Integrand size = 21, antiderivative size = 87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b F^{a+b (c+d x)^2} \log (F)}{4 d (c+d x)^2}+\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^2(F)}{4 d} \] Output:
-1/4*F^(a+b*(d*x+c)^2)/d/(d*x+c)^4-1/4*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c) ^2+1/4*b^2*F^a*Ei(b*(d*x+c)^2*ln(F))*ln(F)^2/d
Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\frac {F^a \left (b^2 \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^2(F)-\frac {F^{b (c+d x)^2} \left (1+b (c+d x)^2 \log (F)\right )}{(c+d x)^4}\right )}{4 d} \] Input:
Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^5,x]
Output:
(F^a*(b^2*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^2 - (F^(b*(c + d*x)^2 )*(1 + b*(c + d*x)^2*Log[F]))/(c + d*x)^4))/(4*d)
Time = 0.63 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2643, 2643, 2639}
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.
\(\displaystyle \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{2} b \log (F) \int \frac {F^{b (c+d x)^2+a}}{(c+d x)^3}dx-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{b (c+d x)^2+a}}{c+d x}dx-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}\right )-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{2} b \log (F) \left (\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right )}{2 d}-\frac {F^{a+b (c+d x)^2}}{2 d (c+d x)^2}\right )-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}\) |
Input:
Int[F^(a + b*(c + d*x)^2)/(c + d*x)^5,x]
Output:
-1/4*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^4) + (b*Log[F]*(-1/2*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^2) + (b*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[ F])/(2*d)))/2
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_ Symbol] :> Simp[F^a*(ExpIntegralEi[b*(c + d*x)^n*Log[F]]/(f*n)), x] /; Free Q[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_ .), x_Symbol] :> Simp[(c + d*x)^(m + 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))) , x] - Simp[b*n*(Log[F]/(m + 1)) Int[(c + d*x)^(m + n)*F^(a + b*(c + d*x) ^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[ -4, (m + 1)/n, 5] && IntegerQ[n] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))
Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{4 d \left (d x +c \right )^{4}}-\frac {b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{4 d \left (d x +c \right )^{2}}-\frac {b^{2} \ln \left (F \right )^{2} F^{a} \operatorname {expIntegral}_{1}\left (-b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{4 d}\) | \(86\) |
Input:
int(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
-1/4/d/(d*x+c)^4*F^(b*(d*x+c)^2)*F^a-1/4/d*b*ln(F)/(d*x+c)^2*F^(b*(d*x+c)^ 2)*F^a-1/4/d*b^2*ln(F)^2*F^a*Ei(1,-b*(d*x+c)^2*ln(F))
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (81) = 162\).
Time = 0.08 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.10 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\frac {{\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{2} - {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 1\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{4 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 6 \, c^{2} d^{3} x^{2} + 4 \, c^{3} d^{2} x + c^{4} d\right )}} \] Input:
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x, algorithm="fricas")
Output:
1/4*((b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x + b^2*c^4)*F^a*Ei((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*log(F)^2 - ((b*d^2 *x^2 + 2*b*c*d*x + b*c^2)*log(F) + 1)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a ))/(d^5*x^4 + 4*c*d^4*x^3 + 6*c^2*d^3*x^2 + 4*c^3*d^2*x + c^4*d)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{5}}\, dx \] Input:
integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**5,x)
Output:
Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**5, x)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x, algorithm="maxima")
Output:
integrate(F^((d*x + c)^2*b + a)/(d*x + c)^5, x)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x, algorithm="giac")
Output:
integrate(F^((d*x + c)^2*b + a)/(d*x + c)^5, x)
Time = 1.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=-\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}{2}+F^{b\,{\left (c+d\,x\right )}^2}\,\left (\frac {1}{2\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}+\frac {1}{2\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^4}\right )\right )}{2\,d} \] Input:
int(F^(a + b*(c + d*x)^2)/(c + d*x)^5,x)
Output:
-(F^a*b^2*log(F)^2*(expint(-b*log(F)*(c + d*x)^2)/2 + F^(b*(c + d*x)^2)*(1 /(2*b*log(F)*(c + d*x)^2) + 1/(2*b^2*log(F)^2*(c + d*x)^4))))/(2*d)
\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\text {too large to display} \] Input:
int(F^(a+b*(d*x+c)^2)/(d*x+c)^5,x)
Output:
(f**(a + b*c**2)*( - f**(2*b*c*d*x + b*d**2*x**2) + 4*int(f**(2*b*c*d*x + b*d**2*x**2)/(log(f)*b*c**7 + 5*log(f)*b*c**6*d*x + 10*log(f)*b*c**5*d**2* x**2 + 10*log(f)*b*c**4*d**3*x**3 + 5*log(f)*b*c**3*d**4*x**4 + log(f)*b*c **2*d**5*x**5 - 2*c**5 - 10*c**4*d*x - 20*c**3*d**2*x**2 - 20*c**2*d**3*x* *3 - 10*c*d**4*x**4 - 2*d**5*x**5),x)*log(f)**2*b**2*c**8*d + 16*int(f**(2 *b*c*d*x + b*d**2*x**2)/(log(f)*b*c**7 + 5*log(f)*b*c**6*d*x + 10*log(f)*b *c**5*d**2*x**2 + 10*log(f)*b*c**4*d**3*x**3 + 5*log(f)*b*c**3*d**4*x**4 + log(f)*b*c**2*d**5*x**5 - 2*c**5 - 10*c**4*d*x - 20*c**3*d**2*x**2 - 20*c **2*d**3*x**3 - 10*c*d**4*x**4 - 2*d**5*x**5),x)*log(f)**2*b**2*c**7*d**2* x + 24*int(f**(2*b*c*d*x + b*d**2*x**2)/(log(f)*b*c**7 + 5*log(f)*b*c**6*d *x + 10*log(f)*b*c**5*d**2*x**2 + 10*log(f)*b*c**4*d**3*x**3 + 5*log(f)*b* c**3*d**4*x**4 + log(f)*b*c**2*d**5*x**5 - 2*c**5 - 10*c**4*d*x - 20*c**3* d**2*x**2 - 20*c**2*d**3*x**3 - 10*c*d**4*x**4 - 2*d**5*x**5),x)*log(f)**2 *b**2*c**6*d**3*x**2 + 16*int(f**(2*b*c*d*x + b*d**2*x**2)/(log(f)*b*c**7 + 5*log(f)*b*c**6*d*x + 10*log(f)*b*c**5*d**2*x**2 + 10*log(f)*b*c**4*d**3 *x**3 + 5*log(f)*b*c**3*d**4*x**4 + log(f)*b*c**2*d**5*x**5 - 2*c**5 - 10* c**4*d*x - 20*c**3*d**2*x**2 - 20*c**2*d**3*x**3 - 10*c*d**4*x**4 - 2*d**5 *x**5),x)*log(f)**2*b**2*c**5*d**4*x**3 + 4*int(f**(2*b*c*d*x + b*d**2*x** 2)/(log(f)*b*c**7 + 5*log(f)*b*c**6*d*x + 10*log(f)*b*c**5*d**2*x**2 + 10* log(f)*b*c**4*d**3*x**3 + 5*log(f)*b*c**3*d**4*x**4 + log(f)*b*c**2*d**...