\(\int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx\) [546]

Optimal result

Integrand size = 48, antiderivative size = 70 \[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx=\frac {2 b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2328, 14, 2209} \[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx=\frac {2 a \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b \operatorname {ExpIntegralEi}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]

[In]

Int[(a + b*F^((c*Sqrt[d + e*x])/Sqrt[f + g*x]))/(d*f + (e*f + d*g)*x + e*g*x^2),x]

[Out]

(2*b*ExpIntegralEi[(c*Sqrt[d + e*x]*Log[F])/Sqrt[f + g*x]])/(e*f - d*g) + (2*a*Log[Sqrt[d + e*x]/Sqrt[f + g*x]
])/(e*f - d*g)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2209

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

Rule 2328

Int[((a_.) + (b_.)*(F_)^(((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]))^(n_.)/((A_.) + (B_.)*(x_)
+ (C_.)*(x_)^2), x_Symbol] :> Dist[2*e*(g/(C*(e*f - d*g))), Subst[Int[(a + b*F^(c*x))^n/x, x], x, Sqrt[d + e*
x]/Sqrt[f + g*x]], x] /; FreeQ[{a, b, c, d, e, f, g, A, B, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[B*e*g - C
*(e*f + d*g), 0] && IGtQ[n, 0]

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

Mathematica [F]

\[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx \]

[In]

Integrate[(a + b*F^((c*Sqrt[d + e*x])/Sqrt[f + g*x]))/(d*f + (e*f + d*g)*x + e*g*x^2),x]

[Out]

Integrate[(a + b*F^((c*Sqrt[d + e*x])/Sqrt[f + g*x]))/(d*f + (e*f + d*g)*x + e*g*x^2), x]

Maple [F]

\[\int \frac {a +b \,F^{\frac {c \sqrt {e x +d}}{\sqrt {g x +f}}}}{d f +\left (d g +e f \right ) x +e g \,x^{2}}d x\]

[In]

int((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))/(d*f+(d*g+e*f)*x+e*g*x^2),x)

[Out]

int((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))/(d*f+(d*g+e*f)*x+e*g*x^2),x)

Fricas [F]

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

[In]

integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="fricas")

[Out]

integral((F^(sqrt(e*x + d)*c/sqrt(g*x + f))*b + a)/(e*g*x^2 + d*f + (e*f + d*g)*x), x)

Sympy [F]

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

[In]

integrate((a+b*F**(c*(e*x+d)**(1/2)/(g*x+f)**(1/2)))/(d*f+(d*g+e*f)*x+e*g*x**2),x)

[Out]

Integral((F**(c*sqrt(d + e*x)/sqrt(f + g*x))*b + a)/((d + e*x)*(f + g*x)), x)

Maxima [F]

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

[In]

integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="maxima")

[Out]

a*(log(e*x + d)/(e*f - d*g) - log(g*x + f)/(e*f - d*g)) + b*integrate(F^(sqrt(e*x + d)*c/sqrt(g*x + f))/(e*g*x
^2 + d*f + (e*f + d*g)*x), x)

Giac [F]

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

[In]

integrate((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="giac")

[Out]

integrate((F^(sqrt(e*x + d)*c/sqrt(g*x + f))*b + a)/(e*g*x^2 + d*f + (e*f + d*g)*x), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx=\int \frac {a+F^{\frac {c\,\sqrt {d+e\,x}}{\sqrt {f+g\,x}}}\,b}{e\,g\,x^2+\left (d\,g+e\,f\right )\,x+d\,f} \,d x \]

[In]

int((a + F^((c*(d + e*x)^(1/2))/(f + g*x)^(1/2))*b)/(d*f + x*(d*g + e*f) + e*g*x^2),x)

[Out]

int((a + F^((c*(d + e*x)^(1/2))/(f + g*x)^(1/2))*b)/(d*f + x*(d*g + e*f) + e*g*x^2), x)