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

Optimal. Leaf size=114 \[ -\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}}-\frac{4 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 g (e f-d g)^{3/2}}+\frac{4 b e n}{3 g \sqrt{f+g x} (e f-d g)} \]

[Out]

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

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Rubi [A]  time = 0.0815045, antiderivative size = 114, normalized size of antiderivative = 1., 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, 51, 63, 208} \[ -\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}}-\frac{4 b e^{3/2} n \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{f+g x}}{\sqrt{e f-d g}}\right )}{3 g (e f-d g)^{3/2}}+\frac{4 b e n}{3 g \sqrt{f+g x} (e f-d g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 51

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

Mathematica [C]  time = 0.0363205, size = 85, normalized size = 0.75 \[ -\frac{2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^{3/2}}-\frac{4 b e n \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{e (f+g x)}{e f-d g}\right )}{3 g \sqrt{f+g x} (d g-e f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*e*n*Hypergeometric2F1[-1/2, 1, 1/2, (e*(f + g*x))/(e*f - d*g)])/(3*g*(-(e*f) + d*g)*Sqrt[f + g*x]) - (2*
(a + b*Log[c*(d + e*x)^n]))/(3*g*(f + g*x)^(3/2))

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Maple [F]  time = 0.916, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) ) \left ( gx+f \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.99567, size = 929, normalized size = 8.15 \begin{align*} \left [-\frac{2 \,{\left ({\left (b e g^{2} n x^{2} + 2 \, b e f g n x + b e f^{2} n\right )} \sqrt{\frac{e}{e f - d g}} \log \left (\frac{e g x + 2 \, e f - d g + 2 \,{\left (e f - d g\right )} \sqrt{g x + f} \sqrt{\frac{e}{e f - d g}}}{e x + d}\right ) -{\left (2 \, b e g n x + 2 \, b e f n - a e f + a d g -{\left (b e f - b d g\right )} n \log \left (e x + d\right ) -{\left (b e f - b d g\right )} \log \left (c\right )\right )} \sqrt{g x + f}\right )}}{3 \,{\left (e f^{3} g - d f^{2} g^{2} +{\left (e f g^{3} - d g^{4}\right )} x^{2} + 2 \,{\left (e f^{2} g^{2} - d f g^{3}\right )} x\right )}}, -\frac{2 \,{\left (2 \,{\left (b e g^{2} n x^{2} + 2 \, b e f g n x + b e f^{2} n\right )} \sqrt{-\frac{e}{e f - d g}} \arctan \left (-\frac{{\left (e f - d g\right )} \sqrt{g x + f} \sqrt{-\frac{e}{e f - d g}}}{e g x + e f}\right ) -{\left (2 \, b e g n x + 2 \, b e f n - a e f + a d g -{\left (b e f - b d g\right )} n \log \left (e x + d\right ) -{\left (b e f - b d g\right )} \log \left (c\right )\right )} \sqrt{g x + f}\right )}}{3 \,{\left (e f^{3} g - d f^{2} g^{2} +{\left (e f g^{3} - d g^{4}\right )} x^{2} + 2 \,{\left (e f^{2} g^{2} - d f g^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.27552, size = 258, normalized size = 2.26 \begin{align*} -\frac{4 \, b g n \arctan \left (\frac{\sqrt{g x + f} e}{\sqrt{d g e - f e^{2}}}\right ) e^{2}}{3 \,{\left (d g^{3} - f g^{2} e\right )} \sqrt{d g e - f e^{2}}} - \frac{2 \,{\left (b d g n \log \left (d g +{\left (g x + f\right )} e - f e\right ) - b f n e \log \left (d g +{\left (g x + f\right )} e - f e\right ) - b d g n \log \left (g\right ) + b f n e \log \left (g\right ) + 2 \,{\left (g x + f\right )} b n e + b d g \log \left (c\right ) - b f e \log \left (c\right ) + a d g - a f e\right )}}{3 \,{\left ({\left (g x + f\right )}^{\frac{3}{2}} d g^{2} -{\left (g x + f\right )}^{\frac{3}{2}} f g e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3*b*g*n*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))*e^2/((d*g^3 - f*g^2*e)*sqrt(d*g*e - f*e^2)) - 2/3*(b*d*
g*n*log(d*g + (g*x + f)*e - f*e) - b*f*n*e*log(d*g + (g*x + f)*e - f*e) - b*d*g*n*log(g) + b*f*n*e*log(g) + 2*
(g*x + f)*b*n*e + b*d*g*log(c) - b*f*e*log(c) + a*d*g - a*f*e)/((g*x + f)^(3/2)*d*g^2 - (g*x + f)^(3/2)*f*g*e)