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

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 = 0.0762581, 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, Rules used = {1250, 448} \[ \frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+3} (a e+b d)}{f^3 (m+3) \left (a+b x^2\right )}+\frac{a d \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+1}}{f (m+1) \left (a+b x^2\right )}+\frac{b e \sqrt{a^2+2 a b x^2+b^2 x^4} (f x)^{m+5}}{f^5 (m+5) \left (a+b x^2\right )} \]

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))

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 448

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

Rubi steps

\begin{align*} \int (f x)^m \left (d+e x^2\right ) \sqrt{a^2+2 a b x^2+b^2 x^4} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int (f x)^m \left (a b+b^2 x^2\right ) \left (d+e x^2\right ) \, dx}{a b+b^2 x^2}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \left (a b d (f x)^m+\frac{b (b d+a e) (f x)^{2+m}}{f^2}+\frac{b^2 e (f x)^{4+m}}{f^4}\right ) \, dx}{a b+b^2 x^2}\\ &=\frac{a d (f x)^{1+m} \sqrt{a^2+2 a b x^2+b^2 x^4}}{f (1+m) \left (a+b x^2\right )}+\frac{(b d+a e) (f x)^{3+m} \sqrt{a^2+2 a b x^2+b^2 x^4}}{f^3 (3+m) \left (a+b x^2\right )}+\frac{b e (f x)^{5+m} \sqrt{a^2+2 a b x^2+b^2 x^4}}{f^5 (5+m) \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0527489, size = 86, normalized size = 0.56 \[ \frac{x \sqrt{\left (a+b x^2\right )^2} (f x)^m \left (a (m+5) \left (d (m+3)+e (m+1) x^2\right )+b (m+1) x^2 \left (d (m+5)+e (m+3) x^2\right )\right )}{(m+1) (m+3) (m+5) \left (a+b x^2\right )} \]

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

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 131, normalized size = 0.9 \begin{align*}{\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}}} \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)^(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 = 1.01512, size = 101, normalized size = 0.66 \begin{align*} \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} \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)^(1/2),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 = 1.58691, size = 216, normalized size = 1.41 \begin{align*} \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} \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)^(1/2),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. \begin{align*} \int \left (f x\right )^{m} \left (d + e x^{2}\right ) \sqrt{\left (a + b x^{2}\right )^{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)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.16549, size = 363, normalized size = 2.37 \begin{align*} \frac{\left (f x\right )^{m} b m^{2} x^{5} e \mathrm{sgn}\left (b x^{2} + a\right ) + 4 \, \left (f x\right )^{m} b m x^{5} e \mathrm{sgn}\left (b x^{2} + a\right ) + \left (f x\right )^{m} b d m^{2} x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + \left (f x\right )^{m} a m^{2} x^{3} e \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, \left (f x\right )^{m} b x^{5} e \mathrm{sgn}\left (b x^{2} + a\right ) + 6 \, \left (f x\right )^{m} b d m x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 6 \, \left (f x\right )^{m} a m x^{3} e \mathrm{sgn}\left (b x^{2} + a\right ) + \left (f x\right )^{m} a d m^{2} x \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, \left (f x\right )^{m} b d x^{3} \mathrm{sgn}\left (b x^{2} + a\right ) + 5 \, \left (f x\right )^{m} a x^{3} e \mathrm{sgn}\left (b x^{2} + a\right ) + 8 \, \left (f x\right )^{m} a d m x \mathrm{sgn}\left (b x^{2} + a\right ) + 15 \, \left (f x\right )^{m} a d x \mathrm{sgn}\left (b x^{2} + a\right )}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \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)^(1/2),x, algorithm="giac")

[Out]

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