3.214 \(\int \frac{d+e x^2}{\sqrt{f x} \sqrt{a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=295 \[ \frac{2 d \sqrt{f x} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{4};\frac{1}{2},\frac{1}{2};\frac{5}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{f \sqrt{a+b x^2+c x^4}}+\frac{2 e (f x)^{5/2} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{5}{4};\frac{1}{2},\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 f^3 \sqrt{a+b x^2+c x^4}} \]

[Out]

(2*d*Sqrt[f*x]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF
1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[a + b*x
^2 + c*x^4]) + (2*e*(f*x)^(5/2)*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 -
4*a*c])]*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]
)/(5*f^3*Sqrt[a + b*x^2 + c*x^4])

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Rubi [A]  time = 0.325269, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {1335, 1141, 510} \[ \frac{2 d \sqrt{f x} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{1}{4};\frac{1}{2},\frac{1}{2};\frac{5}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{f \sqrt{a+b x^2+c x^4}}+\frac{2 e (f x)^{5/2} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{5}{4};\frac{1}{2},\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 f^3 \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)/(Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

(2*d*Sqrt[f*x]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*AppellF
1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[a + b*x
^2 + c*x^4]) + (2*e*(f*x)^(5/2)*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 -
4*a*c])]*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]
)/(5*f^3*Sqrt[a + b*x^2 + c*x^4])

Rule 1335

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0] || IntegersQ[m, q])

Rule 1141

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/((1 + (2*c*x^2)/(b + Rt[b^2 - 4*a*c, 2]))^FracPart[p]*(1 + (2*c*x^2)/(b - Rt[b^2 - 4*a*c,
2]))^FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c
]))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{d+e x^2}{\sqrt{f x} \sqrt{a+b x^2+c x^4}} \, dx &=\int \left (\frac{d}{\sqrt{f x} \sqrt{a+b x^2+c x^4}}+\frac{e (f x)^{3/2}}{f^2 \sqrt{a+b x^2+c x^4}}\right ) \, dx\\ &=d \int \frac{1}{\sqrt{f x} \sqrt{a+b x^2+c x^4}} \, dx+\frac{e \int \frac{(f x)^{3/2}}{\sqrt{a+b x^2+c x^4}} \, dx}{f^2}\\ &=\frac{\left (d \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}\right ) \int \frac{1}{\sqrt{f x} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx}{\sqrt{a+b x^2+c x^4}}+\frac{\left (e \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}\right ) \int \frac{(f x)^{3/2}}{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx}{f^2 \sqrt{a+b x^2+c x^4}}\\ &=\frac{2 d \sqrt{f x} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{1}{4};\frac{1}{2},\frac{1}{2};\frac{5}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{f \sqrt{a+b x^2+c x^4}}+\frac{2 e (f x)^{5/2} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}} F_1\left (\frac{5}{4};\frac{1}{2},\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{5 f^3 \sqrt{a+b x^2+c x^4}}\\ \end{align*}

Mathematica [A]  time = 0.222462, size = 241, normalized size = 0.82 \[ \frac{2 \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \left (5 d x F_1\left (\frac{1}{4};\frac{1}{2},\frac{1}{2};\frac{5}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+e x^3 F_1\left (\frac{5}{4};\frac{1}{2},\frac{1}{2};\frac{9}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )\right )}{5 \sqrt{f x} \sqrt{a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x^2)/(Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

(2*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b +
Sqrt[b^2 - 4*a*c])]*(5*d*x*AppellF1[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + S
qrt[b^2 - 4*a*c])] + e*x^3*AppellF1[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sq
rt[b^2 - 4*a*c])]))/(5*Sqrt[f*x]*Sqrt[a + b*x^2 + c*x^4])

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Maple [F]  time = 0.022, size = 0, normalized size = 0. \begin{align*} \int{(e{x}^{2}+d){\frac{1}{\sqrt{fx}}}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x)

[Out]

int((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x^{2} + d}{\sqrt{c x^{4} + b x^{2} + a} \sqrt{f x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)/(sqrt(c*x^4 + b*x^2 + a)*sqrt(f*x)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )} \sqrt{f x}}{c f x^{5} + b f x^{3} + a f x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)*sqrt(f*x)/(c*f*x^5 + b*f*x^3 + a*f*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x^{2}}{\sqrt{f x} \sqrt{a + b x^{2} + c x^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)/(f*x)**(1/2)/(c*x**4+b*x**2+a)**(1/2),x)

[Out]

Integral((d + e*x**2)/(sqrt(f*x)*sqrt(a + b*x**2 + c*x**4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e x^{2} + d}{\sqrt{c x^{4} + b x^{2} + a} \sqrt{f x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)/(sqrt(c*x^4 + b*x^2 + a)*sqrt(f*x)), x)