\(\int \frac {x^3}{\log (c (a+b x^2)^p)} \, dx\) [102]

Optimal result

Integrand size = 18, antiderivative size = 107 \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b^2 p}+\frac {\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b^2 p} \]

[Out]

-1/2*a*(b*x^2+a)*Ei(ln(c*(b*x^2+a)^p)/p)/b^2/p/((c*(b*x^2+a)^p)^(1/p))+1/2*(b*x^2+a)^2*Ei(2*ln(c*(b*x^2+a)^p)/
p)/b^2/p/((c*(b*x^2+a)^p)^(2/p))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2504, 2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p}-\frac {a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p} \]

[In]

Int[x^3/Log[c*(a + b*x^2)^p],x]

[Out]

-1/2*(a*(a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(b^2*p*(c*(a + b*x^2)^p)^p^(-1)) + ((a + b*x^2)^2*E
xpIntegralEi[(2*Log[c*(a + b*x^2)^p])/p])/(2*b^2*p*(c*(a + b*x^2)^p)^(2/p))

Rule 2209

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

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 2347

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)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, 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 2437

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 2446

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

Rule 2504

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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {a}{b \log \left (c (a+b x)^p\right )}+\frac {a+b x}{b \log \left (c (a+b x)^p\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {a+b x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{2 b}-\frac {a \text {Subst}\left (\int \frac {1}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{2 b} \\ & = \frac {\text {Subst}\left (\int \frac {x}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{2 b^2}-\frac {a \text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{2 b^2} \\ & = \frac {\left (\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {2 x}{p}}}{x} \, dx,x,\log \left (c \left (a+b x^2\right )^p\right )\right )}{2 b^2 p}-\frac {\left (a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{p}}}{x} \, dx,x,\log \left (c \left (a+b x^2\right )^p\right )\right )}{2 b^2 p} \\ & = -\frac {a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b^2 p}+\frac {\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \text {Ei}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b^2 p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.90 \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-2/p} \left (a \left (c \left (a+b x^2\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )-\left (a+b x^2\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right )\right )}{2 b^2 p} \]

[In]

Integrate[x^3/Log[c*(a + b*x^2)^p],x]

[Out]

-1/2*((a + b*x^2)*(a*(c*(a + b*x^2)^p)^p^(-1)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p] - (a + b*x^2)*ExpIntegralE
i[(2*Log[c*(a + b*x^2)^p])/p]))/(b^2*p*(c*(a + b*x^2)^p)^(2/p))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.88 (sec) , antiderivative size = 547, normalized size of antiderivative = 5.11

method	result	size
risch	\(-\frac {\left (b \,x^{2}+a \right )^{2} c^{-\frac {2}{p}} {\left (\left (b \,x^{2}+a \right )^{p}\right )}^{-\frac {2}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right )\right )}{p}} \operatorname {Ei}_{1}\left (-2 \ln \left (b \,x^{2}+a \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )-2 p \ln \left (b \,x^{2}+a \right )}{p}\right )}{2 b^{2} p}+\frac {a \left (b \,x^{2}+a \right ) c^{-\frac {1}{p}} {\left (\left (b \,x^{2}+a \right )^{p}\right )}^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right )\right )}{2 p}} \operatorname {Ei}_{1}\left (-\ln \left (b \,x^{2}+a \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )-2 p \ln \left (b \,x^{2}+a \right )}{2 p}\right )}{2 b^{2} p}\)	\(547\)

[In]

int(x^3/ln(c*(b*x^2+a)^p),x,method=_RETURNVERBOSE)

[Out]

-1/2/b^2/p*(b*x^2+a)^2*c^(-2/p)*((b*x^2+a)^p)^(-2/p)*exp(I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+cs
gn(I*c))*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*(b*x^2+a)^p))/p)*Ei(1,-2*ln(b*x^2+a)-(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*
c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*cs
gn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)+2*ln((b*x^2+a)^p)-2*p*ln(b*x^2+a))/p)+1/2/b^2*a/p*(b*x^2+a)*c^(-1/p)*(
(b*x^2+a)^p)^(-1/p)*exp(1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*c))*(-csgn(I*c*(b*x^2+a)
^p)+csgn(I*(b*x^2+a)^p))/p)*Ei(1,-ln(b*x^2+a)-1/2*(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(
I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(
I*c)+2*ln(c)+2*ln((b*x^2+a)^p)-2*p*ln(b*x^2+a))/p)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {a c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right ) - \operatorname {log\_integral}\left ({\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} c^{\frac {2}{p}}\right )}{2 \, b^{2} c^{\frac {2}{p}} p} \]

[In]

integrate(x^3/log(c*(b*x^2+a)^p),x, algorithm="fricas")

[Out]

-1/2*(a*c^(1/p)*log_integral((b*x^2 + a)*c^(1/p)) - log_integral((b^2*x^4 + 2*a*b*x^2 + a^2)*c^(2/p)))/(b^2*c^
(2/p)*p)

Sympy [F]

\[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x^{3}}{\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}\, dx \]

[In]

integrate(x**3/ln(c*(b*x**2+a)**p),x)

[Out]

Integral(x**3/log(c*(a + b*x**2)**p), x)

Maxima [F]

\[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {x^{3}}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )} \,d x } \]

[In]

integrate(x^3/log(c*(b*x^2+a)^p),x, algorithm="maxima")

[Out]

integrate(x^3/log((b*x^2 + a)^p*c), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.64 \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {a {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right )}{2 \, b^{2} c^{\left (\frac {1}{p}\right )} p} + \frac {{\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right )}{2 \, b^{2} c^{\frac {2}{p}} p} \]

[In]

integrate(x^3/log(c*(b*x^2+a)^p),x, algorithm="giac")

[Out]

-1/2*a*Ei(log(c)/p + log(b*x^2 + a))/(b^2*c^(1/p)*p) + 1/2*Ei(2*log(c)/p + 2*log(b*x^2 + a))/(b^2*c^(2/p)*p)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x^3}{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )} \,d x \]

[In]

int(x^3/log(c*(a + b*x^2)^p),x)

[Out]

int(x^3/log(c*(a + b*x^2)^p), x)