3.5.16 \(\int \frac {1}{(a+b \log (c (d (e+f x)^m)^n))^{5/2}} \, dx\) [416]

Optimal. Leaf size=194 \[ \frac {4 e^{-\frac {a}{b m n}} \sqrt {\pi } (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{3 b^{5/2} f m^{5/2} n^{5/2}}-\frac {2 (e+f x)}{3 b f m n \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}}-\frac {4 (e+f x)}{3 b^2 f m^2 n^2 \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}} \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2436, 2334, 2337, 2211, 2235, 2495} \begin {gather*} \frac {4 \sqrt {\pi } (e+f x) e^{-\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \text {Erfi}\left (\frac {\sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}{\sqrt {b} \sqrt {m} \sqrt {n}}\right )}{3 b^{5/2} f m^{5/2} n^{5/2}}-\frac {4 (e+f x)}{3 b^2 f m^2 n^2 \sqrt {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}}-\frac {2 (e+f x)}{3 b f m n \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 211, normalized size = 1.09 \begin {gather*} -\frac {2 e^{-\frac {a}{b m n}} (e+f x) \left (c \left (d (e+f x)^m\right )^n\right )^{-\frac {1}{m n}} \left (2 b m n \Gamma \left (\frac {1}{2},-\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right ) \left (-\frac {a+b \log \left (c \left (d (e+f x)^m\right )^n\right )}{b m n}\right )^{3/2}+e^{\frac {a}{b m n}} \left (c \left (d (e+f x)^m\right )^n\right )^{\frac {1}{m n}} \left (2 a+b m n+2 b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )\right )}{3 b^2 f m^2 n^2 \left (a+b \log \left (c \left (d (e+f x)^m\right )^n\right )\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +b \ln \left (c \left (d \left (f x +e \right )^{m}\right )^{n}\right )\right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(a+b*ln(c*(d*(f*x+e)^m)^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*(d*(f*x+e)^m)^n))^(5/2),x, algorithm="maxima")

[Out]

integrate((b*log(((f*x + e)^m*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*(d*(f*x+e)^m)^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 \left (e + f x\right )^{m}\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*(d*(f*x+e)**m)**n))**(5/2),x)

[Out]

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

[Out]

integrate((b*log(((f*x + e)^m*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\,{\left (e+f\,x\right )}^m\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 + f*x)^m)^n))^(5/2),x)

[Out]

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

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