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

Optimal. Leaf size=267 \[ \frac {b^2 d^2 f \log ^2(F) F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{2 (d e-c f)^4}-\frac {b d^2 \log (F) F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^3}+\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {b d^2 \log (F) F^{a+\frac {b}{c+d x}}}{2 (d e-c f)^3}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {b d \log (F) F^{a+\frac {b}{c+d x}}}{2 (e+f x) (d e-c f)^2} \]

[Out]

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

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Rubi [A]  time = 1.91, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2223, 6742, 2209, 2210, 2222, 2228, 2178} \[ \frac {b^2 d^2 f \log ^2(F) F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{2 (d e-c f)^4}-\frac {b d^2 \log (F) F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right )}{(d e-c f)^3}+\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {b d^2 \log (F) F^{a+\frac {b}{c+d x}}}{2 (d e-c f)^3}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {b d \log (F) F^{a+\frac {b}{c+d x}}}{2 (e+f x) (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2209

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

Rule 2210

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 2222

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[d/f, Int[F^(a + b/(c + d
*x))/(c + d*x), x], x] - Dist[(d*e - c*f)/f, Int[F^(a + b/(c + d*x))/((c + d*x)*(e + f*x)), x], x] /; FreeQ[{F
, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0]

Rule 2223

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))*((e_.) + (f_.)*(x_))^(m_), x_Symbol] :> Simp[((e + f*x)^(m + 1)*
F^(a + b/(c + d*x)))/(f*(m + 1)), x] + Dist[(b*d*Log[F])/(f*(m + 1)), Int[((e + f*x)^(m + 1)*F^(a + b/(c + d*x
)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && ILtQ[m, -1]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^3} \, dx &=-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {(b d \log (F)) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2 (e+f x)^2} \, dx}{2 f}\\ &=-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {(b d \log (F)) \int \left (\frac {d^2 F^{a+\frac {b}{c+d x}}}{(d e-c f)^2 (c+d x)^2}-\frac {2 d^2 f F^{a+\frac {b}{c+d x}}}{(d e-c f)^3 (c+d x)}+\frac {f^2 F^{a+\frac {b}{c+d x}}}{(d e-c f)^2 (e+f x)^2}+\frac {2 d f^2 F^{a+\frac {b}{c+d x}}}{(d e-c f)^3 (e+f x)}\right ) \, dx}{2 f}\\ &=-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {\left (b d^3 \log (F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x} \, dx}{(d e-c f)^3}-\frac {\left (b d^2 f \log (F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx}{(d e-c f)^3}-\frac {\left (b d^3 \log (F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx}{2 f (d e-c f)^2}-\frac {(b d f \log (F)) \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^2} \, dx}{2 (d e-c f)^2}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^a \text {Ei}\left (\frac {b \log (F)}{c+d x}\right ) \log (F)}{(d e-c f)^3}-\frac {\left (b d^3 \log (F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x} \, dx}{(d e-c f)^3}+\frac {\left (b d^2 \log (F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x) (e+f x)} \, dx}{(d e-c f)^2}+\frac {\left (b^2 d^2 \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2 (e+f x)} \, dx}{2 (d e-c f)^2}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {\left (b d^2 \log (F)\right ) \operatorname {Subst}\left (\int \frac {F^{a-\frac {b f}{d e-c f}+\frac {b d x}{d e-c f}}}{x} \, dx,x,\frac {e+f x}{c+d x}\right )}{(d e-c f)^3}+\frac {\left (b^2 d^2 \log ^2(F)\right ) \int \left (\frac {d F^{a+\frac {b}{c+d x}}}{(d e-c f) (c+d x)^2}-\frac {d f F^{a+\frac {b}{c+d x}}}{(d e-c f)^2 (c+d x)}+\frac {f^2 F^{a+\frac {b}{c+d x}}}{(d e-c f)^2 (e+f x)}\right ) \, dx}{2 (d e-c f)^2}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^3}-\frac {\left (b^2 d^3 f \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x} \, dx}{2 (d e-c f)^4}+\frac {\left (b^2 d^2 f^2 \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{e+f x} \, dx}{2 (d e-c f)^4}+\frac {\left (b^2 d^3 \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x)^2} \, dx}{2 (d e-c f)^3}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {b d^2 F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^3}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^3}+\frac {b^2 d^2 f F^a \text {Ei}\left (\frac {b \log (F)}{c+d x}\right ) \log ^2(F)}{2 (d e-c f)^4}+\frac {\left (b^2 d^3 f \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{c+d x} \, dx}{2 (d e-c f)^4}-\frac {\left (b^2 d^2 f \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{c+d x}}}{(c+d x) (e+f x)} \, dx}{2 (d e-c f)^3}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {b d^2 F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^3}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^3}+\frac {\left (b^2 d^2 f \log ^2(F)\right ) \operatorname {Subst}\left (\int \frac {F^{a-\frac {b f}{d e-c f}+\frac {b d x}{d e-c f}}}{x} \, dx,x,\frac {e+f x}{c+d x}\right )}{2 (d e-c f)^4}\\ &=\frac {d^2 F^{a+\frac {b}{c+d x}}}{2 f (d e-c f)^2}-\frac {F^{a+\frac {b}{c+d x}}}{2 f (e+f x)^2}-\frac {b d^2 F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^3}+\frac {b d F^{a+\frac {b}{c+d x}} \log (F)}{2 (d e-c f)^2 (e+f x)}-\frac {b d^2 F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log (F)}{(d e-c f)^3}+\frac {b^2 d^2 f F^{a-\frac {b f}{d e-c f}} \text {Ei}\left (\frac {b d (e+f x) \log (F)}{(d e-c f) (c+d x)}\right ) \log ^2(F)}{2 (d e-c f)^4}\\ \end {align*}

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Mathematica [F]  time = 0.82, size = 0, normalized size = 0.00 \[ \int \frac {F^{a+\frac {b}{c+d x}}}{(e+f x)^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.47, size = 555, normalized size = 2.08 \[ \frac {{\left ({\left (b^{2} d^{2} f^{3} x^{2} + 2 \, b^{2} d^{2} e f^{2} x + b^{2} d^{2} e^{2} f\right )} \log \relax (F)^{2} - 2 \, {\left (b d^{3} e^{3} - b c d^{2} e^{2} f + {\left (b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x^{2} + 2 \, {\left (b d^{3} e^{2} f - b c d^{2} e f^{2}\right )} x\right )} \log \relax (F)\right )} F^{\frac {a d e - {\left (a c + b\right )} f}{d e - c f}} {\rm Ei}\left (\frac {{\left (b d f x + b d e\right )} \log \relax (F)}{c d e - c^{2} f + {\left (d^{2} e - c d f\right )} x}\right ) + {\left (2 \, c d^{3} e^{3} - 5 \, c^{2} d^{2} e^{2} f + 4 \, c^{3} d e f^{2} - c^{4} f^{3} + {\left (d^{4} e^{2} f - 2 \, c d^{3} e f^{2} + c^{2} d^{2} f^{3}\right )} x^{2} + 2 \, {\left (d^{4} e^{3} - 2 \, c d^{3} e^{2} f + c^{2} d^{2} e f^{2}\right )} x - {\left (b c d^{2} e^{2} f - b c^{2} d e f^{2} + {\left (b d^{3} e f^{2} - b c d^{2} f^{3}\right )} x^{2} + {\left (b d^{3} e^{2} f - b c^{2} d f^{3}\right )} x\right )} \log \relax (F)\right )} F^{\frac {a d x + a c + b}{d x + c}}}{2 \, {\left (d^{4} e^{6} - 4 \, c d^{3} e^{5} f + 6 \, c^{2} d^{2} e^{4} f^{2} - 4 \, c^{3} d e^{3} f^{3} + c^{4} e^{2} f^{4} + {\left (d^{4} e^{4} f^{2} - 4 \, c d^{3} e^{3} f^{3} + 6 \, c^{2} d^{2} e^{2} f^{4} - 4 \, c^{3} d e f^{5} + c^{4} f^{6}\right )} x^{2} + 2 \, {\left (d^{4} e^{5} f - 4 \, c d^{3} e^{4} f^{2} + 6 \, c^{2} d^{2} e^{3} f^{3} - 4 \, c^{3} d e^{2} f^{4} + c^{4} e f^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.19, size = 506, normalized size = 1.90 \[ -\frac {b^{2} d^{2} f \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{2}}{2 \left (c f -d e \right )^{4} \left (-\frac {a c f \ln \relax (F )}{c f -d e}+\frac {a d e \ln \relax (F )}{c f -d e}-\frac {b f \ln \relax (F )}{c f -d e}+a \ln \relax (F )+\frac {b \ln \relax (F )}{d x +c}\right )^{2}}-\frac {b^{2} d^{2} f \,F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )^{2}}{2 \left (c f -d e \right )^{4} \left (-\frac {a c f \ln \relax (F )}{c f -d e}+\frac {a d e \ln \relax (F )}{c f -d e}-\frac {b f \ln \relax (F )}{c f -d e}+a \ln \relax (F )+\frac {b \ln \relax (F )}{d x +c}\right )}-\frac {b^{2} d^{2} f \,F^{\frac {a c f -a d e +b f}{c f -d e}} \Ei \left (1, -a \ln \relax (F )-\frac {b \ln \relax (F )}{d x +c}-\frac {-a c f \ln \relax (F )+a d e \ln \relax (F )-b f \ln \relax (F )}{c f -d e}\right ) \ln \relax (F )^{2}}{2 \left (c f -d e \right )^{4}}-\frac {b \,d^{2} F^{a} F^{\frac {b}{d x +c}} \ln \relax (F )}{\left (c f -d e \right )^{3} \left (-\frac {a c f \ln \relax (F )}{c f -d e}+\frac {a d e \ln \relax (F )}{c f -d e}-\frac {b f \ln \relax (F )}{c f -d e}+a \ln \relax (F )+\frac {b \ln \relax (F )}{d x +c}\right )}-\frac {b \,d^{2} F^{\frac {a c f -a d e +b f}{c f -d e}} \Ei \left (1, -a \ln \relax (F )-\frac {b \ln \relax (F )}{d x +c}-\frac {-a c f \ln \relax (F )+a d e \ln \relax (F )-b f \ln \relax (F )}{c f -d e}\right ) \ln \relax (F )}{\left (c f -d e \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {F^{a+\frac {b}{c+d\,x}}}{{\left (e+f\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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