\(\int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx\) [344]

Optimal result

Integrand size = 21, antiderivative size = 87 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6}{6 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log (F)}{6 d}-\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log ^2(F)}{6 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=-\frac {b^2 F^a \log ^2(F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right )}{6 d}+\frac {(c+d x)^6 F^{a+\frac {b}{(c+d x)^3}}}{6 d}+\frac {b \log (F) (c+d x)^3 F^{a+\frac {b}{(c+d x)^3}}}{6 d} \]

[In]

Int[F^(a + b/(c + d*x)^3)*(c + d*x)^5,x]

[Out]

(F^(a + b/(c + d*x)^3)*(c + d*x)^6)/(6*d) + (b*F^(a + b/(c + d*x)^3)*(c + d*x)^3*Log[F])/(6*d) - (b^2*F^a*ExpI
ntegralEi[(b*Log[F])/(c + d*x)^3]*Log[F]^2)/(6*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+\frac {b}{(c+d x)^3}} (c+d x)^6}{6 d}+\frac {1}{2} (b \log (F)) \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^2 \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6}{6 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log (F)}{6 d}+\frac {1}{2} \left (b^2 \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{(c+d x)^3}}}{c+d x} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^3}} (c+d x)^6}{6 d}+\frac {b F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \log (F)}{6 d}-\frac {b^2 F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log ^2(F)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^3}} (c+d x)^6+b \log (F) \left (F^{\frac {b}{(c+d x)^3}} (c+d x)^3-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^3}\right ) \log (F)\right )\right )}{6 d} \]

[In]

Integrate[F^(a + b/(c + d*x)^3)*(c + d*x)^5,x]

[Out]

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

Maple [F]

\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{5}d x\]

[In]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^5,x)

[Out]

int(F^(a+b/(d*x+c)^3)*(d*x+c)^5,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (81) = 162\).

Time = 0.28 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.45 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=-\frac {F^{a} b^{2} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \left (F\right )^{2} - {\left (d^{6} x^{6} + 6 \, c d^{5} x^{5} + 15 \, c^{2} d^{4} x^{4} + 20 \, c^{3} d^{3} x^{3} + 15 \, c^{4} d^{2} x^{2} + 6 \, c^{5} d x + c^{6} + {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{6 \, d} \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^5,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{5}\, dx \]

[In]

integrate(F**(a+b/(d*x+c)**3)*(d*x+c)**5,x)

[Out]

Integral(F**(a + b/(c + d*x)**3)*(c + d*x)**5, x)

Maxima [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\int { {\left (d x + c\right )}^{5} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^5,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\int { {\left (d x + c\right )}^{5} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]

[In]

integrate(F^(a+b/(d*x+c)^3)*(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((d*x + c)^5*F^(a + b/(d*x + c)^3), x)

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^5 \, dx=\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}{2}+F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,\left (\frac {{\left (c+d\,x\right )}^3}{2\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^6}{2\,b^2\,{\ln \left (F\right )}^2}\right )\right )}{3\,d} \]

[In]

int(F^(a + b/(c + d*x)^3)*(c + d*x)^5,x)

[Out]

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