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

Optimal. Leaf size=317 \[ \frac {d (f x)^{m+1} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {m+1}{2};\frac {1}{2},\frac {1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{m+3} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {m+3}{2};\frac {1}{2},\frac {1}{2};\frac {m+5}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (m+3) \sqrt {a+b x^2+c x^4}} \]

[Out]

d*(f*x)^(1+m)*AppellF1(1/2+1/2*m,1/2,1/2,3/2+1/2*m,-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/(1+m)/(c*x^4+b*x^2+a
)^(1/2)+e*(f*x)^(3+m)*AppellF1(3/2+1/2*m,1/2,1/2,5/2+1/2*m,-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/(3+m)/(c*x
^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.35, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1335, 1141, 510} \[ \frac {d (f x)^{m+1} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {m+1}{2};\frac {1}{2},\frac {1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f (m+1) \sqrt {a+b x^2+c x^4}}+\frac {e (f x)^{m+3} \sqrt {\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {m+3}{2};\frac {1}{2},\frac {1}{2};\frac {m+5}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{f^3 (m+3) \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.26, size = 267, normalized size = 0.84 \[ \frac {x (f x)^m \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \left (d (m+3) F_1\left (\frac {m+1}{2};\frac {1}{2},\frac {1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )+e (m+1) x^2 F_1\left (\frac {m+3}{2};\frac {1}{2},\frac {1}{2};\frac {m+5}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )\right )}{(m+1) (m+3) \sqrt {a+b x^2+c x^4}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(f*x)^m*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])]*(d*(3 + m)*AppellF1[(1 + m)/2, 1/2, 1/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*
a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^2*AppellF1[(3 + m)/2, 1/2, 1/2, (5 + m)/2, (-2*c*x^2)
/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/((1 + m)*(3 + m)*Sqrt[a + b*x^2 + c*x^4])

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {c x^{4} + b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x)^m*(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 \[ \int \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {c x^{4} + b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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