3.122 \(\int \frac{x (d+e x^2)}{\sqrt{a x+b x^3+c x^5}} \, dx\)

Optimal. Leaf size=287 \[ \frac{2 d x^2 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{3}{4};\frac{1}{2},\frac{1}{2};\frac{7}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{3 \sqrt{a x+b x^3+c x^5}}+\frac{2 e x^4 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{7}{4};\frac{1}{2},\frac{1}{2};\frac{11}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{7 \sqrt{a x+b x^3+c x^5}} \]

[Out]

(2*d*x^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[3/4,
1/2, 1/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*Sqrt[a*x + b*x^3 +
c*x^5]) + (2*e*x^4*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])]*Ap
pellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(7*Sqrt[a
*x + b*x^3 + c*x^5])

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Rubi [A]  time = 0.399178, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1954, 1335, 1141, 510} \[ \frac{2 d x^2 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{3}{4};\frac{1}{2},\frac{1}{2};\frac{7}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{3 \sqrt{a x+b x^3+c x^5}}+\frac{2 e x^4 \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{7}{4};\frac{1}{2},\frac{1}{2};\frac{11}{4};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{7 \sqrt{a x+b x^3+c x^5}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(d + e*x^2))/Sqrt[a*x + b*x^3 + c*x^5],x]

[Out]

(2*d*x^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[3/4,
1/2, 1/2, 7/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*Sqrt[a*x + b*x^3 +
c*x^5]) + (2*e*x^4*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])]*Ap
pellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(7*Sqrt[a
*x + b*x^3 + c*x^5])

Rule 1954

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(k_.) + (c_.)*(x_)^(n_.))^(p_)*((A_) + (B_.)*(x_)^(q_)), x_Symbo
l] :> Dist[(a*x^j + b*x^k + c*x^n)^p/(x^(j*p)*(a + b*x^(k - j) + c*x^(2*(k - j)))^p), Int[x^(m + j*p)*(A + B*x
^(k - j))*(a + b*x^(k - j) + c*x^(2*(k - j)))^p, x], x] /; FreeQ[{a, b, c, A, B, j, k, m, p}, x] && EqQ[q, k -
j] && EqQ[n, 2*k - j] &&  !IntegerQ[p] && PosQ[k - j]

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{x \left (d+e x^2\right )}{\sqrt{a x+b x^3+c x^5}} \, dx &=\frac{\left (\sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{\sqrt{x} \left (d+e x^2\right )}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (\sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \left (\frac{d \sqrt{x}}{\sqrt{a+b x^2+c x^4}}+\frac{e x^{5/2}}{\sqrt{a+b x^2+c x^4}}\right ) \, dx}{\sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (d \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a x+b x^3+c x^5}}+\frac{\left (e \sqrt{x} \sqrt{a+b x^2+c x^4}\right ) \int \frac{x^{5/2}}{\sqrt{a+b x^2+c x^4}} \, dx}{\sqrt{a x+b x^3+c x^5}}\\ &=\frac{\left (d \sqrt{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}}}\right ) \int \frac{\sqrt{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 x+b x^3+c x^5}}+\frac{\left (e \sqrt{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}}}\right ) \int \frac{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}}}} \, dx}{\sqrt{a x+b x^3+c x^5}}\\ &=\frac{2 d x^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{3}{4};\frac{1}{2},\frac{1}{2};\frac{7}{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 )}{3 \sqrt{a x+b x^3+c x^5}}+\frac{2 e x^4 \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{7}{4};\frac{1}{2},\frac{1}{2};\frac{11}{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 )}{7 \sqrt{a x+b x^3+c x^5}}\\ \end{align*}

Mathematica [A]  time = 5.13468, size = 239, normalized size = 0.83 \[ \frac{2 \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}} \left (7 d x^2 F_1\left (\frac{3}{4};\frac{1}{2},\frac{1}{2};\frac{7}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+3 e x^4 F_1\left (\frac{7}{4};\frac{1}{2},\frac{1}{2};\frac{11}{4};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )\right )}{21 \sqrt{x \left (a+b x^2+c x^4\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(d + e*x^2))/Sqrt[a*x + b*x^3 + c*x^5],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])]*(7*d*x^2*AppellF1[3/4, 1/2, 1/2, 7/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b +
Sqrt[b^2 - 4*a*c])] + 3*e*x^4*AppellF1[7/4, 1/2, 1/2, 11/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b
+ Sqrt[b^2 - 4*a*c])]))/(21*Sqrt[x*(a + b*x^2 + c*x^4)])

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{x \left ( e{x}^{2}+d \right ){\frac{1}{\sqrt{c{x}^{5}+b{x}^{3}+ax}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x)

[Out]

int(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x**2+d)/(c*x**5+b*x**3+a*x)**(1/2),x)

[Out]

Integral(x*(d + e*x**2)/sqrt(x*(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{{\left (e x^{2} + d\right )} x}{\sqrt{c x^{5} + b x^{3} + a x}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(e*x^2+d)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")

[Out]

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