3.39 \(\int \frac {(a+b x^2) (-a d+4 b c x+3 b d x^2)}{(c+d x)^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {\left (a+b x^2\right )^2}{c+d x} \]

[Out]

(b*x^2+a)^2/(d*x+c)

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {1590} \[ \frac {\left (a+b x^2\right )^2}{c+d x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(-(a*d) + 4*b*c*x + 3*b*d*x^2))/(c + d*x)^2,x]

[Out]

(a + b*x^2)^2/(c + d*x)

Rule 1590

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[(Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*Rr^(n + 1))/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x
, r]), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (-a d+4 b c x+3 b d x^2\right )}{(c+d x)^2} \, dx &=\frac {\left (a+b x^2\right )^2}{c+d x}\\ \end {align*}

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Mathematica [B]  time = 0.04, size = 62, normalized size = 3.65 \[ \frac {a^2 d^4+2 a b d^2 \left (c^2+c d x+d^2 x^2\right )+b^2 \left (c^4+c^3 d x+d^4 x^4\right )}{d^4 (c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(-(a*d) + 4*b*c*x + 3*b*d*x^2))/(c + d*x)^2,x]

[Out]

(a^2*d^4 + 2*a*b*d^2*(c^2 + c*d*x + d^2*x^2) + b^2*(c^4 + c^3*d*x + d^4*x^4))/(d^4*(c + d*x))

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fricas [B]  time = 0.95, size = 78, normalized size = 4.59 \[ \frac {b^{2} d^{4} x^{4} + 2 \, a b d^{4} x^{2} + b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4} + {\left (b^{2} c^{3} d + 2 \, a b c d^{3}\right )} x}{d^{5} x + c d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(3*b*d*x^2+4*b*c*x-a*d)/(d*x+c)^2,x, algorithm="fricas")

[Out]

(b^2*d^4*x^4 + 2*a*b*d^4*x^2 + b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4 + (b^2*c^3*d + 2*a*b*c*d^3)*x)/(d^5*x + c*d^4
)

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giac [B]  time = 0.17, size = 111, normalized size = 6.53 \[ \frac {{\left (b^{2} - \frac {4 \, b^{2} c}{d x + c} + \frac {6 \, b^{2} c^{2}}{{\left (d x + c\right )}^{2}} + \frac {2 \, a b d^{2}}{{\left (d x + c\right )}^{2}}\right )} {\left (d x + c\right )}^{3}}{d^{4}} + \frac {\frac {b^{2} c^{4} d^{3}}{d x + c} + \frac {2 \, a b c^{2} d^{5}}{d x + c} + \frac {a^{2} d^{7}}{d x + c}}{d^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(3*b*d*x^2+4*b*c*x-a*d)/(d*x+c)^2,x, algorithm="giac")

[Out]

(b^2 - 4*b^2*c/(d*x + c) + 6*b^2*c^2/(d*x + c)^2 + 2*a*b*d^2/(d*x + c)^2)*(d*x + c)^3/d^4 + (b^2*c^4*d^3/(d*x
+ c) + 2*a*b*c^2*d^5/(d*x + c) + a^2*d^7/(d*x + c))/d^7

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maple [B]  time = 0.01, size = 76, normalized size = 4.47 \[ \frac {\left (b \,d^{2} x^{3}-b c d \,x^{2}+2 a \,d^{2} x +b \,c^{2} x \right ) b}{d^{3}}-\frac {-a^{2} d^{4}-2 a b \,c^{2} d^{2}-b^{2} c^{4}}{\left (d x +c \right ) d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(3*b*d*x^2+4*b*c*x-a*d)/(d*x+c)^2,x)

[Out]

b/d^3*(b*d^2*x^3-b*c*d*x^2+2*a*d^2*x+b*c^2*x)-(-a^2*d^4-2*a*b*c^2*d^2-b^2*c^4)/d^4/(d*x+c)

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maxima [B]  time = 0.46, size = 82, normalized size = 4.82 \[ \frac {b^{2} c^{4} + 2 \, a b c^{2} d^{2} + a^{2} d^{4}}{d^{5} x + c d^{4}} + \frac {b^{2} d^{2} x^{3} - b^{2} c d x^{2} + {\left (b^{2} c^{2} + 2 \, a b d^{2}\right )} x}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)*(3*b*d*x^2+4*b*c*x-a*d)/(d*x+c)^2,x, algorithm="maxima")

[Out]

(b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)/(d^5*x + c*d^4) + (b^2*d^2*x^3 - b^2*c*d*x^2 + (b^2*c^2 + 2*a*b*d^2)*x)/d^
3

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mupad [B]  time = 0.08, size = 85, normalized size = 5.00 \[ x\,\left (\frac {b^2\,c^2}{d^3}+\frac {2\,a\,b}{d}\right )+\frac {b^2\,x^3}{d}+\frac {a^2\,d^4+2\,a\,b\,c^2\,d^2+b^2\,c^4}{d\,\left (x\,d^4+c\,d^3\right )}-\frac {b^2\,c\,x^2}{d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(4*b*c*x - a*d + 3*b*d*x^2))/(c + d*x)^2,x)

[Out]

x*((b^2*c^2)/d^3 + (2*a*b)/d) + (b^2*x^3)/d + (a^2*d^4 + b^2*c^4 + 2*a*b*c^2*d^2)/(d*(c*d^3 + d^4*x)) - (b^2*c
*x^2)/d^2

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sympy [B]  time = 0.37, size = 73, normalized size = 4.29 \[ - \frac {b^{2} c x^{2}}{d^{2}} + \frac {b^{2} x^{3}}{d} + x \left (\frac {2 a b}{d} + \frac {b^{2} c^{2}}{d^{3}}\right ) + \frac {a^{2} d^{4} + 2 a b c^{2} d^{2} + b^{2} c^{4}}{c d^{4} + d^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(3*b*d*x**2+4*b*c*x-a*d)/(d*x+c)**2,x)

[Out]

-b**2*c*x**2/d**2 + b**2*x**3/d + x*(2*a*b/d + b**2*c**2/d**3) + (a**2*d**4 + 2*a*b*c**2*d**2 + b**2*c**4)/(c*
d**4 + d**5*x)

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