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]

2/3*d*x^2*AppellF1(3/4,1/2,1/2,7/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2
/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(c*x^5+b*x^3+a*x)^(1/2)+2/7*e*x^4*Appe
llF1(7/4,1/2,1/2,11/4,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^2/(b-(-4*a*c+b
^2)^(1/2)))^(1/2)*(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(c*x^5+b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 287, normalized size of antiderivative = 1.00, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 5.12, 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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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{5} + b x^{3} + a x} {\left (e x^{2} + d\right )}}{c x^{4} + b x^{2} + a}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} x}{\sqrt {c x^{5} + b x^{3} + a x}}\,{d x} \]

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)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right ) x}{\sqrt {c \,x^{5}+b \,x^{3}+a x}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} x}{\sqrt {c x^{5} + b x^{3} + a x}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x\,\left (e\,x^2+d\right )}{\sqrt {c\,x^5+b\,x^3+a\,x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (d + e x^{2}\right )}{\sqrt {x \left (a + b x^{2} + c x^{4}\right )}}\, dx \]

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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