\(\int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx\) [265]

Optimal result

Integrand size = 21, antiderivative size = 31 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=-\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^2 \log (F)\right ) \log ^4(F)}{2 d} \]

[Out]

-1/2*F^a/(d*x+c)^8*Ei(5,-b*(d*x+c)^2*ln(F))/d

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2250} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=-\frac {b^4 F^a \log ^4(F) \Gamma \left (-4,-b (c+d x)^2 \log (F)\right )}{2 d} \]

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^9,x]

[Out]

-1/2*(b^4*F^a*Gamma[-4, -(b*(c + d*x)^2*Log[F])]*Log[F]^4)/d

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^2 \log (F)\right ) \log ^4(F)}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=-\frac {b^4 F^a \Gamma \left (-4,-b (c+d x)^2 \log (F)\right ) \log ^4(F)}{2 d} \]

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^9,x]

[Out]

-1/2*(b^4*F^a*Gamma[-4, -(b*(c + d*x)^2*Log[F])]*Log[F]^4)/d

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(151\) vs. \(2(29)=58\).

Time = 0.76 (sec) , antiderivative size = 152, normalized size of antiderivative = 4.90

method	result	size
risch	\(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{8 d \left (d x +c \right )^{8}}-\frac {b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{24 d \left (d x +c \right )^{6}}-\frac {b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{48 d \left (d x +c \right )^{4}}-\frac {b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{48 d \left (d x +c \right )^{2}}-\frac {b^{4} \ln \left (F \right )^{4} F^{a} \operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{48 d}\)	\(152\)

[In]

int(F^(a+b*(d*x+c)^2)/(d*x+c)^9,x,method=_RETURNVERBOSE)

[Out]

-1/8/d/(d*x+c)^8*F^(b*(d*x+c)^2)*F^a-1/24/d*b*ln(F)/(d*x+c)^6*F^(b*(d*x+c)^2)*F^a-1/48/d*b^2*ln(F)^2/(d*x+c)^4
*F^(b*(d*x+c)^2)*F^a-1/48/d*b^3*ln(F)^3/(d*x+c)^2*F^(b*(d*x+c)^2)*F^a-1/48/d*b^4*ln(F)^4*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. 430 vs. \(2 (29) = 58\).

Time = 0.32 (sec) , antiderivative size = 430, normalized size of antiderivative = 13.87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=\frac {{\left (b^{4} d^{8} x^{8} + 8 \, b^{4} c d^{7} x^{7} + 28 \, b^{4} c^{2} d^{6} x^{6} + 56 \, b^{4} c^{3} d^{5} x^{5} + 70 \, b^{4} c^{4} d^{4} x^{4} + 56 \, b^{4} c^{5} d^{3} x^{3} + 28 \, b^{4} c^{6} d^{2} x^{2} + 8 \, b^{4} c^{7} d x + b^{4} c^{8}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{4} - {\left ({\left (b^{3} d^{6} x^{6} + 6 \, b^{3} c d^{5} x^{5} + 15 \, b^{3} c^{2} d^{4} x^{4} + 20 \, b^{3} c^{3} d^{3} x^{3} + 15 \, b^{3} c^{4} d^{2} x^{2} + 6 \, b^{3} c^{5} d x + b^{3} c^{6}\right )} \log \left (F\right )^{3} + {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} + 2 \, {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 6\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{48 \, {\left (d^{9} x^{8} + 8 \, c d^{8} x^{7} + 28 \, c^{2} d^{7} x^{6} + 56 \, c^{3} d^{6} x^{5} + 70 \, c^{4} d^{5} x^{4} + 56 \, c^{5} d^{4} x^{3} + 28 \, c^{6} d^{3} x^{2} + 8 \, c^{7} d^{2} x + c^{8} d\right )}} \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^9,x, algorithm="fricas")

[Out]

1/48*((b^4*d^8*x^8 + 8*b^4*c*d^7*x^7 + 28*b^4*c^2*d^6*x^6 + 56*b^4*c^3*d^5*x^5 + 70*b^4*c^4*d^4*x^4 + 56*b^4*c
^5*d^3*x^3 + 28*b^4*c^6*d^2*x^2 + 8*b^4*c^7*d*x + b^4*c^8)*F^a*Ei((b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F))*log(
F)^4 - ((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)*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)*lo
g(F)^2 + 2*(b*d^2*x^2 + 2*b*c*d*x + b*c^2)*log(F) + 6)*F^(b*d^2*x^2 + 2*b*c*d*x + b*c^2 + a))/(d^9*x^8 + 8*c*d
^8*x^7 + 28*c^2*d^7*x^6 + 56*c^3*d^6*x^5 + 70*c^4*d^5*x^4 + 56*c^5*d^4*x^3 + 28*c^6*d^3*x^2 + 8*c^7*d^2*x + c^
8*d)

Sympy [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{9}}\, dx \]

[In]

integrate(F**(a+b*(d*x+c)**2)/(d*x+c)**9,x)

[Out]

Integral(F**(a + b*(c + d*x)**2)/(c + d*x)**9, x)

Maxima [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{9}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^9,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^9, x)

Giac [F]

\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{9}} \,d x } \]

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^9,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^9, x)

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx=-\frac {F^a\,b^4\,{\ln \left (F\right )}^4\,\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}{48\,d}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^4\,{\ln \left (F\right )}^4\,\left (\frac {1}{24\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}+\frac {1}{24\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^4}+\frac {1}{12\,b^3\,{\ln \left (F\right )}^3\,{\left (c+d\,x\right )}^6}+\frac {1}{4\,b^4\,{\ln \left (F\right )}^4\,{\left (c+d\,x\right )}^8}\right )}{2\,d} \]

[In]

int(F^(a + b*(c + d*x)^2)/(c + d*x)^9,x)

[Out]

- (F^a*b^4*log(F)^4*expint(-b*log(F)*(c + d*x)^2))/(48*d) - (F^a*F^(b*(c + d*x)^2)*b^4*log(F)^4*(1/(24*b*log(F
)*(c + d*x)^2) + 1/(24*b^2*log(F)^2*(c + d*x)^4) + 1/(12*b^3*log(F)^3*(c + d*x)^6) + 1/(4*b^4*log(F)^4*(c + d*
x)^8)))/(2*d)