3.1.29 \(\int \frac {1}{(a+b \log (c (d+e x)^n))^{5/2}} \, dx\) [29]

Optimal. Leaf size=156 \[ \frac {4 e^{-\frac {a}{b n}} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{3 b^{5/2} e n^{5/2}}-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}} \]

[Out]

-2/3*(e*x+d)/b/e/n/(a+b*ln(c*(e*x+d)^n))^(3/2)+4/3*(e*x+d)*erfi((a+b*ln(c*(e*x+d)^n))^(1/2)/b^(1/2)/n^(1/2))*P
i^(1/2)/b^(5/2)/e/exp(a/b/n)/n^(5/2)/((c*(e*x+d)^n)^(1/n))-4/3*(e*x+d)/b^2/e/n^2/(a+b*ln(c*(e*x+d)^n))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2436, 2334, 2337, 2211, 2235} \begin {gather*} \frac {4 \sqrt {\pi } e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {Erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{3 b^{5/2} e n^{5/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[Pi]*(d + e*x)*Erfi[Sqrt[a + b*Log[c*(d + e*x)^n]]/(Sqrt[b]*Sqrt[n])])/(3*b^(5/2)*e*E^(a/(b*n))*n^(5/2)
*(c*(d + e*x)^n)^n^(-1)) - (2*(d + e*x))/(3*b*e*n*(a + b*Log[c*(d + e*x)^n])^(3/2)) - (4*(d + e*x))/(3*b^2*e*n
^2*Sqrt[a + b*Log[c*(d + e*x)^n]])

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[x/(n*(c*x^n)^(1/n)), Subst[Int[E^(x/n)*(a +
b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \log \left (c (d+e x)^n\right )\right )^{5/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^{5/2}} \, dx,x,d+e x\right )}{e}\\ &=-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}+\frac {2 \text {Subst}\left (\int \frac {1}{\left (a+b \log \left (c x^n\right )\right )^{3/2}} \, dx,x,d+e x\right )}{3 b e n}\\ &=-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}+\frac {4 \text {Subst}\left (\int \frac {1}{\sqrt {a+b \log \left (c x^n\right )}} \, dx,x,d+e x\right )}{3 b^2 e n^2}\\ &=-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}+\frac {\left (4 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{n}}}{\sqrt {a+b x}} \, dx,x,\log \left (c (d+e x)^n\right )\right )}{3 b^2 e n^3}\\ &=-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}+\frac {\left (8 (d+e x) \left (c (d+e x)^n\right )^{-1/n}\right ) \text {Subst}\left (\int e^{-\frac {a}{b n}+\frac {x^2}{b n}} \, dx,x,\sqrt {a+b \log \left (c (d+e x)^n\right )}\right )}{3 b^3 e n^3}\\ &=\frac {4 e^{-\frac {a}{b n}} \sqrt {\pi } (d+e x) \left (c (d+e x)^n\right )^{-1/n} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c (d+e x)^n\right )}}{\sqrt {b} \sqrt {n}}\right )}{3 b^{5/2} e n^{5/2}}-\frac {2 (d+e x)}{3 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}}-\frac {4 (d+e x)}{3 b^2 e n^2 \sqrt {a+b \log \left (c (d+e x)^n\right )}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 163, normalized size = 1.04 \begin {gather*} -\frac {2 e^{-\frac {a}{b n}} (d+e x) \left (c (d+e x)^n\right )^{-1/n} \left (2 b n \Gamma \left (\frac {1}{2},-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right ) \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{b n}\right )^{3/2}+e^{\frac {a}{b n}} \left (c (d+e x)^n\right )^{\frac {1}{n}} \left (2 a+b n+2 b \log \left (c (d+e x)^n\right )\right )\right )}{3 b^2 e n^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(d + e*x)*(2*b*n*Gamma[1/2, -((a + b*Log[c*(d + e*x)^n])/(b*n))]*(-((a + b*Log[c*(d + e*x)^n])/(b*n)))^(3/
2) + E^(a/(b*n))*(c*(d + e*x)^n)^n^(-1)*(2*a + b*n + 2*b*Log[c*(d + e*x)^n])))/(3*b^2*e*E^(a/(b*n))*n^2*(c*(d
+ e*x)^n)^n^(-1)*(a + b*Log[c*(d + e*x)^n])^(3/2))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*ln(c*(e*x+d)^n))^(5/2),x)

[Out]

int(1/(a+b*ln(c*(e*x+d)^n))^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*log((x*e + d)^n*c) + a)^(-5/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*log(c*(e*x+d)^n))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*ln(c*(e*x+d)**n))**(5/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))**(-5/2), x)

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

[Out]

integrate((b*log((x*e + d)^n*c) + a)^(-5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*log(c*(d + e*x)^n))^(5/2),x)

[Out]

int(1/(a + b*log(c*(d + e*x)^n))^(5/2), x)

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