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

Optimal. Leaf size=112 \[ \frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {4 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {2290, 2183, 2178} \[ \frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {4 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

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 {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {\left (a+b F^{c x}\right )^2}{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^2}{x}+\frac {2 a b F^{c x}}{x}+\frac {b^2 F^{2 c x}}{x}\right ) \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}\\ &=\frac {2 a^2 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {(4 a 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 {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {F^{2 c x}}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}\\ &=\frac {4 a b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a^2 \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 = 1.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^2}{d f+(e f+d g) x+e g x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

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fricas [F]  time = 130.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, F^{\frac {\sqrt {e x + d} c}{\sqrt {g x + f}}} a b + F^{\frac {2 \, \sqrt {e x + d} c}{\sqrt {g x + f}}} b^{2} + a^{2}}{e g x^{2} + d f + {\left (e f + d g\right )} x}, x\right ) \]

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)))^2/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="fricas")

[Out]

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

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)))^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)^2/(e*g*x^2 + d*f + (e*f + d*g)*x), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,F^{\frac {\sqrt {e x +d}\, c}{\sqrt {g x +f}}}+a \right )^{2}}{e g \,x^{2}+d f +\left (d g +e f \right ) x}\, dx \]

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

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)))^2/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="maxima")

[Out]

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

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

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)))**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)**2/((d + e*x)*(f + g*x)), x)

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