Integrand size = 18, antiderivative size = 74 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=-\frac {2 e F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}+\frac {2 e F^{c (a+b x)} x}{b^2 c^2 \log ^2(F)}+\frac {F^{c (a+b x)} \left (d-e x^2\right )}{b c \log (F)} \] Output:
-2*e*F^(c*(b*x+a))/b^3/c^3/ln(F)^3+2*e*F^(c*(b*x+a))*x/b^2/c^2/ln(F)^2+F^( c*(b*x+a))*(-e*x^2+d)/b/c/ln(F)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.72 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=-\frac {F^{c (a+b x)} \left (2 e-2 b c e x \log (F)+b^2 c^2 \left (-d+e x^2\right ) \log ^2(F)\right )}{b^3 c^3 \log ^3(F)} \] Input:
Integrate[F^(c*(a + b*x))*(d - e*x^2),x]
Output:
-((F^(c*(a + b*x))*(2*e - 2*b*c*e*x*Log[F] + b^2*c^2*(-d + e*x^2)*Log[F]^2 ))/(b^3*c^3*Log[F]^3))
Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2626, 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.
\(\displaystyle \int \left (d-e x^2\right ) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 2626 |
\(\displaystyle \int \left (d F^{c (a+b x)}-e x^2 F^{c (a+b x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e F^{c (a+b x)}}{b^3 c^3 \log ^3(F)}+\frac {2 e x F^{c (a+b x)}}{b^2 c^2 \log ^2(F)}+\frac {d F^{c (a+b x)}}{b c \log (F)}-\frac {e x^2 F^{c (a+b x)}}{b c \log (F)}\) |
Input:
Int[F^(c*(a + b*x))*(d - e*x^2),x]
Output:
(-2*e*F^(c*(a + b*x)))/(b^3*c^3*Log[F]^3) + (2*e*F^(c*(a + b*x))*x)/(b^2*c ^2*Log[F]^2) + (d*F^(c*(a + b*x)))/(b*c*Log[F]) - (e*F^(c*(a + b*x))*x^2)/ (b*c*Log[F])
Int[(F_)^(v_)*(Px_), x_Symbol] :> Int[ExpandIntegrand[F^v, Px, x], x] /; Fr eeQ[F, x] && PolynomialQ[Px, x] && LinearQ[v, x] && !TrueQ[$UseGamma]
Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(\frac {\left (-e \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} d +2 e x \ln \left (F \right ) b c -2 e \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(61\) |
risch | \(\frac {\left (-e \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} d +2 e x \ln \left (F \right ) b c -2 e \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(61\) |
orering | \(\frac {\left (-e \,x^{2} \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} b^{2} c^{2} d +2 e x \ln \left (F \right ) b c -2 e \right ) F^{c \left (b x +a \right )}}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(61\) |
meijerg | \(\frac {F^{a c} e \left (2-\frac {\left (3 b^{2} c^{2} x^{2} \ln \left (F \right )^{2}-6 b c x \ln \left (F \right )+6\right ) {\mathrm e}^{b c x \ln \left (F \right )}}{3}\right )}{c^{3} b^{3} \ln \left (F \right )^{3}}-\frac {F^{a c} d \left (1-{\mathrm e}^{b c x \ln \left (F \right )}\right )}{b c \ln \left (F \right )}\) | \(83\) |
parallelrisch | \(\frac {-x^{2} F^{c \left (b x +a \right )} e \ln \left (F \right )^{2} b^{2} c^{2}+\ln \left (F \right )^{2} F^{c \left (b x +a \right )} b^{2} c^{2} d +2 e \,F^{c \left (b x +a \right )} x \ln \left (F \right ) b c -2 F^{c \left (b x +a \right )} e}{\ln \left (F \right )^{3} b^{3} c^{3}}\) | \(88\) |
norman | \(\frac {\left (\ln \left (F \right )^{2} b^{2} c^{2} d -2 e \right ) {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{c^{3} b^{3} \ln \left (F \right )^{3}}+\frac {2 e x \,{\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right )^{2} b^{2} c^{2}}-\frac {e \,x^{2} {\mathrm e}^{c \left (b x +a \right ) \ln \left (F \right )}}{\ln \left (F \right ) b c}\) | \(89\) |
Input:
int(F^(c*(b*x+a))*(-e*x^2+d),x,method=_RETURNVERBOSE)
Output:
(-e*x^2*ln(F)^2*b^2*c^2+ln(F)^2*b^2*c^2*d+2*e*x*ln(F)*b*c-2*e)*F^(c*(b*x+a ))/ln(F)^3/b^3/c^3
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=\frac {{\left (2 \, b c e x \log \left (F\right ) - {\left (b^{2} c^{2} e x^{2} - b^{2} c^{2} d\right )} \log \left (F\right )^{2} - 2 \, e\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(-e*x^2+d),x, algorithm="fricas")
Output:
(2*b*c*e*x*log(F) - (b^2*c^2*e*x^2 - b^2*c^2*d)*log(F)^2 - 2*e)*F^(b*c*x + a*c)/(b^3*c^3*log(F)^3)
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=\begin {cases} \frac {F^{c \left (a + b x\right )} \left (b^{2} c^{2} d \log {\left (F \right )}^{2} - b^{2} c^{2} e x^{2} \log {\left (F \right )}^{2} + 2 b c e x \log {\left (F \right )} - 2 e\right )}{b^{3} c^{3} \log {\left (F \right )}^{3}} & \text {for}\: b^{3} c^{3} \log {\left (F \right )}^{3} \neq 0 \\d x - \frac {e x^{3}}{3} & \text {otherwise} \end {cases} \] Input:
integrate(F**((b*x+a)*c)*(-e*x**2+d),x)
Output:
Piecewise((F**(c*(a + b*x))*(b**2*c**2*d*log(F)**2 - b**2*c**2*e*x**2*log( F)**2 + 2*b*c*e*x*log(F) - 2*e)/(b**3*c**3*log(F)**3), Ne(b**3*c**3*log(F) **3, 0)), (d*x - e*x**3/3, True))
Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.09 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=\frac {F^{b c x + a c} d}{b c \log \left (F\right )} - \frac {{\left (F^{a c} b^{2} c^{2} x^{2} \log \left (F\right )^{2} - 2 \, F^{a c} b c x \log \left (F\right ) + 2 \, F^{a c}\right )} F^{b c x} e}{b^{3} c^{3} \log \left (F\right )^{3}} \] Input:
integrate(F^((b*x+a)*c)*(-e*x^2+d),x, algorithm="maxima")
Output:
F^(b*c*x + a*c)*d/(b*c*log(F)) - (F^(a*c)*b^2*c^2*x^2*log(F)^2 - 2*F^(a*c) *b*c*x*log(F) + 2*F^(a*c))*F^(b*c*x)*e/(b^3*c^3*log(F)^3)
Result contains complex when optimal does not.
Time = 0.16 (sec) , antiderivative size = 1690, normalized size of antiderivative = 22.84 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=\text {Too large to display} \] Input:
integrate(F^((b*x+a)*c)*(-e*x^2+d),x, algorithm="giac")
Output:
-(((pi^2*b^2*c^2*e*x^2*sgn(F) - pi^2*b^2*c^2*e*x^2 + 2*b^2*c^2*e*x^2*log(a bs(F))^2 - pi^2*b^2*c^2*d*sgn(F) + pi^2*b^2*c^2*d - 2*b^2*c^2*d*log(abs(F) )^2 - 4*b*c*e*x*log(abs(F)) + 4*e)*(3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3* pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)/((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs( F))^2)^2 + (3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)^2) - 2*(pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log (abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2)*(pi*b^2*c^2 *e*x^2*log(abs(F))*sgn(F) - pi*b^2*c^2*e*x^2*log(abs(F)) - pi*b^2*c^2*d*lo g(abs(F))*sgn(F) + pi*b^2*c^2*d*log(abs(F)) - pi*b*c*e*x*sgn(F) + pi*b*c*e *x)/((pi^3*b^3*c^3*sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c ^3 + 3*pi*b^3*c^3*log(abs(F))^2)^2 + (3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) + 2*b^3*c^3*log(abs(F))^3)^2))*cos(-1/2*pi*b*c* x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c) + ((pi^3*b^3*c^3 *sgn(F) - 3*pi*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3* log(abs(F))^2)*(pi^2*b^2*c^2*e*x^2*sgn(F) - pi^2*b^2*c^2*e*x^2 + 2*b^2*c^2 *e*x^2*log(abs(F))^2 - pi^2*b^2*c^2*d*sgn(F) + pi^2*b^2*c^2*d - 2*b^2*c^2* d*log(abs(F))^2 - 4*b*c*e*x*log(abs(F)) + 4*e)/((pi^3*b^3*c^3*sgn(F) - 3*p i*b^3*c^3*log(abs(F))^2*sgn(F) - pi^3*b^3*c^3 + 3*pi*b^3*c^3*log(abs(F))^2 )^2 + (3*pi^2*b^3*c^3*log(abs(F))*sgn(F) - 3*pi^2*b^3*c^3*log(abs(F)) +...
Time = 22.58 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=-\frac {F^{a\,c+b\,c\,x}\,\left (e\,b^2\,c^2\,x^2\,{\ln \left (F\right )}^2-d\,b^2\,c^2\,{\ln \left (F\right )}^2-2\,e\,b\,c\,x\,\ln \left (F\right )+2\,e\right )}{b^3\,c^3\,{\ln \left (F\right )}^3} \] Input:
int(F^(c*(a + b*x))*(d - e*x^2),x)
Output:
-(F^(a*c + b*c*x)*(2*e - b^2*c^2*d*log(F)^2 + b^2*c^2*e*x^2*log(F)^2 - 2*b *c*e*x*log(F)))/(b^3*c^3*log(F)^3)
Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.82 \[ \int F^{c (a+b x)} \left (d-e x^2\right ) \, dx=\frac {f^{b c x +a c} \left (\mathrm {log}\left (f \right )^{2} b^{2} c^{2} d -\mathrm {log}\left (f \right )^{2} b^{2} c^{2} e \,x^{2}+2 \,\mathrm {log}\left (f \right ) b c e x -2 e \right )}{\mathrm {log}\left (f \right )^{3} b^{3} c^{3}} \] Input:
int(F^((b*x+a)*c)*(-e*x^2+d),x)
Output:
(f**(a*c + b*c*x)*(log(f)**2*b**2*c**2*d - log(f)**2*b**2*c**2*e*x**2 + 2* log(f)*b*c*e*x - 2*e))/(log(f)**3*b**3*c**3)