### 3.67 $$\int \frac{(d+e x) (A+B x+C x^2)}{(a+c x^2)^4} \, dx$$

Optimal. Leaf size=165 $-\frac{2 a e (a C+2 A c)-c x (a B e+a C d+5 A c d)}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+a C d+5 A c d)}{16 a^{7/2} c^{3/2}}+\frac{x (a B e+a C d+5 A c d)}{16 a^3 c \left (a+c x^2\right )}-\frac{(d+e x) (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}$

[Out]

-((a*B - (A*c - a*C)*x)*(d + e*x))/(6*a*c*(a + c*x^2)^3) - (2*a*(2*A*c + a*C)*e - c*(5*A*c*d + a*C*d + a*B*e)*
x)/(24*a^2*c^2*(a + c*x^2)^2) + ((5*A*c*d + a*C*d + a*B*e)*x)/(16*a^3*c*(a + c*x^2)) + ((5*A*c*d + a*C*d + a*B
*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))

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Rubi [A]  time = 0.135809, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.16, Rules used = {1645, 639, 199, 205} $-\frac{2 a e (a C+2 A c)-c x (a B e+a C d+5 A c d)}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+a C d+5 A c d)}{16 a^{7/2} c^{3/2}}+\frac{x (a B e+a C d+5 A c d)}{16 a^3 c \left (a+c x^2\right )}-\frac{(d+e x) (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[((d + e*x)*(A + B*x + C*x^2))/(a + c*x^2)^4,x]

[Out]

-((a*B - (A*c - a*C)*x)*(d + e*x))/(6*a*c*(a + c*x^2)^3) - (2*a*(2*A*c + a*C)*e - c*(5*A*c*d + a*C*d + a*B*e)*
x)/(24*a^2*c^2*(a + c*x^2)^2) + ((5*A*c*d + a*C*d + a*B*e)*x)/(16*a^3*c*(a + c*x^2)) + ((5*A*c*d + a*C*d + a*B
*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7/2)*c^(3/2))

Rule 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(p + 1)), x] + Dist[(d*(2*p + 3))/(2*a*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(d+e x) \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx &=-\frac{(a B-(A c-a C) x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac{\int \frac{-5 A c d-a (C d+B e)-2 (2 A c+a C) e x}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac{2 a (2 A c+a C) e-c (5 A c d+a C d+a B e) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{(5 A c d+a C d+a B e) \int \frac{1}{\left (a+c x^2\right )^2} \, dx}{8 a^2 c}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac{2 a (2 A c+a C) e-c (5 A c d+a C d+a B e) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{(5 A c d+a C d+a B e) x}{16 a^3 c \left (a+c x^2\right )}+\frac{(5 A c d+a C d+a B e) \int \frac{1}{a+c x^2} \, dx}{16 a^3 c}\\ &=-\frac{(a B-(A c-a C) x) (d+e x)}{6 a c \left (a+c x^2\right )^3}-\frac{2 a (2 A c+a C) e-c (5 A c d+a C d+a B e) x}{24 a^2 c^2 \left (a+c x^2\right )^2}+\frac{(5 A c d+a C d+a B e) x}{16 a^3 c \left (a+c x^2\right )}+\frac{(5 A c d+a C d+a B e) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.161578, size = 171, normalized size = 1.04 $\frac{\frac{8 a^{5/2} \left (a^2 C e-a c (A e+B (d+e x)+C d x)+A c^2 d x\right )}{\left (a+c x^2\right )^3}+\frac{2 a^{3/2} \left (-6 a^2 C e+a c x (B e+C d)+5 A c^2 d x\right )}{\left (a+c x^2\right )^2}+\frac{3 \sqrt{a} c x (a B e+a C d+5 A c d)}{a+c x^2}+3 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) (a B e+a C d+5 A c d)}{48 a^{7/2} c^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((d + e*x)*(A + B*x + C*x^2))/(a + c*x^2)^4,x]

[Out]

((2*a^(3/2)*(-6*a^2*C*e + 5*A*c^2*d*x + a*c*(C*d + B*e)*x))/(a + c*x^2)^2 + (3*Sqrt[a]*c*(5*A*c*d + a*C*d + a*
B*e)*x)/(a + c*x^2) + (8*a^(5/2)*(a^2*C*e + A*c^2*d*x - a*c*(A*e + C*d*x + B*(d + e*x))))/(a + c*x^2)^3 + 3*Sq
rt[c]*(5*A*c*d + a*C*d + a*B*e)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(48*a^(7/2)*c^2)

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Maple [A]  time = 0.049, size = 182, normalized size = 1.1 \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{3}} \left ({\frac{ \left ( 5\,Acd+aBe+Cad \right ) c{x}^{5}}{16\,{a}^{3}}}+{\frac{ \left ( 5\,Acd+aBe+Cad \right ){x}^{3}}{6\,{a}^{2}}}-{\frac{Ce{x}^{2}}{4\,c}}+{\frac{ \left ( 11\,Acd-aBe-Cad \right ) x}{16\,ac}}-{\frac{2\,Ace+2\,Bcd+aCe}{12\,{c}^{2}}} \right ) }+{\frac{5\,Ad}{16\,{a}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{Be}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{Cd}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(C*x^2+B*x+A)/(c*x^2+a)^4,x)

[Out]

(1/16*(5*A*c*d+B*a*e+C*a*d)/a^3*c*x^5+1/6/a^2*(5*A*c*d+B*a*e+C*a*d)*x^3-1/4*C*e*x^2/c+1/16*(11*A*c*d-B*a*e-C*a
*d)/a/c*x-1/12*(2*A*c*e+2*B*c*d+C*a*e)/c^2)/(c*x^2+a)^3+5/16/a^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d+1/16/
a^2/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*e+1/16/a^2/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*C*d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.4554, size = 1341, normalized size = 8.13 \begin{align*} \left [-\frac{24 \, C a^{4} c e x^{2} + 16 \, B a^{4} c d - 6 \,{\left (B a^{2} c^{3} e +{\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} d\right )} x^{5} - 16 \,{\left (B a^{3} c^{2} e +{\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} d\right )} x^{3} + 3 \,{\left ({\left (B a c^{3} e +{\left (C a c^{3} + 5 \, A c^{4}\right )} d\right )} x^{6} + B a^{4} e + 3 \,{\left (B a^{2} c^{2} e +{\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} d\right )} x^{4} + 3 \,{\left (B a^{3} c e +{\left (C a^{3} c + 5 \, A a^{2} c^{2}\right )} d\right )} x^{2} +{\left (C a^{4} + 5 \, A a^{3} c\right )} d\right )} \sqrt{-a c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) + 8 \,{\left (C a^{5} + 2 \, A a^{4} c\right )} e + 6 \,{\left (B a^{4} c e +{\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} d\right )} x}{96 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, -\frac{12 \, C a^{4} c e x^{2} + 8 \, B a^{4} c d - 3 \,{\left (B a^{2} c^{3} e +{\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} d\right )} x^{5} - 8 \,{\left (B a^{3} c^{2} e +{\left (C a^{3} c^{2} + 5 \, A a^{2} c^{3}\right )} d\right )} x^{3} - 3 \,{\left ({\left (B a c^{3} e +{\left (C a c^{3} + 5 \, A c^{4}\right )} d\right )} x^{6} + B a^{4} e + 3 \,{\left (B a^{2} c^{2} e +{\left (C a^{2} c^{2} + 5 \, A a c^{3}\right )} d\right )} x^{4} + 3 \,{\left (B a^{3} c e +{\left (C a^{3} c + 5 \, A a^{2} c^{2}\right )} d\right )} x^{2} +{\left (C a^{4} + 5 \, A a^{3} c\right )} d\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) + 4 \,{\left (C a^{5} + 2 \, A a^{4} c\right )} e + 3 \,{\left (B a^{4} c e +{\left (C a^{4} c - 11 \, A a^{3} c^{2}\right )} d\right )} x}{48 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="fricas")

[Out]

[-1/96*(24*C*a^4*c*e*x^2 + 16*B*a^4*c*d - 6*(B*a^2*c^3*e + (C*a^2*c^3 + 5*A*a*c^4)*d)*x^5 - 16*(B*a^3*c^2*e +
(C*a^3*c^2 + 5*A*a^2*c^3)*d)*x^3 + 3*((B*a*c^3*e + (C*a*c^3 + 5*A*c^4)*d)*x^6 + B*a^4*e + 3*(B*a^2*c^2*e + (C*
a^2*c^2 + 5*A*a*c^3)*d)*x^4 + 3*(B*a^3*c*e + (C*a^3*c + 5*A*a^2*c^2)*d)*x^2 + (C*a^4 + 5*A*a^3*c)*d)*sqrt(-a*c
)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 8*(C*a^5 + 2*A*a^4*c)*e + 6*(B*a^4*c*e + (C*a^4*c - 11*A*a^3
*c^2)*d)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2), -1/48*(12*C*a^4*c*e*x^2 + 8*B*a^4*c*d - 3
*(B*a^2*c^3*e + (C*a^2*c^3 + 5*A*a*c^4)*d)*x^5 - 8*(B*a^3*c^2*e + (C*a^3*c^2 + 5*A*a^2*c^3)*d)*x^3 - 3*((B*a*c
^3*e + (C*a*c^3 + 5*A*c^4)*d)*x^6 + B*a^4*e + 3*(B*a^2*c^2*e + (C*a^2*c^2 + 5*A*a*c^3)*d)*x^4 + 3*(B*a^3*c*e +
(C*a^3*c + 5*A*a^2*c^2)*d)*x^2 + (C*a^4 + 5*A*a^3*c)*d)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 4*(C*a^5 + 2*A*a^4*
c)*e + 3*(B*a^4*c*e + (C*a^4*c - 11*A*a^3*c^2)*d)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2)]

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Sympy [A]  time = 127.102, size = 298, normalized size = 1.81 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{7} c^{3}}} \left (5 A c d + B a e + C a d\right ) \log{\left (- a^{4} c \sqrt{- \frac{1}{a^{7} c^{3}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{a^{7} c^{3}}} \left (5 A c d + B a e + C a d\right ) \log{\left (a^{4} c \sqrt{- \frac{1}{a^{7} c^{3}}} + x \right )}}{32} + \frac{- 8 A a^{3} c e - 8 B a^{3} c d - 4 C a^{4} e - 12 C a^{3} c e x^{2} + x^{5} \left (15 A c^{4} d + 3 B a c^{3} e + 3 C a c^{3} d\right ) + x^{3} \left (40 A a c^{3} d + 8 B a^{2} c^{2} e + 8 C a^{2} c^{2} d\right ) + x \left (33 A a^{2} c^{2} d - 3 B a^{3} c e - 3 C a^{3} c d\right )}{48 a^{6} c^{2} + 144 a^{5} c^{3} x^{2} + 144 a^{4} c^{4} x^{4} + 48 a^{3} c^{5} x^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(C*x**2+B*x+A)/(c*x**2+a)**4,x)

[Out]

-sqrt(-1/(a**7*c**3))*(5*A*c*d + B*a*e + C*a*d)*log(-a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + sqrt(-1/(a**7*c**3)
)*(5*A*c*d + B*a*e + C*a*d)*log(a**4*c*sqrt(-1/(a**7*c**3)) + x)/32 + (-8*A*a**3*c*e - 8*B*a**3*c*d - 4*C*a**4
*e - 12*C*a**3*c*e*x**2 + x**5*(15*A*c**4*d + 3*B*a*c**3*e + 3*C*a*c**3*d) + x**3*(40*A*a*c**3*d + 8*B*a**2*c*
*2*e + 8*C*a**2*c**2*d) + x*(33*A*a**2*c**2*d - 3*B*a**3*c*e - 3*C*a**3*c*d))/(48*a**6*c**2 + 144*a**5*c**3*x*
*2 + 144*a**4*c**4*x**4 + 48*a**3*c**5*x**6)

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Giac [A]  time = 1.15865, size = 262, normalized size = 1.59 \begin{align*} \frac{{\left (C a d + 5 \, A c d + B a e\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3} c} + \frac{3 \, C a c^{3} d x^{5} + 15 \, A c^{4} d x^{5} + 3 \, B a c^{3} x^{5} e + 8 \, C a^{2} c^{2} d x^{3} + 40 \, A a c^{3} d x^{3} + 8 \, B a^{2} c^{2} x^{3} e - 12 \, C a^{3} c x^{2} e - 3 \, C a^{3} c d x + 33 \, A a^{2} c^{2} d x - 3 \, B a^{3} c x e - 8 \, B a^{3} c d - 4 \, C a^{4} e - 8 \, A a^{3} c e}{48 \,{\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="giac")

[Out]

1/16*(C*a*d + 5*A*c*d + B*a*e)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(3*C*a*c^3*d*x^5 + 15*A*c^4*d*x^
5 + 3*B*a*c^3*x^5*e + 8*C*a^2*c^2*d*x^3 + 40*A*a*c^3*d*x^3 + 8*B*a^2*c^2*x^3*e - 12*C*a^3*c*x^2*e - 3*C*a^3*c*
d*x + 33*A*a^2*c^2*d*x - 3*B*a^3*c*x*e - 8*B*a^3*c*d - 4*C*a^4*e - 8*A*a^3*c*e)/((c*x^2 + a)^3*a^3*c^2)