3.207 \(\int \frac {(d+e x^2) \sqrt {a+b x^2+c x^4}}{(f x)^{3/2}} \, dx\)

Optimal. Leaf size=295 \[ \frac {2 e (f x)^{3/2} \sqrt {a+b x^2+c x^4} F_1\left (\frac {3}{4};-\frac {1}{2},-\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f^3 \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}-\frac {2 d \sqrt {a+b x^2+c x^4} F_1\left (-\frac {1}{4};-\frac {1}{2},-\frac {1}{2};\frac {3}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {f x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.32, antiderivative size = 295, 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} \[ \frac {2 e (f x)^{3/2} \sqrt {a+b x^2+c x^4} F_1\left (\frac {3}{4};-\frac {1}{2},-\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{3 f^3 \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}}-\frac {2 d \sqrt {a+b x^2+c x^4} F_1\left (-\frac {1}{4};-\frac {1}{2},-\frac {1}{2};\frac {3}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f \sqrt {f x} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(f*x)^(3/2),x]

[Out]

(-2*d*Sqrt[a + b*x^2 + c*x^4]*AppellF1[-1/4, -1/2, -1/2, 3/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(
b + Sqrt[b^2 - 4*a*c])])/(f*Sqrt[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])]) + (2*e*(f*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4]*AppellF1[3/4, -1/2, -1/2, 7/4, (-2*c*x^2)/(b - Sqr
t[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(3*f^3*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])])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.86, size = 370, normalized size = 1.25 \[ \frac {x \left (28 x^2 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} (2 a e+7 b d) F_1\left (\frac {3}{4};\frac {1}{2},\frac {1}{2};\frac {7}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )+12 x^4 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} (b e+14 c d) F_1\left (\frac {7}{4};\frac {1}{2},\frac {1}{2};\frac {11}{4};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )-42 \left (7 d-e x^2\right ) \left (a+b x^2+c x^4\right )\right )}{147 (f x)^{3/2} \sqrt {a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4])/(f*x)^(3/2),x]

[Out]

(x*(-42*(7*d - e*x^2)*(a + b*x^2 + c*x^4) + 28*(7*b*d + 2*a*e)*x^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])]*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])] + 12*(14*c*d + b*e)*x^4*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])]*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])])
)/(147*(f*x)^(3/2)*Sqrt[a + b*x^2 + c*x^4])

________________________________________________________________________________________

fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )} \sqrt {f x}}{f^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(1/2)/(f*x)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}{\left (f x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(1/2)/(f*x)^(3/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{\left (f x \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )}}{\left (f x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(c*x^4+b*x^2+a)^(1/2)/(f*x)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (e\,x^2+d\right )\,\sqrt {c\,x^4+b\,x^2+a}}{{\left (f\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x^{2}\right ) \sqrt {a + b x^{2} + c x^{4}}}{\left (f x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(c*x**4+b*x**2+a)**(1/2)/(f*x)**(3/2),x)

[Out]

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

________________________________________________________________________________________