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

Optimal. Leaf size=153 \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+3} (a e+b d)}{f^3 (m+3) \left (a+b x^2\right )}+\frac{a d \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+1}}{f (m+1) \left (a+b x^2\right )}+\frac{b e \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+5}}{f^5 (m+5) \left (a+b x^2\right )} \]

[Out]

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

_______________________________________________________________________________________

Rubi [A]  time = 0.221173, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057 \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+3} (a e+b d)}{f^3 (m+3) \left (a+b x^2\right )}+\frac{a d \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+1}}{f (m+1) \left (a+b x^2\right )}+\frac{b e \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+5}}{f^5 (m+5) \left (a+b x^2\right )} \]

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]

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

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 34.6397, size = 138, normalized size = 0.9 \[ \frac{a d \left (f x\right )^{m + 1} \sqrt{a^{2} + 2 a b x^{2} + b^{2} x^{4}}}{f \left (a + b x^{2}\right ) \left (m + 1\right )} + \frac{b e \left (f x\right )^{m + 5} \sqrt{a^{2} + 2 a b x^{2} + b^{2} x^{4}}}{f^{5} \left (a + b x^{2}\right ) \left (m + 5\right )} + \frac{\left (f x\right )^{m + 3} \left (a e + b d\right ) \sqrt{a^{2} + 2 a b x^{2} + b^{2} x^{4}}}{f^{3} \left (a + b x^{2}\right ) \left (m + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x)**m*(e*x**2+d)*(b**2*x**4+2*a*b*x**2+a**2)**(1/2),x)

[Out]

a*d*(f*x)**(m + 1)*sqrt(a**2 + 2*a*b*x**2 + b**2*x**4)/(f*(a + b*x**2)*(m + 1))
+ b*e*(f*x)**(m + 5)*sqrt(a**2 + 2*a*b*x**2 + b**2*x**4)/(f**5*(a + b*x**2)*(m +
5)) + (f*x)**(m + 3)*(a*e + b*d)*sqrt(a**2 + 2*a*b*x**2 + b**2*x**4)/(f**3*(a +
b*x**2)*(m + 3))

_______________________________________________________________________________________

Mathematica [A]  time = 0.10254, size = 65, normalized size = 0.42 \[ \frac{\sqrt{\left (a+b x^2\right )^2} (f x)^m \left (\frac{x^3 (a e+b d)}{m+3}+\frac{a d x}{m+1}+\frac{b e x^5}{m+5}\right )}{a+b x^2} \]

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]

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

_______________________________________________________________________________________

Maple [A]  time = 0.007, size = 131, normalized size = 0.9 \[{\frac{ \left ( be{m}^{2}{x}^{4}+4\,bem{x}^{4}+ae{m}^{2}{x}^{2}+bd{m}^{2}{x}^{2}+3\,be{x}^{4}+6\,aem{x}^{2}+6\,bdm{x}^{2}+ad{m}^{2}+5\,ae{x}^{2}+5\,bd{x}^{2}+8\,adm+15\,ad \right ) x \left ( fx \right ) ^{m}}{ \left ( 5+m \right ) \left ( 3+m \right ) \left ( 1+m \right ) \left ( b{x}^{2}+a \right ) }\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}} \]

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]

x*(b*e*m^2*x^4+4*b*e*m*x^4+a*e*m^2*x^2+b*d*m^2*x^2+3*b*e*x^4+6*a*e*m*x^2+6*b*d*m
*x^2+a*d*m^2+5*a*e*x^2+5*b*d*x^2+8*a*d*m+15*a*d)*(f*x)^m*((b*x^2+a)^2)^(1/2)/(5+
m)/(3+m)/(1+m)/(b*x^2+a)

_______________________________________________________________________________________

Maxima [A]  time = 0.719245, size = 101, normalized size = 0.66 \[ \frac{{\left (b f^{m}{\left (m + 1\right )} x^{3} + a f^{m}{\left (m + 3\right )} x\right )} d x^{m}}{m^{2} + 4 \, m + 3} + \frac{{\left (b f^{m}{\left (m + 3\right )} x^{5} + a f^{m}{\left (m + 5\right )} x^{3}\right )} e x^{m}}{m^{2} + 8 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x^2 + d)*(f*x)^m,x, algorithm="maxima")

[Out]

(b*f^m*(m + 1)*x^3 + a*f^m*(m + 3)*x)*d*x^m/(m^2 + 4*m + 3) + (b*f^m*(m + 3)*x^5
+ a*f^m*(m + 5)*x^3)*e*x^m/(m^2 + 8*m + 15)

_______________________________________________________________________________________

Fricas [A]  time = 0.285681, size = 127, normalized size = 0.83 \[ \frac{{\left ({\left (b e m^{2} + 4 \, b e m + 3 \, b e\right )} x^{5} +{\left ({\left (b d + a e\right )} m^{2} + 5 \, b d + 5 \, a e + 6 \,{\left (b d + a e\right )} m\right )} x^{3} +{\left (a d m^{2} + 8 \, a d m + 15 \, a d\right )} x\right )} \left (f x\right )^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x^2 + d)*(f*x)^m,x, algorithm="fricas")

[Out]

((b*e*m^2 + 4*b*e*m + 3*b*e)*x^5 + ((b*d + a*e)*m^2 + 5*b*d + 5*a*e + 6*(b*d + a
*e)*m)*x^3 + (a*d*m^2 + 8*a*d*m + 15*a*d)*x)*(f*x)^m/(m^3 + 9*m^2 + 23*m + 15)

_______________________________________________________________________________________

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

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)

_______________________________________________________________________________________

GIAC/XCAS [A]  time = 0.270117, size = 396, normalized size = 2.59 \[ \frac{b m^{2} x^{5} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + 4 \, b m x^{5} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + b d m^{2} x^{3} e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right ) + a m^{2} x^{3} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + 3 \, b x^{5} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + 6 \, b d m x^{3} e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right ) + 6 \, a m x^{3} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + a d m^{2} x e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right ) + 5 \, b d x^{3} e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right ) + 5 \, a x^{3} e^{\left (m{\rm ln}\left (f x\right ) + 1\right )}{\rm sign}\left (b x^{2} + a\right ) + 8 \, a d m x e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right ) + 15 \, a d x e^{\left (m{\rm ln}\left (f x\right )\right )}{\rm sign}\left (b x^{2} + a\right )}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(b^2*x^4 + 2*a*b*x^2 + a^2)*(e*x^2 + d)*(f*x)^m,x, algorithm="giac")

[Out]

(b*m^2*x^5*e^(m*ln(f*x) + 1)*sign(b*x^2 + a) + 4*b*m*x^5*e^(m*ln(f*x) + 1)*sign(
b*x^2 + a) + b*d*m^2*x^3*e^(m*ln(f*x))*sign(b*x^2 + a) + a*m^2*x^3*e^(m*ln(f*x)
+ 1)*sign(b*x^2 + a) + 3*b*x^5*e^(m*ln(f*x) + 1)*sign(b*x^2 + a) + 6*b*d*m*x^3*e
^(m*ln(f*x))*sign(b*x^2 + a) + 6*a*m*x^3*e^(m*ln(f*x) + 1)*sign(b*x^2 + a) + a*d
*m^2*x*e^(m*ln(f*x))*sign(b*x^2 + a) + 5*b*d*x^3*e^(m*ln(f*x))*sign(b*x^2 + a) +
5*a*x^3*e^(m*ln(f*x) + 1)*sign(b*x^2 + a) + 8*a*d*m*x*e^(m*ln(f*x))*sign(b*x^2
+ a) + 15*a*d*x*e^(m*ln(f*x))*sign(b*x^2 + a))/(m^3 + 9*m^2 + 23*m + 15)