3.303 \(\int F^{a+\frac {b}{c+d x}} (c+d x)^3 \, dx\)

Optimal. Leaf size=28 \[ \frac {b^4 F^a \log ^4(F) \Gamma \left (-4,-\frac {b \log (F)}{c+d x}\right )}{d} \]

[Out]

F^a*(d*x+c)^4*Ei(5,-b*ln(F)/(d*x+c))/d

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2218} \[ \frac {b^4 F^a \log ^4(F) \text {Gamma}\left (-4,-\frac {b \log (F)}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2218

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

Rubi steps

\begin {align*} \int F^{a+\frac {b}{c+d x}} (c+d x)^3 \, dx &=\frac {b^4 F^a \Gamma \left (-4,-\frac {b \log (F)}{c+d x}\right ) \log ^4(F)}{d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {b^4 F^a \log ^4(F) \Gamma \left (-4,-\frac {b \log (F)}{c+d x}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 175, normalized size = 6.25 \[ -\frac {F^{a} b^{4} {\rm Ei}\left (\frac {b \log \relax (F)}{d x + c}\right ) \log \relax (F)^{4} - {\left (6 \, d^{4} x^{4} + 24 \, c d^{3} x^{3} + 36 \, c^{2} d^{2} x^{2} + 24 \, c^{3} d x + 6 \, c^{4} + {\left (b^{3} d x + b^{3} c\right )} \log \relax (F)^{3} + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \relax (F)^{2} + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \relax (F)\right )} F^{\frac {a d x + a c + b}{d x + c}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(F^a*b^4*Ei(b*log(F)/(d*x + c))*log(F)^4 - (6*d^4*x^4 + 24*c*d^3*x^3 + 36*c^2*d^2*x^2 + 24*c^3*d*x + 6*c
^4 + (b^3*d*x + b^3*c)*log(F)^3 + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*log(F)^2 + 2*(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*x + a*c + b)/(d*x + c)))/d

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} F^{a + \frac {b}{d x + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.13, size = 368, normalized size = 13.14 \[ \frac {b^{4} F^{a} \Ei \left (1, -\frac {b \ln \relax (F )}{d x +c}\right ) \ln \relax (F )^{4}}{24 d}+\frac {b^{3} x \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{3}}{24}+\frac {b^{2} d \,x^{2} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{2}}{24}+\frac {b \,d^{2} x^{3} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )}{12}+\frac {d^{3} x^{4} F^{a} F^{\frac {b}{d x +c}}}{4}+\frac {b^{3} c \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{3}}{24 d}+\frac {b^{2} c x \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{2}}{12}+\frac {b c d \,x^{2} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )}{4}+c \,d^{2} x^{3} F^{a} F^{\frac {b}{d x +c}}+\frac {b^{2} c^{2} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{2}}{24 d}+\frac {b \,c^{2} x \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )}{4}+\frac {3 c^{2} d \,x^{2} F^{a} F^{\frac {b}{d x +c}}}{2}+\frac {b \,c^{3} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )}{12 d}+c^{3} x \,F^{a} F^{\frac {b}{d x +c}}+\frac {c^{4} F^{a} F^{\frac {b}{d x +c}}}{4 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+1/(d*x+c)*b)*(d*x+c)^3,x)

[Out]

1/4*d^3*F^a*F^(1/(d*x+c)*b)*x^4+d^2*F^a*F^(1/(d*x+c)*b)*c*x^3+3/2*d*F^a*F^(1/(d*x+c)*b)*c^2*x^2+F^a*F^(1/(d*x+
c)*b)*c^3*x+1/4/d*F^a*F^(1/(d*x+c)*b)*c^4+1/12*d^2*b*ln(F)*F^a*F^(1/(d*x+c)*b)*x^3+1/4*d*b*ln(F)*F^a*F^(1/(d*x
+c)*b)*c*x^2+1/4*b*ln(F)*F^a*F^(1/(d*x+c)*b)*c^2*x+1/12/d*b*ln(F)*F^a*F^(1/(d*x+c)*b)*c^3+1/24*d*b^2*ln(F)^2*F
^a*F^(1/(d*x+c)*b)*x^2+1/12*b^2*ln(F)^2*F^a*F^(1/(d*x+c)*b)*c*x+1/24/d*b^2*ln(F)^2*F^a*F^(1/(d*x+c)*b)*c^2+1/2
4*b^3*ln(F)^3*F^a*F^(1/(d*x+c)*b)*x+1/24/d*b^3*ln(F)^3*F^a*F^(1/(d*x+c)*b)*c+1/24/d*b^4*ln(F)^4*F^a*Ei(1,-1/(d
*x+c)*b*ln(F))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{24} \, {\left (6 \, F^{a} d^{3} x^{4} + 2 \, {\left (F^{a} b d^{2} \log \relax (F) + 12 \, F^{a} c d^{2}\right )} x^{3} + {\left (F^{a} b^{2} d \log \relax (F)^{2} + 6 \, F^{a} b c d \log \relax (F) + 36 \, F^{a} c^{2} d\right )} x^{2} + {\left (F^{a} b^{3} \log \relax (F)^{3} + 2 \, F^{a} b^{2} c \log \relax (F)^{2} + 6 \, F^{a} b c^{2} \log \relax (F) + 24 \, F^{a} c^{3}\right )} x\right )} F^{\frac {b}{d x + c}} + \int \frac {{\left (F^{a} b^{4} d x \log \relax (F)^{4} - F^{a} b^{3} c^{2} \log \relax (F)^{3} - 2 \, F^{a} b^{2} c^{3} \log \relax (F)^{2} - 6 \, F^{a} b c^{4} \log \relax (F)\right )} F^{\frac {b}{d x + c}}}{24 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(6*F^a*d^3*x^4 + 2*(F^a*b*d^2*log(F) + 12*F^a*c*d^2)*x^3 + (F^a*b^2*d*log(F)^2 + 6*F^a*b*c*d*log(F) + 36*
F^a*c^2*d)*x^2 + (F^a*b^3*log(F)^3 + 2*F^a*b^2*c*log(F)^2 + 6*F^a*b*c^2*log(F) + 24*F^a*c^3)*x)*F^(b/(d*x + c)
) + integrate(1/24*(F^a*b^4*d*x*log(F)^4 - F^a*b^3*c^2*log(F)^3 - 2*F^a*b^2*c^3*log(F)^2 - 6*F^a*b*c^4*log(F))
*F^(b/(d*x + c))/(d^2*x^2 + 2*c*d*x + c^2), x)

________________________________________________________________________________________

mupad [B]  time = 3.70, size = 148, normalized size = 5.29 \[ \frac {F^a\,F^{\frac {b}{c+d\,x}}\,{\left (c+d\,x\right )}^4}{4\,d}+\frac {F^a\,b^4\,{\ln \relax (F)}^4\,\mathrm {expint}\left (-\frac {b\,\ln \relax (F)}{c+d\,x}\right )}{24\,d}+\frac {F^a\,F^{\frac {b}{c+d\,x}}\,b^2\,{\ln \relax (F)}^2\,{\left (c+d\,x\right )}^2}{24\,d}+\frac {F^a\,F^{\frac {b}{c+d\,x}}\,b\,\ln \relax (F)\,{\left (c+d\,x\right )}^3}{12\,d}+\frac {F^a\,F^{\frac {b}{c+d\,x}}\,b^3\,{\ln \relax (F)}^3\,\left (c+d\,x\right )}{24\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{a + \frac {b}{c + d x}} \left (c + d x\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________