Integrand size = 20, antiderivative size = 110 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f p x+\frac {d g p x^2}{4 e}-\frac {1}{8} g p x^4+\frac {2 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {d^2 g p \log \left (d+e x^2\right )}{4 e^2}+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} g x^4 \log \left (c \left (d+e x^2\right )^p\right ) \] Output:
-2*f*p*x+1/4*d*g*p*x^2/e-1/8*g*p*x^4+2*d^(1/2)*f*p*arctan(e^(1/2)*x/d^(1/2 ))/e^(1/2)-1/4*d^2*g*p*ln(e*x^2+d)/e^2+f*x*ln(c*(e*x^2+d)^p)+1/4*g*x^4*ln( c*(e*x^2+d)^p)
Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-2 f p x+\frac {d g p x^2}{4 e}-\frac {1}{8} g p x^4+\frac {2 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {d^2 g p \log \left (d+e x^2\right )}{4 e^2}+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} g x^4 \log \left (c \left (d+e x^2\right )^p\right ) \] Input:
Integrate[(f + g*x^3)*Log[c*(d + e*x^2)^p],x]
Output:
-2*f*p*x + (d*g*p*x^2)/(4*e) - (g*p*x^4)/8 + (2*Sqrt[d]*f*p*ArcTan[(Sqrt[e ]*x)/Sqrt[d]])/Sqrt[e] - (d^2*g*p*Log[d + e*x^2])/(4*e^2) + f*x*Log[c*(d + e*x^2)^p] + (g*x^4*Log[c*(d + e*x^2)^p])/4
Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2921, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx\) |
\(\Big \downarrow \) 2921 |
\(\displaystyle \int \left (f \log \left (c \left (d+e x^2\right )^p\right )+g x^3 \log \left (c \left (d+e x^2\right )^p\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 \sqrt {d} f p \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}+f x \log \left (c \left (d+e x^2\right )^p\right )+\frac {1}{4} g x^4 \log \left (c \left (d+e x^2\right )^p\right )-\frac {d^2 g p \log \left (d+e x^2\right )}{4 e^2}+\frac {d g p x^2}{4 e}-2 f p x-\frac {1}{8} g p x^4\) |
Input:
Int[(f + g*x^3)*Log[c*(d + e*x^2)^p],x]
Output:
-2*f*p*x + (d*g*p*x^2)/(4*e) - (g*p*x^4)/8 + (2*Sqrt[d]*f*p*ArcTan[(Sqrt[e ]*x)/Sqrt[d]])/Sqrt[e] - (d^2*g*p*Log[d + e*x^2])/(4*e^2) + f*x*Log[c*(d + e*x^2)^p] + (g*x^4*Log[c*(d + e*x^2)^p])/4
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> With[{t = ExpandIntegrand[(a + b*Log[ c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; FreeQ[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && Integ erQ[r] && IntegerQ[s] && (EqQ[q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))
Time = 1.86 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95
method | result | size |
parts | \(\frac {g \,x^{4} \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{4}+f x \ln \left (c \left (e \,x^{2}+d \right )^{p}\right )-\frac {e p \left (-\frac {-\frac {1}{4} e g \,x^{4}+\frac {1}{2} d g \,x^{2}-4 e f x}{e^{2}}+\frac {d \left (\frac {d g \ln \left (e \,x^{2}+d \right )}{2 e}-\frac {4 e f \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{\sqrt {d e}}\right )}{e^{2}}\right )}{2}\) | \(104\) |
risch | \(\left (\frac {1}{4} g \,x^{4}+f x \right ) \ln \left (\left (e \,x^{2}+d \right )^{p}\right )+\frac {i \operatorname {csgn}\left (i c \right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} x^{4} g \pi }{8}-\frac {i \pi g \,x^{4} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{8}+\frac {i x \pi f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i x \pi f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}}{2}-\frac {i x \pi f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i x \pi f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{2}+\frac {i {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) x^{4} g \pi }{8}-\frac {i \pi g \,x^{4} {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}}{8}+\frac {\ln \left (c \right ) g \,x^{4}}{4}-\frac {g p \,x^{4}}{8}+\frac {d g p \,x^{2}}{4 e}+\frac {p \ln \left (-\sqrt {-d e}\, x +d \right ) f \sqrt {-d e}}{e}-\frac {p \ln \left (\sqrt {-d e}\, x +d \right ) f \sqrt {-d e}}{e}+\ln \left (c \right ) f x -\frac {p \ln \left (-\sqrt {-d e}\, x +d \right ) d^{2} g}{4 e^{2}}-\frac {p \ln \left (\sqrt {-d e}\, x +d \right ) d^{2} g}{4 e^{2}}-2 f p x\) | \(402\) |
Input:
int((g*x^3+f)*ln(c*(e*x^2+d)^p),x,method=_RETURNVERBOSE)
Output:
1/4*g*x^4*ln(c*(e*x^2+d)^p)+f*x*ln(c*(e*x^2+d)^p)-1/2*e*p*(-1/e^2*(-1/4*e* g*x^4+1/2*d*g*x^2-4*e*f*x)+d/e^2*(1/2*d*g/e*ln(e*x^2+d)-4*e*f/(d*e)^(1/2)* arctan(x*e/(d*e)^(1/2))))
Time = 0.08 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.27 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\left [-\frac {e^{2} g p x^{4} - 2 \, d e g p x^{2} - 8 \, e^{2} f p \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 16 \, e^{2} f p x - 2 \, {\left (e^{2} g p x^{4} + 4 \, e^{2} f p x - d^{2} g p\right )} \log \left (e x^{2} + d\right ) - 2 \, {\left (e^{2} g x^{4} + 4 \, e^{2} f x\right )} \log \left (c\right )}{8 \, e^{2}}, -\frac {e^{2} g p x^{4} - 2 \, d e g p x^{2} - 16 \, e^{2} f p \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 16 \, e^{2} f p x - 2 \, {\left (e^{2} g p x^{4} + 4 \, e^{2} f p x - d^{2} g p\right )} \log \left (e x^{2} + d\right ) - 2 \, {\left (e^{2} g x^{4} + 4 \, e^{2} f x\right )} \log \left (c\right )}{8 \, e^{2}}\right ] \] Input:
integrate((g*x^3+f)*log(c*(e*x^2+d)^p),x, algorithm="fricas")
Output:
[-1/8*(e^2*g*p*x^4 - 2*d*e*g*p*x^2 - 8*e^2*f*p*sqrt(-d/e)*log((e*x^2 + 2*e *x*sqrt(-d/e) - d)/(e*x^2 + d)) + 16*e^2*f*p*x - 2*(e^2*g*p*x^4 + 4*e^2*f* p*x - d^2*g*p)*log(e*x^2 + d) - 2*(e^2*g*x^4 + 4*e^2*f*x)*log(c))/e^2, -1/ 8*(e^2*g*p*x^4 - 2*d*e*g*p*x^2 - 16*e^2*f*p*sqrt(d/e)*arctan(e*x*sqrt(d/e) /d) + 16*e^2*f*p*x - 2*(e^2*g*p*x^4 + 4*e^2*f*p*x - d^2*g*p)*log(e*x^2 + d ) - 2*(e^2*g*x^4 + 4*e^2*f*x)*log(c))/e^2]
Time = 16.78 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.95 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\begin {cases} \left (f x + \frac {g x^{4}}{4}\right ) \log {\left (0^{p} c \right )} & \text {for}\: d = 0 \wedge e = 0 \\\left (f x + \frac {g x^{4}}{4}\right ) \log {\left (c d^{p} \right )} & \text {for}\: e = 0 \\- 2 f p x + f x \log {\left (c \left (e x^{2}\right )^{p} \right )} - \frac {g p x^{4}}{8} + \frac {g x^{4} \log {\left (c \left (e x^{2}\right )^{p} \right )}}{4} & \text {for}\: d = 0 \\- \frac {d^{2} g \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4 e^{2}} + \frac {2 d f p \log {\left (x - \sqrt {- \frac {d}{e}} \right )}}{e \sqrt {- \frac {d}{e}}} - \frac {d f \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{e \sqrt {- \frac {d}{e}}} + \frac {d g p x^{2}}{4 e} - 2 f p x + f x \log {\left (c \left (d + e x^{2}\right )^{p} \right )} - \frac {g p x^{4}}{8} + \frac {g x^{4} \log {\left (c \left (d + e x^{2}\right )^{p} \right )}}{4} & \text {otherwise} \end {cases} \] Input:
integrate((g*x**3+f)*ln(c*(e*x**2+d)**p),x)
Output:
Piecewise(((f*x + g*x**4/4)*log(0**p*c), Eq(d, 0) & Eq(e, 0)), ((f*x + g*x **4/4)*log(c*d**p), Eq(e, 0)), (-2*f*p*x + f*x*log(c*(e*x**2)**p) - g*p*x* *4/8 + g*x**4*log(c*(e*x**2)**p)/4, Eq(d, 0)), (-d**2*g*log(c*(d + e*x**2) **p)/(4*e**2) + 2*d*f*p*log(x - sqrt(-d/e))/(e*sqrt(-d/e)) - d*f*log(c*(d + e*x**2)**p)/(e*sqrt(-d/e)) + d*g*p*x**2/(4*e) - 2*f*p*x + f*x*log(c*(d + e*x**2)**p) - g*p*x**4/8 + g*x**4*log(c*(d + e*x**2)**p)/4, True))
Exception generated. \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x^3+f)*log(c*(e*x^2+d)^p),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.12 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.89 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=-\frac {1}{8} \, {\left (g p - 2 \, g \log \left (c\right )\right )} x^{4} + \frac {d g p x^{2}}{4 \, e} + \frac {2 \, d f p \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}} - \frac {d^{2} g p \log \left (e x^{2} + d\right )}{4 \, e^{2}} - {\left (2 \, f p - f \log \left (c\right )\right )} x + \frac {1}{4} \, {\left (g p x^{4} + 4 \, f p x\right )} \log \left (e x^{2} + d\right ) \] Input:
integrate((g*x^3+f)*log(c*(e*x^2+d)^p),x, algorithm="giac")
Output:
-1/8*(g*p - 2*g*log(c))*x^4 + 1/4*d*g*p*x^2/e + 2*d*f*p*arctan(e*x/sqrt(d* e))/sqrt(d*e) - 1/4*d^2*g*p*log(e*x^2 + d)/e^2 - (2*f*p - f*log(c))*x + 1/ 4*(g*p*x^4 + 4*f*p*x)*log(e*x^2 + d)
Time = 26.49 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=f\,x\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )-\frac {g\,p\,x^4}{8}-2\,f\,p\,x+\frac {g\,x^4\,\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{4}+\frac {d\,g\,p\,x^2}{4\,e}+\frac {2\,\sqrt {d}\,f\,p\,\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {e}}-\frac {d^2\,g\,p\,\ln \left (e\,x^2+d\right )}{4\,e^2} \] Input:
int(log(c*(d + e*x^2)^p)*(f + g*x^3),x)
Output:
f*x*log(c*(d + e*x^2)^p) - (g*p*x^4)/8 - 2*f*p*x + (g*x^4*log(c*(d + e*x^2 )^p))/4 + (d*g*p*x^2)/(4*e) + (2*d^(1/2)*f*p*atan((e^(1/2)*x)/d^(1/2)))/e^ (1/2) - (d^2*g*p*log(d + e*x^2))/(4*e^2)
Time = 0.15 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02 \[ \int \left (f+g x^3\right ) \log \left (c \left (d+e x^2\right )^p\right ) \, dx=\frac {16 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {e x}{\sqrt {e}\, \sqrt {d}}\right ) e f p -2 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d^{2} g +8 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{2} f x +2 \,\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e^{2} g \,x^{4}+2 d e g p \,x^{2}-16 e^{2} f p x -e^{2} g p \,x^{4}}{8 e^{2}} \] Input:
int((g*x^3+f)*log(c*(e*x^2+d)^p),x)
Output:
(16*sqrt(e)*sqrt(d)*atan((e*x)/(sqrt(e)*sqrt(d)))*e*f*p - 2*log((d + e*x** 2)**p*c)*d**2*g + 8*log((d + e*x**2)**p*c)*e**2*f*x + 2*log((d + e*x**2)** p*c)*e**2*g*x**4 + 2*d*e*g*p*x**2 - 16*e**2*f*p*x - e**2*g*p*x**4)/(8*e**2 )