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

Optimal. Leaf size=134 \[ \frac {\left (a+b x^2\right ) (f x)^{m+1} (b d-a e) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b f (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {e \left (a+b x^2\right ) (f x)^{m+1}}{b f (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 134, normalized size of antiderivative = 1.00, 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, 459, 364} \[ \frac {\left (a+b x^2\right ) (f x)^{m+1} (b d-a e) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b f (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {e \left (a+b x^2\right ) (f x)^{m+1}}{b f (m+1) \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 78, normalized size = 0.58 \[ -\frac {x \left (a+b x^2\right ) (f x)^m \left ((a e-b d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )-a e\right )}{a b (m+1) \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{\sqrt {b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}}, x\right ) \]

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)^(1/2),x, algorithm="fricas")

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right ) \left (f x \right )^{m}}{\sqrt {b^{2} x^{4}+2 a b \,x^{2}+a^{2}}}\, dx \]

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)^(1/2),x)

[Out]

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

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

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)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (f\,x\right )}^m\,\left (e\,x^2+d\right )}{\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)**(1/2),x)

[Out]

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

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