3.1.89 \(\int (f x)^m (d+e x^2) \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [89]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 = {1264, 459} \begin {gather*} \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 )} \end {gather*}

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 459

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]

Rule 1264

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 (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*}

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Mathematica [A]
time = 0.06, size = 86, normalized size = 0.56 \begin {gather*} \frac {x (f x)^m \sqrt {\left (a+b x^2\right )^2} \left (a (5+m) \left (d (3+m)+e (1+m) x^2\right )+b (1+m) x^2 \left (d (5+m)+e (3+m) x^2\right )\right )}{(1+m) (3+m) (5+m) \left (a+b x^2\right )} \end {gather*}

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))

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Maple [A]
time = 0.02, size = 131, normalized size = 0.86

method	result	size
gosper	\(\frac {x \left (b e \,m^{2} x^{4}+4 b e m \,x^{4}+a e \,m^{2} x^{2}+b d \,m^{2} x^{2}+3 b e \,x^{4}+6 a e m \,x^{2}+6 b d m \,x^{2}+a d \,m^{2}+5 a e \,x^{2}+5 b d \,x^{2}+8 a d m +15 a d \right ) \left (f x \right )^{m} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{\left (5+m \right ) \left (3+m \right ) \left (1+m \right ) \left (b \,x^{2}+a \right )}\)	\(131\)
risch	\(\frac {x \left (b e \,m^{2} x^{4}+4 b e m \,x^{4}+a e \,m^{2} x^{2}+b d \,m^{2} x^{2}+3 b e \,x^{4}+6 a e m \,x^{2}+6 b d m \,x^{2}+a d \,m^{2}+5 a e \,x^{2}+5 b d \,x^{2}+8 a d m +15 a d \right ) \left (f x \right )^{m} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{\left (5+m \right ) \left (3+m \right ) \left (1+m \right ) \left (b \,x^{2}+a \right )}\)	\(131\)

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,method=_RETURNVERBOSE)

[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)

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Maxima [A]
time = 0.29, size = 78, normalized size = 0.51 \begin {gather*} \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^{\left (m \log \left (x\right ) + 1\right )}}{m^{2} + 8 \, m + 15} \end {gather*}

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^(m*log
(x) + 1)/(m^2 + 8*m + 15)

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Fricas [A]
time = 0.39, size = 98, normalized size = 0.64 \begin {gather*} \frac {{\left ({\left (b d m^{2} + 6 \, b d m + 5 \, b d\right )} x^{3} + {\left (a d m^{2} + 8 \, a d m + 15 \, a d\right )} x + {\left ({\left (b m^{2} + 4 \, b m + 3 \, b\right )} x^{5} + {\left (a m^{2} + 6 \, a m + 5 \, a\right )} x^{3}\right )} e\right )} \left (f x\right )^{m}}{m^{3} + 9 \, m^{2} + 23 \, m + 15} \end {gather*}

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

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (122) = 244\).
time = 2.79, size = 269, normalized size = 1.76 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\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 \end {gather*}

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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