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

Optimal. Leaf size=154 \[ \frac{\left (a+b x^2\right ) (f x)^{m+1} (a e (m+1)+b d (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b f (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(f x)^{m+1} (b d-a e)}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

((b*d - a*e)*(f*x)^(1 + m))/(4*a*b*f*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((b*d*(3 - m) + a*e*(1 + m
))*(f*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*f*(1 + m)*Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.119163, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {1250, 457, 364} \[ \frac{\left (a+b x^2\right ) (f x)^{m+1} (a e (m+1)+b d (3-m)) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{4 a^3 b f (m+1) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(f x)^{m+1} (b d-a e)}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)*(f*x)^(1 + m))/(4*a*b*f*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + ((b*d*(3 - m) + a*e*(1 + m
))*(f*x)^(1 + m)*(a + b*x^2)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(4*a^3*b*f*(1 + m)*Sqrt
[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(f x)^m \left (d+e x^2\right )}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac{(f x)^m \left (d+e x^2\right )}{\left (a b+b^2 x^2\right )^3} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(b d-a e) (f x)^{1+m}}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left ((b d (3-m)+a e (1+m)) \left (a b+b^2 x^2\right )\right ) \int \frac{(f x)^m}{\left (a b+b^2 x^2\right )^2} \, dx}{4 a \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(b d-a e) (f x)^{1+m}}{4 a b f \left (a+b x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{(b d (3-m)+a e (1+m)) (f x)^{1+m} \left (a+b x^2\right ) \, _2F_1\left (2,\frac{1+m}{2};\frac{3+m}{2};-\frac{b x^2}{a}\right )}{4 a^3 b f (1+m) \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0844478, size = 101, normalized size = 0.66 \[ \frac{x \left (a+b x^2\right ) (f x)^m \left ((b d-a e) \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+a e \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^3 b (m+1) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(f*x)^m*(a + b*x^2)*(a*e*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + (b*d - a*e)*Hypergeomet
ric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^3*b*(1 + m)*Sqrt[(a + b*x^2)^2])

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx \right ) ^{m} \left ( e{x}^{2}+d \right ) \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \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)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

int((f*x)^m*(e*x^2+d)/(b^2*x^4+2*a*b*x^2+a^2)^(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 (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\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)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x^2 + d)*(f*x)^m/(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*
x^2 + a^4), 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 (\left (a + b x^{2}\right )^{2}\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)/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((f*x)**m*(d + e*x**2)/((a + b*x**2)**2)**(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 (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\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)/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

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