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

Optimal. Leaf size=70 \[ \frac {2 a \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]

[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)

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Rubi [A]  time = 0.12, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2290, 14, 2178} \[ \frac {2 a \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]

Antiderivative was successfully verified.

[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 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 2290

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*} \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 \operatorname {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 \operatorname {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) \operatorname {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*}

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Mathematica [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int \frac {a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}}{d f+(e f+d g) x+e g x^2} \, dx \]

Verification is Not applicable to the result.

[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]

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fricas [F]  time = 58.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a}{e g x^{2} + d f + {\left (e f + d g\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} b + a}{e g x^{2} + d f + {\left (e f + d g\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \frac {b \,F^{\frac {\sqrt {e x +d}\, c}{\sqrt {g x +f}}}+a}{e g \,x^{2}+d f +\left (d g +e f \right ) x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {\log \left (e x + d\right )}{e f - d g} - \frac {\log \left (g x + f\right )}{e f - d g}\right )} + b \int \frac {F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}}}{e g x^{2} + d f + {\left (e f + d g\right )} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+F^{\frac {c\,\sqrt {d+e\,x}}{\sqrt {f+g\,x}}}\,b}{e\,g\,x^2+\left (d\,g+e\,f\right )\,x+d\,f} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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