\(\int \log ^{\frac {5}{2}}(c (d+e x)) \, dx\) [9]

Optimal result

Integrand size = 12, antiderivative size = 98 \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\log (c (d+e x))}\right )}{8 c e}+\frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e} \]

[Out]

-5/2*(e*x+d)*ln(c*(e*x+d))^(3/2)/e+(e*x+d)*ln(c*(e*x+d))^(5/2)/e-15/8*erfi(ln(c*(e*x+d))^(1/2))*Pi^(1/2)/c/e+1
5/4*(e*x+d)*ln(c*(e*x+d))^(1/2)/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2436, 2333, 2336, 2211, 2235} \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=-\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\log (c (d+e x))}\right )}{8 c e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e} \]

[In]

Int[Log[c*(d + e*x)]^(5/2),x]

[Out]

(-15*Sqrt[Pi]*Erfi[Sqrt[Log[c*(d + e*x)]]])/(8*c*e) + (15*(d + e*x)*Sqrt[Log[c*(d + e*x)]])/(4*e) - (5*(d + e*
x)*Log[c*(d + e*x)]^(3/2))/(2*e) + ((d + e*x)*Log[c*(d + e*x)]^(5/2))/e

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 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && 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*} \text {integral}& = \frac {\text {Subst}\left (\int \log ^{\frac {5}{2}}(c x) \, dx,x,d+e x\right )}{e} \\ & = \frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}-\frac {5 \text {Subst}\left (\int \log ^{\frac {3}{2}}(c x) \, dx,x,d+e x\right )}{2 e} \\ & = -\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}+\frac {15 \text {Subst}\left (\int \sqrt {\log (c x)} \, dx,x,d+e x\right )}{4 e} \\ & = \frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}-\frac {15 \text {Subst}\left (\int \frac {1}{\sqrt {\log (c x)}} \, dx,x,d+e x\right )}{8 e} \\ & = \frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}-\frac {15 \text {Subst}\left (\int \frac {e^x}{\sqrt {x}} \, dx,x,\log (c (d+e x))\right )}{8 c e} \\ & = \frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e}-\frac {15 \text {Subst}\left (\int e^{x^2} \, dx,x,\sqrt {\log (c (d+e x))}\right )}{4 c e} \\ & = -\frac {15 \sqrt {\pi } \text {erfi}\left (\sqrt {\log (c (d+e x))}\right )}{8 c e}+\frac {15 (d+e x) \sqrt {\log (c (d+e x))}}{4 e}-\frac {5 (d+e x) \log ^{\frac {3}{2}}(c (d+e x))}{2 e}+\frac {(d+e x) \log ^{\frac {5}{2}}(c (d+e x))}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.77 \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\frac {-15 \sqrt {\pi } \text {erfi}\left (\sqrt {\log (c (d+e x))}\right )+2 c (d+e x) \sqrt {\log (c (d+e x))} \left (15-10 \log (c (d+e x))+4 \log ^2(c (d+e x))\right )}{8 c e} \]

[In]

Integrate[Log[c*(d + e*x)]^(5/2),x]

[Out]

(-15*Sqrt[Pi]*Erfi[Sqrt[Log[c*(d + e*x)]]] + 2*c*(d + e*x)*Sqrt[Log[c*(d + e*x)]]*(15 - 10*Log[c*(d + e*x)] +
4*Log[c*(d + e*x)]^2))/(8*c*e)

Maple [F]

\[\int \ln \left (c \left (e x +d \right )\right )^{\frac {5}{2}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(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)

Sympy [F(-1)]

Timed out. \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.80 \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\frac {2 \, {\left (c e x + c d\right )} {\left (4 \, \log \left (c e x + c d\right )^{\frac {5}{2}} - 10 \, \log \left (c e x + c d\right )^{\frac {3}{2}} + 15 \, \sqrt {\log \left (c e x + c d\right )}\right )} + 15 i \, \sqrt {\pi } \operatorname {erf}\left (i \, \sqrt {\log \left (c e x + c d\right )}\right )}{8 \, c e} \]

[In]

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

[Out]

1/8*(2*(c*e*x + c*d)*(4*log(c*e*x + c*d)^(5/2) - 10*log(c*e*x + c*d)^(3/2) + 15*sqrt(log(c*e*x + c*d))) + 15*I
*sqrt(pi)*erf(I*sqrt(log(c*e*x + c*d))))/(c*e)

Giac [F]

\[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\int { \log \left ({\left (e x + d\right )} c\right )^{\frac {5}{2}} \,d x } \]

[In]

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

[Out]

integrate(log((e*x + d)*c)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 1.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.98 \[ \int \log ^{\frac {5}{2}}(c (d+e x)) \, dx=\frac {{\ln \left (c\,\left (d+e\,x\right )\right )}^{5/2}\,\left (\frac {15\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\ln \left (c\,\left (d+e\,x\right )\right )}\right )}{8}+c\,\left (d+e\,x\right )\,\left (\frac {15\,\sqrt {-\ln \left (c\,\left (d+e\,x\right )\right )}}{4}+\frac {5\,{\left (-\ln \left (c\,\left (d+e\,x\right )\right )\right )}^{3/2}}{2}+{\left (-\ln \left (c\,\left (d+e\,x\right )\right )\right )}^{5/2}\right )\right )}{c\,e\,{\left (-\ln \left (c\,\left (d+e\,x\right )\right )\right )}^{5/2}} \]

[In]

int(log(c*(d + e*x))^(5/2),x)

[Out]

(log(c*(d + e*x))^(5/2)*((15*pi^(1/2)*erfc((-log(c*(d + e*x)))^(1/2)))/8 + c*(d + e*x)*((15*(-log(c*(d + e*x))
)^(1/2))/4 + (5*(-log(c*(d + e*x)))^(3/2))/2 + (-log(c*(d + e*x)))^(5/2))))/(c*e*(-log(c*(d + e*x)))^(5/2))