3.1.14 \(\int \frac {1}{\log ^{\frac {5}{2}}(c (d+e x))} \, dx\) [14]

Optimal. Leaf size=77 \[ \frac {4 \sqrt {\pi } \text {erfi}\left (\sqrt {\log (c (d+e x))}\right )}{3 c e}-\frac {2 (d+e x)}{3 e \log ^{\frac {3}{2}}(c (d+e x))}-\frac {4 (d+e x)}{3 e \sqrt {\log (c (d+e x))}} \]

[Out]

-2/3*(e*x+d)/e/ln(c*(e*x+d))^(3/2)+4/3*erfi(ln(c*(e*x+d))^(1/2))*Pi^(1/2)/c/e-4/3*(e*x+d)/e/ln(c*(e*x+d))^(1/2
)

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Rubi [A]
time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2436, 2334, 2336, 2211, 2235} \begin {gather*} \frac {4 \sqrt {\pi } \text {Erfi}\left (\sqrt {\log (c (d+e x))}\right )}{3 c e}-\frac {2 (d+e x)}{3 e \log ^{\frac {3}{2}}(c (d+e x))}-\frac {4 (d+e x)}{3 e \sqrt {\log (c (d+e x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Log[c*(d + e*x)]^(-5/2),x]

[Out]

(4*Sqrt[Pi]*Erfi[Sqrt[Log[c*(d + e*x)]]])/(3*c*e) - (2*(d + e*x))/(3*e*Log[c*(d + e*x)]^(3/2)) - (4*(d + e*x))
/(3*e*Sqrt[Log[c*(d + e*x)]])

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 2336

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

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

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Mathematica [A]
time = 0.06, size = 72, normalized size = 0.94 \begin {gather*} -\frac {2 \left (2 \Gamma \left (\frac {1}{2},-\log (c (d+e x))\right ) (-\log (c (d+e x)))^{3/2}+c (d+e x) (1+2 \log (c (d+e x)))\right )}{3 c e \log ^{\frac {3}{2}}(c (d+e x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(d + e*x)]^(-5/2),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 47, normalized size = 0.61 \begin {gather*} -\frac {\left (-\log \left (c x e + c d\right )\right )^{\frac {3}{2}} e^{\left (-1\right )} \Gamma \left (-\frac {3}{2}, -\log \left (c x e + c d\right )\right )}{c \log \left (c x e + c d\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-log(c*x*e + c*d))^(3/2)*e^(-1)*gamma(-3/2, -log(c*x*e + c*d))/(c*log(c*x*e + c*d)^(3/2))

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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/log(c*(e*x+d))^(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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate(log((x*e + d)*c)^(-5/2), x)

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Mupad [B]
time = 0.18, size = 113, normalized size = 1.47 \begin {gather*} \frac {4\,\sqrt {\pi }\,{\left (-\ln \left (c\,\left (d+e\,x\right )\right )\right )}^{5/2}\,\mathrm {erfc}\left (\sqrt {-\ln \left (c\,\left (d+e\,x\right )\right )}\right )}{3\,c\,e\,{\ln \left (c\,\left (d+e\,x\right )\right )}^{5/2}}-\frac {4\,d\,{\ln \left (c\,\left (d+e\,x\right )\right )}^2+2\,d\,\ln \left (c\,\left (d+e\,x\right )\right )+2\,e\,x\,\ln \left (c\,\left (d+e\,x\right )\right )+4\,e\,x\,{\ln \left (c\,\left (d+e\,x\right )\right )}^2}{3\,e\,{\ln \left (c\,\left (d+e\,x\right )\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/log(c*(d + e*x))^(5/2),x)

[Out]

(4*pi^(1/2)*(-log(c*(d + e*x)))^(5/2)*erfc((-log(c*(d + e*x)))^(1/2)))/(3*c*e*log(c*(d + e*x))^(5/2)) - (4*d*l
og(c*(d + e*x))^2 + 2*d*log(c*(d + e*x)) + 2*e*x*log(c*(d + e*x)) + 4*e*x*log(c*(d + e*x))^2)/(3*e*log(c*(d +
e*x))^(5/2))

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