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

Optimal. Leaf size=297 \[ \frac {2 d (f x)^{3/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {3}{4};\frac {1}{2},\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{7/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {7}{4};\frac {1}{2},\frac {1}{2};\frac {11}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 f^3 \sqrt {a+b x^2+c x^4}} \]

[Out]

2/3*d*(f*x)^(3/2)*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)/f/(c*x^4+b*x^2+a)^(1/2)+2/7*e*
(f*x)^(7/2)*AppellF1(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)/f^3/(c*x^4+b*x^2+a)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 297, normalized size of antiderivative = 1.00, 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} \[ \frac {2 d (f x)^{3/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {3}{4};\frac {1}{2},\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{7/2} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {7}{4};\frac {1}{2},\frac {1}{2};\frac {11}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{7 f^3 \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(f*x)^(3/2)*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])]*Appel
lF1[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*f*Sqrt[a +
b*x^2 + c*x^4]) + (2*e*(f*x)^(7/2)*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[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*f^3*Sqrt[a + b*x^2 + c*x^4])

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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 5.14, size = 242, normalized size = 0.81 \[ \frac {2 \sqrt {f x} \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \left (7 d x F_1\left (\frac {3}{4};\frac {1}{2},\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )+3 e x^3 F_1\left (\frac {7}{4};\frac {1}{2},\frac {1}{2};\frac {11}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )\right )}{21 \sqrt {a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[f*x]*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*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^3*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[a + b*x^2 + c*x^4])

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )} \sqrt {f x}}{\sqrt {c x^{4} + b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {f x}\, \left (e \,x^{2}+d \right )}{\sqrt {c \,x^{4}+b \,x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {f\,x}\,\left (e\,x^2+d\right )}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________