3.553 \(\int \frac {(a+b \log (c (d+\frac {e}{\sqrt {x}})^2))^p}{x^2} \, dx\)

Optimal. Leaf size=216 \[ \frac {d 2^{p+1} e^{-\frac {a}{2 b}} \left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \Gamma \left (p+1,\frac {-a-b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{2 b}\right )}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}-\frac {e^{-\frac {a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \Gamma \left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )}{c e^2} \]

[Out]

-GAMMA(1+p,(-a-b*ln(c*(d+e/x^(1/2))^2))/b)*(a+b*ln(c*(d+e/x^(1/2))^2))^p/c/e^2/exp(a/b)/(((-a-b*ln(c*(d+e/x^(1
/2))^2))/b)^p)+2^(1+p)*d*GAMMA(1+p,1/2*(-a-b*ln(c*(d+e/x^(1/2))^2))/b)*(a+b*ln(c*(d+e/x^(1/2))^2))^p*(d+e/x^(1
/2))/e^2/exp(1/2*a/b)/(((-a-b*ln(c*(d+e/x^(1/2))^2))/b)^p)/(c*(d+e/x^(1/2))^2)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 213, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {2454, 2401, 2389, 2300, 2181, 2390, 2310} \[ \frac {d 2^{p+1} e^{-\frac {a}{2 b}} \left (d+\frac {e}{\sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \text {Gamma}\left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{2 b}\right )}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}-\frac {e^{-\frac {a}{b}} \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p} \text {Gamma}\left (p+1,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )}{c e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e/Sqrt[x])^2])^p/x^2,x]

[Out]

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

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2300

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 2310

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

Rule 2389

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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 0] && IGtQ[q, 0]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx &=-\left (2 \operatorname {Subst}\left (\int x \left (a+b \log \left (c (d+e x)^2\right )\right )^p \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \left (-\frac {d \left (a+b \log \left (c (d+e x)^2\right )\right )^p}{e}+\frac {(d+e x) \left (a+b \log \left (c (d+e x)^2\right )\right )^p}{e}\right ) \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\frac {2 \operatorname {Subst}\left (\int (d+e x) \left (a+b \log \left (c (d+e x)^2\right )\right )^p \, dx,x,\frac {1}{\sqrt {x}}\right )}{e}+\frac {(2 d) \operatorname {Subst}\left (\int \left (a+b \log \left (c (d+e x)^2\right )\right )^p \, dx,x,\frac {1}{\sqrt {x}}\right )}{e}\\ &=-\frac {2 \operatorname {Subst}\left (\int x \left (a+b \log \left (c x^2\right )\right )^p \, dx,x,d+\frac {e}{\sqrt {x}}\right )}{e^2}+\frac {(2 d) \operatorname {Subst}\left (\int \left (a+b \log \left (c x^2\right )\right )^p \, dx,x,d+\frac {e}{\sqrt {x}}\right )}{e^2}\\ &=-\frac {\operatorname {Subst}\left (\int e^x (a+b x)^p \, dx,x,\log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )}{c e^2}+\frac {\left (d \left (d+\frac {e}{\sqrt {x}}\right )\right ) \operatorname {Subst}\left (\int e^{x/2} (a+b x)^p \, dx,x,\log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}\\ &=-\frac {e^{-\frac {a}{b}} \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p}}{c e^2}+\frac {2^{1+p} d e^{-\frac {a}{2 b}} \left (d+\frac {e}{\sqrt {x}}\right ) \Gamma \left (1+p,-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{2 b}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p \left (-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )}{b}\right )^{-p}}{e^2 \sqrt {c \left (d+\frac {e}{\sqrt {x}}\right )^2}}\\ \end {align*}

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Mathematica [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^2\right )\right )^p}{x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Log[c*(d + e/Sqrt[x])^2])^p/x^2,x]

[Out]

Integrate[(a + b*Log[c*(d + e/Sqrt[x])^2])^p/x^2, x]

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \log \left (\frac {c d^{2} x + 2 \, c d e \sqrt {x} + c e^{2}}{x}\right ) + a\right )}^{p}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((c*d^2*x + 2*c*d*e*sqrt(x) + c*e^2)/x) + a)^p/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{2}\right ) + a\right )}^{p}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*(d + e/sqrt(x))^2) + a)^p/x^2, x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (\left (d +\frac {e}{\sqrt {x}}\right )^{2} c \right )+a \right )^{p}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c {\left (d + \frac {e}{\sqrt {x}}\right )}^{2}\right ) + a\right )}^{p}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*(d + e/sqrt(x))^2) + a)^p/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{\sqrt {x}}\right )}^2\right )\right )}^p}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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