2*b*f*(2*e*n*atan(1/(sqrt(e/(d*g -
e*f))*sqrt(f + g*x)))/(sqrt(e/(d*g - e*f))*(d*g - e*f)) + log(c*(d + e*x)**n)/sqrt(f + g*x)) - 2*b*(-2*e*n*(-
g*sqrt(f + g*x)/e - g*atan(1/(sqrt(e/(d*g - e*f))*sqrt(f + g*x)))/(e*sqrt(e/(d*g - e*f))))/g - f*(2*e*n*atan(1
/(sqrt(e/(d*g - e*f))*sqrt(f + g*x)))/(sqrt(e/(d*g - e*f))*(d*g - e*f)) + log(c*(d - e*f/g + e*(f + g*x)/g)**n
)/sqrt(f + g*x)) - sqrt(f + g*x)*log(c*(d - e*f/g + e*(f + g*x)/g)**n)))/g, Ne(g, 0)), ((a*x + b*(-e*n*(-d*Pie
cewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/e + x/e) + x*log(c*(d + e*x)**n)))/sqrt(f), True))

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