Integrand size = 21, antiderivative size = 87 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^3} \log (F)}{6 d (c+d x)^3}+\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log ^2(F)}{6 d} \] Output:
-1/6*F^(a+b*(d*x+c)^3)/d/(d*x+c)^6-1/6*b*F^(a+b*(d*x+c)^3)*ln(F)/d/(d*x+c) ^3+1/6*b^2*F^a*Ei(b*(d*x+c)^3*ln(F))*ln(F)^2/d
Time = 0.37 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\frac {F^a \left (b^2 \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right ) \log ^2(F)-\frac {F^{b (c+d x)^3} \left (1+b (c+d x)^3 \log (F)\right )}{(c+d x)^6}\right )}{6 d} \] Input:
Integrate[F^(a + b*(c + d*x)^3)/(c + d*x)^7,x]
Output:
(F^a*(b^2*ExpIntegralEi[b*(c + d*x)^3*Log[F]]*Log[F]^2 - (F^(b*(c + d*x)^3 )*(1 + b*(c + d*x)^3*Log[F]))/(c + d*x)^6))/(6*d)
Time = 0.65 (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.
\(\displaystyle \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{2} b \log (F) \int \frac {F^{b (c+d x)^3+a}}{(c+d x)^4}dx-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\) |
\(\Big \downarrow \) 2643 |
\(\displaystyle \frac {1}{2} b \log (F) \left (b \log (F) \int \frac {F^{b (c+d x)^3+a}}{c+d x}dx-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\) |
\(\Big \downarrow \) 2639 |
\(\displaystyle \frac {1}{2} b \log (F) \left (\frac {b F^a \log (F) \operatorname {ExpIntegralEi}\left (b (c+d x)^3 \log (F)\right )}{3 d}-\frac {F^{a+b (c+d x)^3}}{3 d (c+d x)^3}\right )-\frac {F^{a+b (c+d x)^3}}{6 d (c+d x)^6}\) |
Input:
Int[F^(a + b*(c + d*x)^3)/(c + d*x)^7,x]
Output:
-1/6*F^(a + b*(c + d*x)^3)/(d*(c + d*x)^6) + (b*Log[F]*(-1/3*F^(a + b*(c + d*x)^3)/(d*(c + d*x)^3) + (b*F^a*ExpIntegralEi[b*(c + d*x)^3*Log[F]]*Log[ F])/(3*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]))
\[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{7}}d x\]
Input:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x)
Output:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x)
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (81) = 162\).
Time = 0.08 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.09 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\frac {{\left (b^{2} d^{6} x^{6} + 6 \, b^{2} c d^{5} x^{5} + 15 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{3} d^{3} x^{3} + 15 \, b^{2} c^{4} d^{2} x^{2} + 6 \, b^{2} c^{5} d x + b^{2} c^{6}\right )} F^{a} {\rm Ei}\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right )^{2} - {\left ({\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right ) + 1\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{6 \, {\left (d^{7} x^{6} + 6 \, c d^{6} x^{5} + 15 \, c^{2} d^{5} x^{4} + 20 \, c^{3} d^{4} x^{3} + 15 \, c^{4} d^{3} x^{2} + 6 \, c^{5} d^{2} x + c^{6} d\right )}} \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x, algorithm="fricas")
Output:
1/6*((b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + 20*b^2*c^3*d^3* x^3 + 15*b^2*c^4*d^2*x^2 + 6*b^2*c^5*d*x + b^2*c^6)*F^a*Ei((b*d^3*x^3 + 3* b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F))*log(F)^2 - ((b*d^3*x^3 + 3*b*c* d^2*x^2 + 3*b*c^2*d*x + b*c^3)*log(F) + 1)*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a))/(d^7*x^6 + 6*c*d^6*x^5 + 15*c^2*d^5*x^4 + 20*c^3 *d^4*x^3 + 15*c^4*d^3*x^2 + 6*c^5*d^2*x + c^6*d)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{3}}}{\left (c + d x\right )^{7}}\, dx \] Input:
integrate(F**(a+b*(d*x+c)**3)/(d*x+c)**7,x)
Output:
Integral(F**(a + b*(c + d*x)**3)/(c + d*x)**7, x)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x, algorithm="maxima")
Output:
integrate(F^((d*x + c)^3*b + a)/(d*x + c)^7, x)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \] Input:
integrate(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x, algorithm="giac")
Output:
integrate(F^((d*x + c)^3*b + a)/(d*x + c)^7, x)
Time = 0.75 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=-\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}{2}+F^{b\,{\left (c+d\,x\right )}^3}\,\left (\frac {1}{2\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}+\frac {1}{2\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^6}\right )\right )}{3\,d} \] Input:
int(F^(a + b*(c + d*x)^3)/(c + d*x)^7,x)
Output:
-(F^a*b^2*log(F)^2*(expint(-b*log(F)*(c + d*x)^3)/2 + F^(b*(c + d*x)^3)*(1 /(2*b*log(F)*(c + d*x)^3) + 1/(2*b^2*log(F)^2*(c + d*x)^6))))/(3*d)
\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^7} \, dx=\text {too large to display} \] Input:
int(F^(a+b*(d*x+c)^3)/(d*x+c)^7,x)
Output:
(f**(a + b*c**3)*( - 2*f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3) + 9*int(f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(log(f)*b*c**10 + 7*log(f)*b*c**9*d*x + 21*log(f)*b*c**8*d**2*x**2 + 35*log(f)*b*c**7*d**3* x**3 + 35*log(f)*b*c**6*d**4*x**4 + 21*log(f)*b*c**5*d**5*x**5 + 7*log(f)* b*c**4*d**6*x**6 + log(f)*b*c**3*d**7*x**7 - 2*c**7 - 14*c**6*d*x - 42*c** 5*d**2*x**2 - 70*c**4*d**3*x**3 - 70*c**3*d**4*x**4 - 42*c**2*d**5*x**5 - 14*c*d**6*x**6 - 2*d**7*x**7),x)*log(f)**2*b**2*c**12*d + 54*int(f**(3*b*c **2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(log(f)*b*c**10 + 7*log(f)*b*c**9 *d*x + 21*log(f)*b*c**8*d**2*x**2 + 35*log(f)*b*c**7*d**3*x**3 + 35*log(f) *b*c**6*d**4*x**4 + 21*log(f)*b*c**5*d**5*x**5 + 7*log(f)*b*c**4*d**6*x**6 + log(f)*b*c**3*d**7*x**7 - 2*c**7 - 14*c**6*d*x - 42*c**5*d**2*x**2 - 70 *c**4*d**3*x**3 - 70*c**3*d**4*x**4 - 42*c**2*d**5*x**5 - 14*c*d**6*x**6 - 2*d**7*x**7),x)*log(f)**2*b**2*c**11*d**2*x + 135*int(f**(3*b*c**2*d*x + 3*b*c*d**2*x**2 + b*d**3*x**3)/(log(f)*b*c**10 + 7*log(f)*b*c**9*d*x + 21* log(f)*b*c**8*d**2*x**2 + 35*log(f)*b*c**7*d**3*x**3 + 35*log(f)*b*c**6*d* *4*x**4 + 21*log(f)*b*c**5*d**5*x**5 + 7*log(f)*b*c**4*d**6*x**6 + log(f)* b*c**3*d**7*x**7 - 2*c**7 - 14*c**6*d*x - 42*c**5*d**2*x**2 - 70*c**4*d**3 *x**3 - 70*c**3*d**4*x**4 - 42*c**2*d**5*x**5 - 14*c*d**6*x**6 - 2*d**7*x* *7),x)*log(f)**2*b**2*c**10*d**3*x**2 + 180*int(f**(3*b*c**2*d*x + 3*b*c*d **2*x**2 + b*d**3*x**3)/(log(f)*b*c**10 + 7*log(f)*b*c**9*d*x + 21*log(...