3.2.44 \(\int \frac {a+b \log (c (d+e x)^n)}{(f+g x)^{9/2}} \, dx\) [144]

Optimal. Leaf size=176 \[ \frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {4 b e^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}} \]

[Out]

4/35*b*e*n/g/(-d*g+e*f)/(g*x+f)^(5/2)+4/21*b*e^2*n/g/(-d*g+e*f)^2/(g*x+f)^(3/2)-4/7*b*e^(7/2)*n*arctanh(e^(1/2
)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/g/(-d*g+e*f)^(7/2)-2/7*(a+b*ln(c*(e*x+d)^n))/g/(g*x+f)^(7/2)+4/7*b*e^3*n/g/(
-d*g+e*f)^3/(g*x+f)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 176, normalized size of antiderivative = 1.00, 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 = {2442, 53, 65, 214} \begin {gather*} -\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}-\frac {4 b e^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}+\frac {4 b e^3 n}{7 g \sqrt {f+g x} (e f-d g)^3}+\frac {4 b e^2 n}{21 g (f+g x)^{3/2} (e f-d g)^2}+\frac {4 b e n}{35 g (f+g x)^{5/2} (e f-d g)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(9/2),x]

[Out]

(4*b*e*n)/(35*g*(e*f - d*g)*(f + g*x)^(5/2)) + (4*b*e^2*n)/(21*g*(e*f - d*g)^2*(f + g*x)^(3/2)) + (4*b*e^3*n)/
(7*g*(e*f - d*g)^3*Sqrt[f + g*x]) - (4*b*e^(7/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]])/(7*g*(e*f
- d*g)^(7/2)) - (2*(a + b*Log[c*(d + e*x)^n]))/(7*g*(f + g*x)^(7/2))

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)^{9/2}} \, dx &=-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {(2 b e n) \int \frac {1}{(d+e x) (f+g x)^{7/2}} \, dx}{7 g}\\ &=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^2 n\right ) \int \frac {1}{(d+e x) (f+g x)^{5/2}} \, dx}{7 g (e f-d g)}\\ &=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^3 n\right ) \int \frac {1}{(d+e x) (f+g x)^{3/2}} \, dx}{7 g (e f-d g)^2}\\ &=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (2 b e^4 n\right ) \int \frac {1}{(d+e x) \sqrt {f+g x}} \, dx}{7 g (e f-d g)^3}\\ &=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}+\frac {\left (4 b e^4 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 )}{7 g^2 (e f-d g)^3}\\ &=\frac {4 b e n}{35 g (e f-d g) (f+g x)^{5/2}}+\frac {4 b e^2 n}{21 g (e f-d g)^2 (f+g x)^{3/2}}+\frac {4 b e^3 n}{7 g (e f-d g)^3 \sqrt {f+g x}}-\frac {4 b e^{7/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{7 g (e f-d g)^{7/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{7 g (f+g x)^{7/2}}\\ \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 = 78, normalized size = 0.44 \begin {gather*} \frac {2 \left (\frac {2 b e n (f+g x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};\frac {e (f+g x)}{e f-d g}\right )}{e f-d g}-5 \left (a+b \log \left (c (d+e x)^n\right )\right )\right )}{35 g (f+g x)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/(f + g*x)^(9/2),x]

[Out]

(2*((2*b*e*n*(f + g*x)*Hypergeometric2F1[-5/2, 1, -3/2, (e*(f + g*x))/(e*f - d*g)])/(e*f - d*g) - 5*(a + b*Log
[c*(d + e*x)^n])))/(35*g*(f + g*x)^(7/2))

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (e x +d \right )^{n}\right )}{\left (g x +f \right )^{\frac {9}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(9/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(g*x+f)^(9/2),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)^(9/2),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 [B] Leaf count of result is larger than twice the leaf count of optimal. 566 vs. \(2 (151) = 302\).
time = 0.40, size = 1177, normalized size = 6.69 \begin {gather*} \left [-\frac {2 \, {\left (15 \, {\left (b g^{4} n x^{4} + 4 \, b f g^{3} n x^{3} + 6 \, b f^{2} g^{2} n x^{2} + 4 \, b f^{3} g n x + b f^{4} n\right )} \sqrt {-\frac {e}{d g - f e}} e^{3} \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 (15 \, a d^{3} g^{3} + {\left (30 \, b g^{3} n x^{3} + 100 \, b f g^{2} n x^{2} + 116 \, b f^{2} g n x + 46 \, b f^{3} n - 15 \, a f^{3}\right )} e^{3} - {\left (10 \, b d g^{3} n x^{2} + 32 \, b d f g^{2} n x + 22 \, b d f^{2} g n - 45 \, a d f^{2} g\right )} e^{2} + 3 \, {\left (2 \, b d^{2} g^{3} n x + 2 \, b d^{2} f g^{2} n - 15 \, a d^{2} f g^{2}\right )} e + 15 \, {\left (b d^{3} g^{3} n - 3 \, b d^{2} f g^{2} n e + 3 \, b d f^{2} g n e^{2} - b f^{3} n e^{3}\right )} \log \left (x e + d\right ) + 15 \, {\left (b d^{3} g^{3} - 3 \, b d^{2} f g^{2} e + 3 \, b d f^{2} g e^{2} - b f^{3} e^{3}\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{105 \, {\left (d^{3} g^{8} x^{4} + 4 \, d^{3} f g^{7} x^{3} + 6 \, d^{3} f^{2} g^{6} x^{2} + 4 \, d^{3} f^{3} g^{5} x + d^{3} f^{4} g^{4} - {\left (f^{3} g^{5} x^{4} + 4 \, f^{4} g^{4} x^{3} + 6 \, f^{5} g^{3} x^{2} + 4 \, f^{6} g^{2} x + f^{7} g\right )} e^{3} + 3 \, {\left (d f^{2} g^{6} x^{4} + 4 \, d f^{3} g^{5} x^{3} + 6 \, d f^{4} g^{4} x^{2} + 4 \, d f^{5} g^{3} x + d f^{6} g^{2}\right )} e^{2} - 3 \, {\left (d^{2} f g^{7} x^{4} + 4 \, d^{2} f^{2} g^{6} x^{3} + 6 \, d^{2} f^{3} g^{5} x^{2} + 4 \, d^{2} f^{4} g^{4} x + d^{2} f^{5} g^{3}\right )} e\right )}}, -\frac {2 \, {\left (\frac {30 \, {\left (b g^{4} n x^{4} + 4 \, b f g^{3} n x^{3} + 6 \, b f^{2} g^{2} n x^{2} + 4 \, b f^{3} g n x + b f^{4} n\right )} \arctan \left (-\frac {\sqrt {d g - f e} e^{\left (-\frac {1}{2}\right )}}{\sqrt {g x + f}}\right ) e^{\frac {7}{2}}}{\sqrt {d g - f e}} + {\left (15 \, a d^{3} g^{3} + {\left (30 \, b g^{3} n x^{3} + 100 \, b f g^{2} n x^{2} + 116 \, b f^{2} g n x + 46 \, b f^{3} n - 15 \, a f^{3}\right )} e^{3} - {\left (10 \, b d g^{3} n x^{2} + 32 \, b d f g^{2} n x + 22 \, b d f^{2} g n - 45 \, a d f^{2} g\right )} e^{2} + 3 \, {\left (2 \, b d^{2} g^{3} n x + 2 \, b d^{2} f g^{2} n - 15 \, a d^{2} f g^{2}\right )} e + 15 \, {\left (b d^{3} g^{3} n - 3 \, b d^{2} f g^{2} n e + 3 \, b d f^{2} g n e^{2} - b f^{3} n e^{3}\right )} \log \left (x e + d\right ) + 15 \, {\left (b d^{3} g^{3} - 3 \, b d^{2} f g^{2} e + 3 \, b d f^{2} g e^{2} - b f^{3} e^{3}\right )} \log \left (c\right )\right )} \sqrt {g x + f}\right )}}{105 \, {\left (d^{3} g^{8} x^{4} + 4 \, d^{3} f g^{7} x^{3} + 6 \, d^{3} f^{2} g^{6} x^{2} + 4 \, d^{3} f^{3} g^{5} x + d^{3} f^{4} g^{4} - {\left (f^{3} g^{5} x^{4} + 4 \, f^{4} g^{4} x^{3} + 6 \, f^{5} g^{3} x^{2} + 4 \, f^{6} g^{2} x + f^{7} g\right )} e^{3} + 3 \, {\left (d f^{2} g^{6} x^{4} + 4 \, d f^{3} g^{5} x^{3} + 6 \, d f^{4} g^{4} x^{2} + 4 \, d f^{5} g^{3} x + d f^{6} g^{2}\right )} e^{2} - 3 \, {\left (d^{2} f g^{7} x^{4} + 4 \, d^{2} f^{2} g^{6} x^{3} + 6 \, d^{2} f^{3} g^{5} x^{2} + 4 \, d^{2} f^{4} g^{4} x + d^{2} f^{5} g^{3}\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)^(9/2),x, algorithm="fricas")

[Out]

[-2/105*(15*(b*g^4*n*x^4 + 4*b*f*g^3*n*x^3 + 6*b*f^2*g^2*n*x^2 + 4*b*f^3*g*n*x + b*f^4*n)*sqrt(-e/(d*g - f*e))
*e^3*log(-(d*g - 2*(d*g - f*e)*sqrt(g*x + f)*sqrt(-e/(d*g - f*e)) - (g*x + 2*f)*e)/(x*e + d)) + (15*a*d^3*g^3
+ (30*b*g^3*n*x^3 + 100*b*f*g^2*n*x^2 + 116*b*f^2*g*n*x + 46*b*f^3*n - 15*a*f^3)*e^3 - (10*b*d*g^3*n*x^2 + 32*
b*d*f*g^2*n*x + 22*b*d*f^2*g*n - 45*a*d*f^2*g)*e^2 + 3*(2*b*d^2*g^3*n*x + 2*b*d^2*f*g^2*n - 15*a*d^2*f*g^2)*e
+ 15*(b*d^3*g^3*n - 3*b*d^2*f*g^2*n*e + 3*b*d*f^2*g*n*e^2 - b*f^3*n*e^3)*log(x*e + d) + 15*(b*d^3*g^3 - 3*b*d^
2*f*g^2*e + 3*b*d*f^2*g*e^2 - b*f^3*e^3)*log(c))*sqrt(g*x + f))/(d^3*g^8*x^4 + 4*d^3*f*g^7*x^3 + 6*d^3*f^2*g^6
*x^2 + 4*d^3*f^3*g^5*x + d^3*f^4*g^4 - (f^3*g^5*x^4 + 4*f^4*g^4*x^3 + 6*f^5*g^3*x^2 + 4*f^6*g^2*x + f^7*g)*e^3
+ 3*(d*f^2*g^6*x^4 + 4*d*f^3*g^5*x^3 + 6*d*f^4*g^4*x^2 + 4*d*f^5*g^3*x + d*f^6*g^2)*e^2 - 3*(d^2*f*g^7*x^4 +
4*d^2*f^2*g^6*x^3 + 6*d^2*f^3*g^5*x^2 + 4*d^2*f^4*g^4*x + d^2*f^5*g^3)*e), -2/105*(30*(b*g^4*n*x^4 + 4*b*f*g^3
*n*x^3 + 6*b*f^2*g^2*n*x^2 + 4*b*f^3*g*n*x + b*f^4*n)*arctan(-sqrt(d*g - f*e)*e^(-1/2)/sqrt(g*x + f))*e^(7/2)/
sqrt(d*g - f*e) + (15*a*d^3*g^3 + (30*b*g^3*n*x^3 + 100*b*f*g^2*n*x^2 + 116*b*f^2*g*n*x + 46*b*f^3*n - 15*a*f^
3)*e^3 - (10*b*d*g^3*n*x^2 + 32*b*d*f*g^2*n*x + 22*b*d*f^2*g*n - 45*a*d*f^2*g)*e^2 + 3*(2*b*d^2*g^3*n*x + 2*b*
d^2*f*g^2*n - 15*a*d^2*f*g^2)*e + 15*(b*d^3*g^3*n - 3*b*d^2*f*g^2*n*e + 3*b*d*f^2*g*n*e^2 - b*f^3*n*e^3)*log(x
*e + d) + 15*(b*d^3*g^3 - 3*b*d^2*f*g^2*e + 3*b*d*f^2*g*e^2 - b*f^3*e^3)*log(c))*sqrt(g*x + f))/(d^3*g^8*x^4 +
4*d^3*f*g^7*x^3 + 6*d^3*f^2*g^6*x^2 + 4*d^3*f^3*g^5*x + d^3*f^4*g^4 - (f^3*g^5*x^4 + 4*f^4*g^4*x^3 + 6*f^5*g^
3*x^2 + 4*f^6*g^2*x + f^7*g)*e^3 + 3*(d*f^2*g^6*x^4 + 4*d*f^3*g^5*x^3 + 6*d*f^4*g^4*x^2 + 4*d*f^5*g^3*x + d*f^
6*g^2)*e^2 - 3*(d^2*f*g^7*x^4 + 4*d^2*f^2*g^6*x^3 + 6*d^2*f^3*g^5*x^2 + 4*d^2*f^4*g^4*x + d^2*f^5*g^3)*e)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**(9/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4847 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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)^(9/2),x, algorithm="giac")

[Out]

integrate((b*log((x*e + d)^n*c) + a)/(g*x + f)^(9/2), x)

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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 )}^{9/2}} \,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)^(9/2),x)

[Out]

int((a + b*log(c*(d + e*x)^n))/(f + g*x)^(9/2), x)

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