\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx\) [264]

Optimal result

Integrand size = 21, antiderivative size = 121 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac {b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac {b^3 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)}{12 d} \]

[Out]

-1/6*F^(a+b*(d*x+c)^2)/d/(d*x+c)^6-1/12*b*F^(a+b*(d*x+c)^2)*ln(F)/d/(d*x+c)^4-1/12*b^2*F^(a+b*(d*x+c)^2)*ln(F)
^2/d/(d*x+c)^2+1/12*b^3*F^a*Ei(b*(d*x+c)^2*ln(F))*ln(F)^3/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\frac {b^3 F^a \log ^3(F) \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right )}{12 d}-\frac {b^2 \log ^2(F) F^{a+b (c+d x)^2}}{12 d (c+d x)^2}-\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b \log (F) F^{a+b (c+d x)^2}}{12 d (c+d x)^4} \]

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]

[Out]

-1/6*F^(a + b*(c + d*x)^2)/(d*(c + d*x)^6) - (b*F^(a + b*(c + d*x)^2)*Log[F])/(12*d*(c + d*x)^4) - (b^2*F^(a +
b*(c + d*x)^2)*Log[F]^2)/(12*d*(c + d*x)^2) + (b^3*F^a*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3)/(12*d)

Rule 2241

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] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[d*e - c*f, 0]

Rule 2245

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] - Dist[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]))

Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}+\frac {1}{3} (b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}+\frac {1}{6} \left (b^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^3} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac {b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac {1}{6} \left (b^3 \log ^3(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{c+d x} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{6 d (c+d x)^6}-\frac {b F^{a+b (c+d x)^2} \log (F)}{12 d (c+d x)^4}-\frac {b^2 F^{a+b (c+d x)^2} \log ^2(F)}{12 d (c+d x)^2}+\frac {b^3 F^a \text {Ei}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)}{12 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.65 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\frac {F^a \left (b^3 \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^3(F)-\frac {F^{b (c+d x)^2} \left (2+b (c+d x)^2 \log (F)+b^2 (c+d x)^4 \log ^2(F)\right )}{(c+d x)^6}\right )}{12 d} \]

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^7,x]

[Out]

(F^a*(b^3*ExpIntegralEi[b*(c + d*x)^2*Log[F]]*Log[F]^3 - (F^(b*(c + d*x)^2)*(2 + b*(c + d*x)^2*Log[F] + b^2*(c
+ d*x)^4*Log[F]^2))/(c + d*x)^6))/(12*d)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{6 d \left (d x +c \right )^{6}}-\frac {b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{12 d \left (d x +c \right )^{4}}-\frac {b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{12 d \left (d x +c \right )^{2}}-\frac {b^{3} \ln \left (F \right )^{3} F^{a} \operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{12 d}\)	\(119\)

[In]

int(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x,method=_RETURNVERBOSE)

[Out]

-1/6/d/(d*x+c)^6*F^(b*(d*x+c)^2)*F^a-1/12/d*b*ln(F)/(d*x+c)^4*F^(b*(d*x+c)^2)*F^a-1/12/d*b^2*ln(F)^2/(d*x+c)^2
*F^(b*(d*x+c)^2)*F^a-1/12/d*b^3*ln(F)^3*F^a*Ei(1,-b*(d*x+c)^2*ln(F))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (113) = 226\).

Time = 0.28 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.41 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\frac {{\left (b^{3} d^{6} x^{6} + 6 \, b^{3} c d^{5} x^{5} + 15 \, b^{3} c^{2} d^{4} x^{4} + 20 \, b^{3} c^{3} d^{3} x^{3} + 15 \, b^{3} c^{4} d^{2} x^{2} + 6 \, b^{3} c^{5} d x + b^{3} c^{6}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{3} - {\left ({\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 2\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{12 \, {\left (d^{7} x^{6} + 6 \, c d^{6} x^{5} + 15 \, c^{2} d^{5} x^{4} + 20 \, c^{3} d^{4} x^{3} + 15 \, c^{4} d^{3} x^{2} + 6 \, c^{5} d^{2} x + c^{6} d\right )}} \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="fricas")

[Out]

1/12*((b^3*d^6*x^6 + 6*b^3*c*d^5*x^5 + 15*b^3*c^2*d^4*x^4 + 20*b^3*c^3*d^3*x^3 + 15*b^3*c^4*d^2*x^2 + 6*b^3*c^
5*d*x + b^3*c^6)*F^a*Ei((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*log(F)^3 - ((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)*log(F)^2 + (b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F) + 2)*F^(b*d^2*x^
2 + 2*b*c*d*x + b*c^2 + 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)

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{7}}\, dx \]

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**7,x)

[Out]

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**7, x)

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{7}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^7,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^7, x)

Mupad [B] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.86 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^7} \, dx=-\frac {F^a\,b^3\,{\ln \left (F\right )}^3\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}{12\,d}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3\,\left (\frac {1}{6\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}+\frac {1}{6\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^4}+\frac {1}{3\,b^3\,{\ln \left (F\right )}^3\,{\left (c+d\,x\right )}^6}\right )}{2\,d} \]

[In]

int(F^(a + b*(c + d*x)^2)/(c + d*x)^7,x)

[Out]

- (F^a*b^3*log(F)^3*expint(-b*log(F)*(c + d*x)^2))/(12*d) - (F^a*F^(b*(c + d*x)^2)*b^3*log(F)^3*(1/(6*b*log(F)
*(c + d*x)^2) + 1/(6*b^2*log(F)^2*(c + d*x)^4) + 1/(3*b^3*log(F)^3*(c + d*x)^6)))/(2*d)