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

Optimal. Leaf size=276 \[ \frac {b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+7} (3 a e+b d)}{f^7 (m+7) \left (a+b x^2\right )}+\frac {3 a b \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+5} (a e+b d)}{f^5 (m+5) \left (a+b x^2\right )}+\frac {a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+3} (a e+3 b d)}{f^3 (m+3) \left (a+b x^2\right )}+\frac {b^3 e \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+9}}{f^9 (m+9) \left (a+b x^2\right )}+\frac {a^3 d \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+1}}{f (m+1) \left (a+b x^2\right )} \]

[Out]

a^3*d*(f*x)^(1+m)*((b*x^2+a)^2)^(1/2)/f/(1+m)/(b*x^2+a)+a^2*(a*e+3*b*d)*(f*x)^(3+m)*((b*x^2+a)^2)^(1/2)/f^3/(3
+m)/(b*x^2+a)+3*a*b*(a*e+b*d)*(f*x)^(5+m)*((b*x^2+a)^2)^(1/2)/f^5/(5+m)/(b*x^2+a)+b^2*(3*a*e+b*d)*(f*x)^(7+m)*
((b*x^2+a)^2)^(1/2)/f^7/(7+m)/(b*x^2+a)+b^3*e*(f*x)^(9+m)*((b*x^2+a)^2)^(1/2)/f^9/(9+m)/(b*x^2+a)

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Rubi [A]  time = 0.15, antiderivative size = 276, 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 = {1250, 448} \[ \frac {a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+3} (a e+3 b d)}{f^3 (m+3) \left (a+b x^2\right )}+\frac {3 a b \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+5} (a e+b d)}{f^5 (m+5) \left (a+b x^2\right )}+\frac {b^2 \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+7} (3 a e+b d)}{f^7 (m+7) \left (a+b x^2\right )}+\frac {a^3 d \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+1}}{f (m+1) \left (a+b x^2\right )}+\frac {b^3 e \sqrt {a^2+2 a b x^2+b^2 x^4} (f x)^{m+9}}{f^9 (m+9) \left (a+b x^2\right )} \]

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]

(a^3*d*(f*x)^(1 + m)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(f*(1 + m)*(a + b*x^2)) + (a^2*(3*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)) + (3*a*b*(b*d + a*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)) + (b^2*(b*d + 3*a*e)*(f*x)^(7 + m)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^
4])/(f^7*(7 + m)*(a + b*x^2)) + (b^3*e*(f*x)^(9 + m)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(f^9*(9 + m)*(a + b*x^2)
)

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]

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 (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 {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (f x)^m \left (a b+b^2 x^2\right )^3 \left (d+e x^2\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^3 b^3 d (f x)^m+\frac {a^2 b^3 (3 b d+a e) (f x)^{2+m}}{f^2}+\frac {3 a b^4 (b d+a e) (f x)^{4+m}}{f^4}+\frac {b^5 (b d+3 a e) (f x)^{6+m}}{f^6}+\frac {b^6 e (f x)^{8+m}}{f^8}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac {a^3 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 {a^2 (3 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 {3 a b (b d+a 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 )}+\frac {b^2 (b d+3 a e) (f x)^{7+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{f^7 (7+m) \left (a+b x^2\right )}+\frac {b^3 e (f x)^{9+m} \sqrt {a^2+2 a b x^2+b^2 x^4}}{f^9 (9+m) \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 112, normalized size = 0.41 \[ \frac {x \left (\left (a+b x^2\right )^2\right )^{3/2} (f x)^m \left (\frac {a^3 d}{m+1}+\frac {a^2 x^2 (a e+3 b d)}{m+3}+\frac {b^2 x^6 (3 a e+b d)}{m+7}+\frac {3 a b x^4 (a e+b d)}{m+5}+\frac {b^3 e x^8}{m+9}\right )}{\left (a+b x^2\right )^3} \]

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)^2)^(3/2)*((a^3*d)/(1 + m) + (a^2*(3*b*d + a*e)*x^2)/(3 + m) + (3*a*b*(b*d + a*e)*x^4)/
(5 + m) + (b^2*(b*d + 3*a*e)*x^6)/(7 + m) + (b^3*e*x^8)/(9 + m)))/(a + b*x^2)^3

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fricas [A]  time = 1.03, size = 381, normalized size = 1.38 \[ \frac {{\left ({\left (b^{3} e m^{4} + 16 \, b^{3} e m^{3} + 86 \, b^{3} e m^{2} + 176 \, b^{3} e m + 105 \, b^{3} e\right )} x^{9} + {\left ({\left (b^{3} d + 3 \, a b^{2} e\right )} m^{4} + 135 \, b^{3} d + 405 \, a b^{2} e + 18 \, {\left (b^{3} d + 3 \, a b^{2} e\right )} m^{3} + 104 \, {\left (b^{3} d + 3 \, a b^{2} e\right )} m^{2} + 222 \, {\left (b^{3} d + 3 \, a b^{2} e\right )} m\right )} x^{7} + 3 \, {\left ({\left (a b^{2} d + a^{2} b e\right )} m^{4} + 189 \, a b^{2} d + 189 \, a^{2} b e + 20 \, {\left (a b^{2} d + a^{2} b e\right )} m^{3} + 130 \, {\left (a b^{2} d + a^{2} b e\right )} m^{2} + 300 \, {\left (a b^{2} d + a^{2} b e\right )} m\right )} x^{5} + {\left ({\left (3 \, a^{2} b d + a^{3} e\right )} m^{4} + 945 \, a^{2} b d + 315 \, a^{3} e + 22 \, {\left (3 \, a^{2} b d + a^{3} e\right )} m^{3} + 164 \, {\left (3 \, a^{2} b d + a^{3} e\right )} m^{2} + 458 \, {\left (3 \, a^{2} b d + a^{3} e\right )} m\right )} x^{3} + {\left (a^{3} d m^{4} + 24 \, a^{3} d m^{3} + 206 \, a^{3} d m^{2} + 744 \, a^{3} d m + 945 \, a^{3} d\right )} x\right )} \left (f x\right )^{m}}{m^{5} + 25 \, m^{4} + 230 \, m^{3} + 950 \, m^{2} + 1689 \, m + 945} \]

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]

((b^3*e*m^4 + 16*b^3*e*m^3 + 86*b^3*e*m^2 + 176*b^3*e*m + 105*b^3*e)*x^9 + ((b^3*d + 3*a*b^2*e)*m^4 + 135*b^3*
d + 405*a*b^2*e + 18*(b^3*d + 3*a*b^2*e)*m^3 + 104*(b^3*d + 3*a*b^2*e)*m^2 + 222*(b^3*d + 3*a*b^2*e)*m)*x^7 +
3*((a*b^2*d + a^2*b*e)*m^4 + 189*a*b^2*d + 189*a^2*b*e + 20*(a*b^2*d + a^2*b*e)*m^3 + 130*(a*b^2*d + a^2*b*e)*
m^2 + 300*(a*b^2*d + a^2*b*e)*m)*x^5 + ((3*a^2*b*d + a^3*e)*m^4 + 945*a^2*b*d + 315*a^3*e + 22*(3*a^2*b*d + a^
3*e)*m^3 + 164*(3*a^2*b*d + a^3*e)*m^2 + 458*(3*a^2*b*d + a^3*e)*m)*x^3 + (a^3*d*m^4 + 24*a^3*d*m^3 + 206*a^3*
d*m^2 + 744*a^3*d*m + 945*a^3*d)*x)*(f*x)^m/(m^5 + 25*m^4 + 230*m^3 + 950*m^2 + 1689*m + 945)

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giac [B]  time = 0.52, size = 1013, normalized size = 3.67 \[ \text {result too large to display} \]

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]

((f*x)^m*b^3*m^4*x^9*e*sgn(b*x^2 + a) + 16*(f*x)^m*b^3*m^3*x^9*e*sgn(b*x^2 + a) + (f*x)^m*b^3*d*m^4*x^7*sgn(b*
x^2 + a) + 3*(f*x)^m*a*b^2*m^4*x^7*e*sgn(b*x^2 + a) + 86*(f*x)^m*b^3*m^2*x^9*e*sgn(b*x^2 + a) + 18*(f*x)^m*b^3
*d*m^3*x^7*sgn(b*x^2 + a) + 54*(f*x)^m*a*b^2*m^3*x^7*e*sgn(b*x^2 + a) + 176*(f*x)^m*b^3*m*x^9*e*sgn(b*x^2 + a)
+ 3*(f*x)^m*a*b^2*d*m^4*x^5*sgn(b*x^2 + a) + 104*(f*x)^m*b^3*d*m^2*x^7*sgn(b*x^2 + a) + 3*(f*x)^m*a^2*b*m^4*x
^5*e*sgn(b*x^2 + a) + 312*(f*x)^m*a*b^2*m^2*x^7*e*sgn(b*x^2 + a) + 105*(f*x)^m*b^3*x^9*e*sgn(b*x^2 + a) + 60*(
f*x)^m*a*b^2*d*m^3*x^5*sgn(b*x^2 + a) + 222*(f*x)^m*b^3*d*m*x^7*sgn(b*x^2 + a) + 60*(f*x)^m*a^2*b*m^3*x^5*e*sg
n(b*x^2 + a) + 666*(f*x)^m*a*b^2*m*x^7*e*sgn(b*x^2 + a) + 3*(f*x)^m*a^2*b*d*m^4*x^3*sgn(b*x^2 + a) + 390*(f*x)
^m*a*b^2*d*m^2*x^5*sgn(b*x^2 + a) + 135*(f*x)^m*b^3*d*x^7*sgn(b*x^2 + a) + (f*x)^m*a^3*m^4*x^3*e*sgn(b*x^2 + a
) + 390*(f*x)^m*a^2*b*m^2*x^5*e*sgn(b*x^2 + a) + 405*(f*x)^m*a*b^2*x^7*e*sgn(b*x^2 + a) + 66*(f*x)^m*a^2*b*d*m
^3*x^3*sgn(b*x^2 + a) + 900*(f*x)^m*a*b^2*d*m*x^5*sgn(b*x^2 + a) + 22*(f*x)^m*a^3*m^3*x^3*e*sgn(b*x^2 + a) + 9
00*(f*x)^m*a^2*b*m*x^5*e*sgn(b*x^2 + a) + (f*x)^m*a^3*d*m^4*x*sgn(b*x^2 + a) + 492*(f*x)^m*a^2*b*d*m^2*x^3*sgn
(b*x^2 + a) + 567*(f*x)^m*a*b^2*d*x^5*sgn(b*x^2 + a) + 164*(f*x)^m*a^3*m^2*x^3*e*sgn(b*x^2 + a) + 567*(f*x)^m*
a^2*b*x^5*e*sgn(b*x^2 + a) + 24*(f*x)^m*a^3*d*m^3*x*sgn(b*x^2 + a) + 1374*(f*x)^m*a^2*b*d*m*x^3*sgn(b*x^2 + a)
+ 458*(f*x)^m*a^3*m*x^3*e*sgn(b*x^2 + a) + 206*(f*x)^m*a^3*d*m^2*x*sgn(b*x^2 + a) + 945*(f*x)^m*a^2*b*d*x^3*s
gn(b*x^2 + a) + 315*(f*x)^m*a^3*x^3*e*sgn(b*x^2 + a) + 744*(f*x)^m*a^3*d*m*x*sgn(b*x^2 + a) + 945*(f*x)^m*a^3*
d*x*sgn(b*x^2 + a))/(m^5 + 25*m^4 + 230*m^3 + 950*m^2 + 1689*m + 945)

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maple [B]  time = 0.01, size = 495, normalized size = 1.79 \[ \frac {\left (b^{3} e \,m^{4} x^{8}+16 b^{3} e \,m^{3} x^{8}+3 a \,b^{2} e \,m^{4} x^{6}+b^{3} d \,m^{4} x^{6}+86 b^{3} e \,m^{2} x^{8}+54 a \,b^{2} e \,m^{3} x^{6}+18 b^{3} d \,m^{3} x^{6}+176 b^{3} e m \,x^{8}+3 a^{2} b e \,m^{4} x^{4}+3 a \,b^{2} d \,m^{4} x^{4}+312 a \,b^{2} e \,m^{2} x^{6}+104 b^{3} d \,m^{2} x^{6}+105 b^{3} e \,x^{8}+60 a^{2} b e \,m^{3} x^{4}+60 a \,b^{2} d \,m^{3} x^{4}+666 a \,b^{2} e m \,x^{6}+222 b^{3} d m \,x^{6}+a^{3} e \,m^{4} x^{2}+3 a^{2} b d \,m^{4} x^{2}+390 a^{2} b e \,m^{2} x^{4}+390 a \,b^{2} d \,m^{2} x^{4}+405 a \,b^{2} e \,x^{6}+135 b^{3} d \,x^{6}+22 a^{3} e \,m^{3} x^{2}+66 a^{2} b d \,m^{3} x^{2}+900 a^{2} b e m \,x^{4}+900 a \,b^{2} d m \,x^{4}+a^{3} d \,m^{4}+164 a^{3} e \,m^{2} x^{2}+492 a^{2} b d \,m^{2} x^{2}+567 a^{2} b e \,x^{4}+567 a \,b^{2} d \,x^{4}+24 a^{3} d \,m^{3}+458 a^{3} e m \,x^{2}+1374 a^{2} b d m \,x^{2}+206 a^{3} d \,m^{2}+315 a^{3} e \,x^{2}+945 a^{2} b d \,x^{2}+744 a^{3} d m +945 a^{3} d \right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x \left (f x \right )^{m}}{\left (m +9\right ) \left (m +7\right ) \left (m +5\right ) \left (m +3\right ) \left (m +1\right ) \left (b \,x^{2}+a \right )^{3}} \]

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]

x*(b^3*e*m^4*x^8+16*b^3*e*m^3*x^8+3*a*b^2*e*m^4*x^6+b^3*d*m^4*x^6+86*b^3*e*m^2*x^8+54*a*b^2*e*m^3*x^6+18*b^3*d
*m^3*x^6+176*b^3*e*m*x^8+3*a^2*b*e*m^4*x^4+3*a*b^2*d*m^4*x^4+312*a*b^2*e*m^2*x^6+104*b^3*d*m^2*x^6+105*b^3*e*x
^8+60*a^2*b*e*m^3*x^4+60*a*b^2*d*m^3*x^4+666*a*b^2*e*m*x^6+222*b^3*d*m*x^6+a^3*e*m^4*x^2+3*a^2*b*d*m^4*x^2+390
*a^2*b*e*m^2*x^4+390*a*b^2*d*m^2*x^4+405*a*b^2*e*x^6+135*b^3*d*x^6+22*a^3*e*m^3*x^2+66*a^2*b*d*m^3*x^2+900*a^2
*b*e*m*x^4+900*a*b^2*d*m*x^4+a^3*d*m^4+164*a^3*e*m^2*x^2+492*a^2*b*d*m^2*x^2+567*a^2*b*e*x^4+567*a*b^2*d*x^4+2
4*a^3*d*m^3+458*a^3*e*m*x^2+1374*a^2*b*d*m*x^2+206*a^3*d*m^2+315*a^3*e*x^2+945*a^2*b*d*x^2+744*a^3*d*m+945*a^3
*d)*(f*x)^m*((b*x^2+a)^2)^(3/2)/(m+9)/(m+7)/(m+5)/(m+3)/(m+1)/(b*x^2+a)^3

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maxima [A]  time = 0.96, size = 243, normalized size = 0.88 \[ \frac {{\left ({\left (m^{3} + 9 \, m^{2} + 23 \, m + 15\right )} b^{3} f^{m} x^{7} + 3 \, {\left (m^{3} + 11 \, m^{2} + 31 \, m + 21\right )} a b^{2} f^{m} x^{5} + 3 \, {\left (m^{3} + 13 \, m^{2} + 47 \, m + 35\right )} a^{2} b f^{m} x^{3} + {\left (m^{3} + 15 \, m^{2} + 71 \, m + 105\right )} a^{3} f^{m} x\right )} d x^{m}}{m^{4} + 16 \, m^{3} + 86 \, m^{2} + 176 \, m + 105} + \frac {{\left ({\left (m^{3} + 15 \, m^{2} + 71 \, m + 105\right )} b^{3} f^{m} x^{9} + 3 \, {\left (m^{3} + 17 \, m^{2} + 87 \, m + 135\right )} a b^{2} f^{m} x^{7} + 3 \, {\left (m^{3} + 19 \, m^{2} + 111 \, m + 189\right )} a^{2} b f^{m} x^{5} + {\left (m^{3} + 21 \, m^{2} + 143 \, m + 315\right )} a^{3} f^{m} x^{3}\right )} e x^{m}}{m^{4} + 24 \, m^{3} + 206 \, m^{2} + 744 \, m + 945} \]

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]

((m^3 + 9*m^2 + 23*m + 15)*b^3*f^m*x^7 + 3*(m^3 + 11*m^2 + 31*m + 21)*a*b^2*f^m*x^5 + 3*(m^3 + 13*m^2 + 47*m +
35)*a^2*b*f^m*x^3 + (m^3 + 15*m^2 + 71*m + 105)*a^3*f^m*x)*d*x^m/(m^4 + 16*m^3 + 86*m^2 + 176*m + 105) + ((m^
3 + 15*m^2 + 71*m + 105)*b^3*f^m*x^9 + 3*(m^3 + 17*m^2 + 87*m + 135)*a*b^2*f^m*x^7 + 3*(m^3 + 19*m^2 + 111*m +
189)*a^2*b*f^m*x^5 + (m^3 + 21*m^2 + 143*m + 315)*a^3*f^m*x^3)*e*x^m/(m^4 + 24*m^3 + 206*m^2 + 744*m + 945)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (f\,x\right )}^m\,\left (e\,x^2+d\right )\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \]

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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{m} \left (d + e x^{2}\right ) \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

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)**(3/2),x)

[Out]

Integral((f*x)**m*(d + e*x**2)*((a + b*x**2)**2)**(3/2), x)

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