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

Optimal. Leaf size=323 \[ \frac{d (f x)^{m+1} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+1}{2};\frac{3}{2},\frac{3}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{a f (m+1) \sqrt{a+b x^2+c x^4}}+\frac{e (f x)^{m+3} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+3}{2};\frac{3}{2},\frac{3}{2};\frac{m+5}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{a f^3 (m+3) \sqrt{a+b x^2+c x^4}} \]

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

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Rubi [A]  time = 0.388058, antiderivative size = 323, normalized size of antiderivative = 1., 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} \[ \frac{d (f x)^{m+1} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+1}{2};\frac{3}{2},\frac{3}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{a f (m+1) \sqrt{a+b x^2+c x^4}}+\frac{e (f x)^{m+3} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1} F_1\left (\frac{m+3}{2};\frac{3}{2},\frac{3}{2};\frac{m+5}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{a f^3 (m+3) \sqrt{a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

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])

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 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])

Rubi steps

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

Mathematica [A]  time = 0.42826, size = 307, normalized size = 0.95 \[ \frac{x (f x)^m \left (\sqrt{b^2-4 a c}-b-2 c x^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^2}{b-\sqrt{b^2-4 a c}}} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x^2}{\sqrt{b^2-4 a c}+b}\right )^{3/2} \left (d (m+3) F_1\left (\frac{m+1}{2};\frac{3}{2},\frac{3}{2};\frac{m+3}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )+e (m+1) x^2 F_1\left (\frac{m+3}{2};\frac{3}{2},\frac{3}{2};\frac{m+5}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}},\frac{2 c x^2}{\sqrt{b^2-4 a c}-b}\right )\right )}{(m+1) (m+3) \left (\sqrt{b^2-4 a c}-b\right ) \left (a+b x^2+c x^4\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.01, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) \left ( c{x}^{4}+b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

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)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2} + a}{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{c^{2} x^{8} + 2 \, b c x^{6} +{\left (b^{2} + 2 \, a c\right )} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2 + a)*(e*x^2 + d)*(f*x)^m/(c^2*x^8 + 2*b*c*x^6 + (b^2 + 2*a*c)*x^4 + 2*a*b*x^2 + a^
2), x)

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

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)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)^(3/2),x, algorithm="giac")

[Out]

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