3.2.42 \(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^{5/2}} \, dx\) [142]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 114, 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 = {2442, 53, 65, 214} \begin {gather*} -\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)} \end {gather*}

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 53

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 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*} \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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 85, normalized size = 0.75 \begin {gather*} -\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 (-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}} \end {gather*}

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 = 0.13, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\left (g x +f \right )^{\frac {5}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*%e^2*f-4*%e*d*g>0)', see `as
sume?` for m

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Fricas [A]
time = 0.41, size = 414, normalized size = 3.63 \begin {gather*} \left [-\frac {2 \, {\left ({\left (b g^{2} n x^{2} + 2 \, b f g n x + b f^{2} n\right )} \sqrt {-\frac {e}{d g - f e}} e \log \left (-\frac {d g - 2 \, {\left (d g - f e\right )} \sqrt {g x + f} \sqrt {-\frac {e}{d g - f e}} - {\left (g x + 2 \, f\right )} e}{x e + d}\right ) + {\left (a d g + {\left (2 \, b g n x + 2 \, b f n - a f\right )} e + {\left (b d g n - b f n e\right )} \log \left (x e + d\right ) + {\left (b d g - b f e\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{3 \, {\left (d g^{4} x^{2} + 2 \, d f g^{3} x + d f^{2} g^{2} - {\left (f g^{3} x^{2} + 2 \, f^{2} g^{2} x + f^{3} g\right )} e\right )}}, -\frac {2 \, {\left (\frac {2 \, {\left (b g^{2} n x^{2} + 2 \, b f g n x + b f^{2} n\right )} \arctan \left (-\frac {\sqrt {d g - f e} e^{\left (-\frac {1}{2}\right )}}{\sqrt {g x + f}}\right ) e^{\frac {3}{2}}}{\sqrt {d g - f e}} + {\left (a d g + {\left (2 \, b g n x + 2 \, b f n - a f\right )} e + {\left (b d g n - b f n e\right )} \log \left (x e + d\right ) + {\left (b d g - b f e\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{3 \, {\left (d g^{4} x^{2} + 2 \, d f g^{3} x + d f^{2} g^{2} - {\left (f g^{3} x^{2} + 2 \, f^{2} g^{2} x + f^{3} g\right )} e\right )}}\right ] \end {gather*}

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*g^2*n*x^2 + 2*b*f*g*n*x + b*f^2*n)*sqrt(-e/(d*g - f*e))*e*log(-(d*g - 2*(d*g - f*e)*sqrt(g*x + f)*sq
rt(-e/(d*g - f*e)) - (g*x + 2*f)*e)/(x*e + d)) + (a*d*g + (2*b*g*n*x + 2*b*f*n - a*f)*e + (b*d*g*n - b*f*n*e)*
log(x*e + d) + (b*d*g - b*f*e)*log(c))*sqrt(g*x + f))/(d*g^4*x^2 + 2*d*f*g^3*x + d*f^2*g^2 - (f*g^3*x^2 + 2*f^
2*g^2*x + f^3*g)*e), -2/3*(2*(b*g^2*n*x^2 + 2*b*f*g*n*x + b*f^2*n)*arctan(-sqrt(d*g - f*e)*e^(-1/2)/sqrt(g*x +
f))*e^(3/2)/sqrt(d*g - f*e) + (a*d*g + (2*b*g*n*x + 2*b*f*n - a*f)*e + (b*d*g*n - b*f*n*e)*log(x*e + d) + (b*
d*g - b*f*e)*log(c))*sqrt(g*x + f))/(d*g^4*x^2 + 2*d*f*g^3*x + d*f^2*g^2 - (f*g^3*x^2 + 2*f^2*g^2*x + f^3*g)*e
)]

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Sympy [A]
time = 34.24, size = 117, normalized size = 1.03 \begin {gather*} \frac {- \frac {2 a}{3 \left (f + g x\right )^{\frac {3}{2}}} + 2 b \left (\frac {2 e n \left (- \frac {g}{\sqrt {f + g x} \left (d g - e f\right )} - \frac {g \operatorname {atan}{\left (\frac {\sqrt {f + g x}}{\sqrt {\frac {d g - e f}{e}}} \right )}}{\sqrt {\frac {d g - e f}{e}} \left (d g - e f\right )}\right )}{3 g} - \frac {\log {\left (c \left (d - \frac {e f}{g} + \frac {e \left (f + g x\right )}{g}\right )^{n} \right )}}{3 \left (f + g x\right )^{\frac {3}{2}}}\right )}{g} \end {gather*}

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]

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

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Giac [A]
time = 6.00, size = 188, normalized size = 1.65 \begin {gather*} -\frac {4 \, b n \arctan \left (\frac {\sqrt {g x + f} e}{\sqrt {d g e - f e^{2}}}\right ) e^{2}}{3 \, {\left (d g^{2} - f g 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 {gather*}

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*n*arctan(sqrt(g*x + f)*e/sqrt(d*g*e - f*e^2))*e^2/((d*g^2 - f*g*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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{{\left (f+g\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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