3.3.15 \(\int \frac {d+e x^2}{(f x)^{3/2} \sqrt {a+b x^2+c x^4}} \, dx\) [215]

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

[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)))*(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)-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)))*(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/(f*x)^(1/2)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]
time = 0.23, 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 = {1349, 1155, 524} \begin {gather*} \frac {2 e (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^3 \sqrt {a+b x^2+c x^4}}-\frac {2 d \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 {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 {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 1155

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 1349

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

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Mathematica [A]
time = 10.42, size = 356, normalized size = 1.21 \begin {gather*} \frac {2 x \left (-21 d \left (a+b x^2+c x^4\right )+7 (b d+a e) x^2 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+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 )+9 c d x^4 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+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 )\right )}{21 a (f x)^{3/2} \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-21*d*(a + b*x^2 + c*x^4) + 7*(b*d + 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])] + 9*c*d*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])]))/(21*a*(f*x)^(3/2)*Sqrt[a
+ b*x^2 + c*x^4])

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {e \,x^{2}+d}{\left (f x \right )^{\frac {3}{2}} \sqrt {c \,x^{4}+b \,x^{2}+a}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*(x^2*e + d)*sqrt(f*x)/(c*f^2*x^6 + b*f^2*x^4 + a*f^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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