3.3.12 \(\int \frac {(f x)^{3/2} (d+e x^2)}{\sqrt {a+b x^2+c x^4}} \, dx\) [212]

Optimal. Leaf size=297 \[ \frac {2 d (f x)^{5/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{5 f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{9/2} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {9}{4};\frac {1}{2},\frac {1}{2};\frac {13}{4};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{9 f^3 \sqrt {a+b x^2+c x^4}} \]

[Out]

2/5*d*(f*x)^(5/2)*AppellF1(5/4,1/2,1/2,9/4,-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)+2/9*e*
(f*x)^(9/2)*AppellF1(9/4,1/2,1/2,13/4,-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)

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Rubi [A]
time = 0.22, 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 = {1349, 1155, 524} \begin {gather*} \frac {2 d (f x)^{5/2} \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 {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{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 )}{5 f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{9/2} \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 {9}{4};\frac {1}{2},\frac {1}{2};\frac {13}{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 f^3 \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(f*x)^(5/2)*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[5/4, 1/2, 1/2, 9/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(5*f*Sqrt[a +
b*x^2 + c*x^4]) + (2*e*(f*x)^(9/2)*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[9/4, 1/2, 1/2, 13/4, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^2)/(b + Sqrt[b^2 - 4*a
*c])])/(9*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 {(f x)^{3/2} \left (d+e x^2\right )}{\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 (f x)^{7/2}}{f^2 \sqrt {a+b x^2+c x^4}}\right ) \, dx\\ &=d \int \frac {(f x)^{3/2}}{\sqrt {a+b x^2+c x^4}} \, dx+\frac {e \int \frac {(f x)^{7/2}}{\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 {(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 {(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}}}} \, dx}{f^2 \sqrt {a+b x^2+c x^4}}\\ &=\frac {2 d (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}}} F_1\left (\frac {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{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 )}{5 f \sqrt {a+b x^2+c x^4}}+\frac {2 e (f x)^{9/2} \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 {9}{4};\frac {1}{2},\frac {1}{2};\frac {13}{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 f^3 \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [A]
time = 10.40, size = 354, normalized size = 1.19 \begin {gather*} \frac {2 f \sqrt {f x} \left (5 e \left (a+b x^2+c x^4\right )-5 a e \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 {1}{4};\frac {1}{2},\frac {1}{2};\frac {5}{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 )+(5 c d-3 b 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 {5}{4};\frac {1}{2},\frac {1}{2};\frac {9}{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 )}{25 c \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*f*Sqrt[f*x]*(5*e*(a + b*x^2 + c*x^4) - 5*a*e*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[1/4, 1/2, 1/2, 5/4, (-2*c*x^2)/(b +
Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + (5*c*d - 3*b*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[5/4,
1/2, 1/2, 9/4, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(25*c*Sqrt[a + b*x^2
+ c*x^4])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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