3.6.44 \(\int \frac {(a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}})^3}{d f+(e f+d g) x+e g x^2} \, dx\) [544]

Optimal. Leaf size=154 \[ \frac {6 a^2 b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {6 a b^2 \text {Ei}\left (\frac {2 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 b^3 \text {Ei}\left (\frac {3 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a^3 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.060, Rules used = {2328, 2214, 2209} \begin {gather*} \frac {2 a^3 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {6 a^2 b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {6 a 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 b^3 \text {Ei}\left (\frac {3 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*F^((c*Sqrt[d + e*x])/Sqrt[f + g*x]))^3/(d*f + (e*f + d*g)*x + e*g*x^2),x]

[Out]

(6*a^2*b*ExpIntegralEi[(c*Sqrt[d + e*x]*Log[F])/Sqrt[f + g*x]])/(e*f - d*g) + (6*a*b^2*ExpIntegralEi[(2*c*Sqrt
[d + e*x]*Log[F])/Sqrt[f + g*x]])/(e*f - d*g) + (2*b^3*ExpIntegralEi[(3*c*Sqrt[d + e*x]*Log[F])/Sqrt[f + g*x]]
)/(e*f - d*g) + (2*a^3*Log[Sqrt[d + e*x]/Sqrt[f + g*x]])/(e*f - d*g)

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 2214

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 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*} \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^3}{d f+(e f+d g) x+e g x^2} \, dx &=\frac {2 \text {Subst}\left (\int \frac {\left (a+b F^{c x}\right )^3}{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^3}{x}+\frac {3 a^2 b F^{c x}}{x}+\frac {3 a b^2 F^{2 c x}}{x}+\frac {b^3 F^{3 c x}}{x}\right ) \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}\\ &=\frac {2 a^3 \log \left (\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {\left (6 a^2 b\right ) \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 {\left (6 a b^2\right ) \text {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 {\left (2 b^3\right ) \text {Subst}\left (\int \frac {F^{3 c x}}{x} \, dx,x,\frac {\sqrt {d+e x}}{\sqrt {f+g x}}\right )}{e f-d g}\\ &=\frac {6 a^2 b \text {Ei}\left (\frac {c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {6 a 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 b^3 \text {Ei}\left (\frac {3 c \sqrt {d+e x} \log (F)}{\sqrt {f+g x}}\right )}{e f-d g}+\frac {2 a^3 \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 = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b F^{\frac {c \sqrt {d+e x}}{\sqrt {f+g x}}}\right )^3}{d f+(e f+d g) x+e g x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(a + b*F^((c*Sqrt[d + e*x])/Sqrt[f + g*x]))^3/(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]))^3/(d*f + (e*f + d*g)*x + e*g*x^2), x]

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \,F^{\frac {c \sqrt {e x +d}}{\sqrt {g x +f}}}\right )^{3}}{d f +\left (d g +e f \right ) x +e g \,x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*F^(c*(e*x+d)^(1/2)/(g*x+f)^(1/2)))^3/(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)))^3/(d*f+(d*g+e*f)*x+e*g*x^2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)))^3/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="maxima")

[Out]

a^3*(log(e*x + d)/(e*f - d*g) - log(g*x + f)/(e*f - d*g)) + b^3*integrate(F^(3*sqrt(e*x + d)*c/sqrt(g*x + f))/
(e*g*x^2 + d*f + (e*f + d*g)*x), x) + 3*a*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) + 3*a^2*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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)))^3/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="fricas")

[Out]

integral((3*F^(sqrt(x*e + d)*c/sqrt(g*x + f))*a^2*b + 3*F^(2*sqrt(x*e + d)*c/sqrt(g*x + f))*a*b^2 + F^(3*sqrt(
x*e + d)*c/sqrt(g*x + f))*b^3 + a^3)/(d*g*x + d*f + (g*x^2 + f*x)*e), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (F^{\frac {c \sqrt {d + e x}}{\sqrt {f + g x}}} b + a\right )^{3}}{\left (d + e x\right ) \left (f + g x\right )}\, dx \end {gather*}

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)))**3/(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)**3/((d + e*x)*(f + g*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)))^3/(d*f+(d*g+e*f)*x+e*g*x^2),x, algorithm="giac")

[Out]

integrate((F^(sqrt(x*e + d)*c/sqrt(g*x + f))*b + a)^3/(g*x^2*e + d*f + (d*g + f*e)*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+F^{\frac {c\,\sqrt {d+e\,x}}{\sqrt {f+g\,x}}}\,b\right )}^3}{e\,g\,x^2+\left (d\,g+e\,f\right )\,x+d\,f} \,d x \end {gather*}

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)^3/(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)^3/(d*f + x*(d*g + e*f) + e*g*x^2), x)

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