3.24 \(\int \frac {(a+c x^2) (A+B x+C x^2)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=156 \[ \frac {a e^2 (2 C d-B e)+c d \left (4 C d^2-e (3 B d-2 A e)\right )}{e^5 (d+e x)}-\frac {\left (a e^2+c d^2\right ) \left (A e^2-B d e+C d^2\right )}{2 e^5 (d+e x)^2}+\frac {\log (d+e x) \left (a C e^2+c \left (6 C d^2-e (3 B d-A e)\right )\right )}{e^5}-\frac {c x (3 C d-B e)}{e^4}+\frac {c C x^2}{2 e^3} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 154, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {1628} \[ \frac {a e^2 (2 C d-B e)-c d e (3 B d-2 A e)+4 c C d^3}{e^5 (d+e x)}-\frac {\left (a e^2+c d^2\right ) \left (A e^2-B d e+C d^2\right )}{2 e^5 (d+e x)^2}+\frac {\log (d+e x) \left (a C e^2-c e (3 B d-A e)+6 c C d^2\right )}{e^5}-\frac {c x (3 C d-B e)}{e^4}+\frac {c C x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + c*x^2)*(A + B*x + C*x^2))/(d + e*x)^3,x]

[Out]

-((c*(3*C*d - B*e)*x)/e^4) + (c*C*x^2)/(2*e^3) - ((c*d^2 + a*e^2)*(C*d^2 - B*d*e + A*e^2))/(2*e^5*(d + e*x)^2)
+ (4*c*C*d^3 - c*d*e*(3*B*d - 2*A*e) + a*e^2*(2*C*d - B*e))/(e^5*(d + e*x)) + ((6*c*C*d^2 + a*C*e^2 - c*e*(3*
B*d - A*e))*Log[d + e*x])/e^5

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^2\right ) \left (A+B x+C x^2\right )}{(d+e x)^3} \, dx &=\int \left (\frac {c (-3 C d+B e)}{e^4}+\frac {c C x}{e^3}+\frac {\left (c d^2+a e^2\right ) \left (C d^2-B d e+A e^2\right )}{e^4 (d+e x)^3}+\frac {-4 c C d^3+c d e (3 B d-2 A e)-a e^2 (2 C d-B e)}{e^4 (d+e x)^2}+\frac {6 c C d^2+a C e^2-c e (3 B d-A e)}{e^4 (d+e x)}\right ) \, dx\\ &=-\frac {c (3 C d-B e) x}{e^4}+\frac {c C x^2}{2 e^3}-\frac {\left (c d^2+a e^2\right ) \left (C d^2-B d e+A e^2\right )}{2 e^5 (d+e x)^2}+\frac {4 c C d^3-c d e (3 B d-2 A e)+a e^2 (2 C d-B e)}{e^5 (d+e x)}+\frac {\left (6 c C d^2+a C e^2-c e (3 B d-A e)\right ) \log (d+e x)}{e^5}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 176, normalized size = 1.13 \[ \frac {\log (d+e x) \left (a C e^2+A c e^2-3 B c d e+6 c C d^2\right )}{e^5}+\frac {-a B e^3+2 a C d e^2+2 A c d e^2-3 B c d^2 e+4 c C d^3}{e^5 (d+e x)}+\frac {-a A e^4+a B d e^3-a C d^2 e^2-A c d^2 e^2+B c d^3 e-c C d^4}{2 e^5 (d+e x)^2}+\frac {c x (B e-3 C d)}{e^4}+\frac {c C x^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + c*x^2)*(A + B*x + C*x^2))/(d + e*x)^3,x]

[Out]

(c*(-3*C*d + B*e)*x)/e^4 + (c*C*x^2)/(2*e^3) + (-(c*C*d^4) + B*c*d^3*e - A*c*d^2*e^2 - a*C*d^2*e^2 + a*B*d*e^3
- a*A*e^4)/(2*e^5*(d + e*x)^2) + (4*c*C*d^3 - 3*B*c*d^2*e + 2*A*c*d*e^2 + 2*a*C*d*e^2 - a*B*e^3)/(e^5*(d + e*
x)) + ((6*c*C*d^2 - 3*B*c*d*e + A*c*e^2 + a*C*e^2)*Log[d + e*x])/e^5

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fricas [A]  time = 0.79, size = 273, normalized size = 1.75 \[ \frac {C c e^{4} x^{4} + 7 \, C c d^{4} - 5 \, B c d^{3} e - B a d e^{3} - A a e^{4} + 3 \, {\left (C a + A c\right )} d^{2} e^{2} - 2 \, {\left (2 \, C c d e^{3} - B c e^{4}\right )} x^{3} - {\left (11 \, C c d^{2} e^{2} - 4 \, B c d e^{3}\right )} x^{2} + 2 \, {\left (C c d^{3} e - 2 \, B c d^{2} e^{2} - B a e^{4} + 2 \, {\left (C a + A c\right )} d e^{3}\right )} x + 2 \, {\left (6 \, C c d^{4} - 3 \, B c d^{3} e + {\left (C a + A c\right )} d^{2} e^{2} + {\left (6 \, C c d^{2} e^{2} - 3 \, B c d e^{3} + {\left (C a + A c\right )} e^{4}\right )} x^{2} + 2 \, {\left (6 \, C c d^{3} e - 3 \, B c d^{2} e^{2} + {\left (C a + A c\right )} d e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)*(C*x^2+B*x+A)/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*(C*c*e^4*x^4 + 7*C*c*d^4 - 5*B*c*d^3*e - B*a*d*e^3 - A*a*e^4 + 3*(C*a + A*c)*d^2*e^2 - 2*(2*C*c*d*e^3 - B*
c*e^4)*x^3 - (11*C*c*d^2*e^2 - 4*B*c*d*e^3)*x^2 + 2*(C*c*d^3*e - 2*B*c*d^2*e^2 - B*a*e^4 + 2*(C*a + A*c)*d*e^3
)*x + 2*(6*C*c*d^4 - 3*B*c*d^3*e + (C*a + A*c)*d^2*e^2 + (6*C*c*d^2*e^2 - 3*B*c*d*e^3 + (C*a + A*c)*e^4)*x^2 +
2*(6*C*c*d^3*e - 3*B*c*d^2*e^2 + (C*a + A*c)*d*e^3)*x)*log(e*x + d))/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

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giac [A]  time = 0.15, size = 167, normalized size = 1.07 \[ {\left (6 \, C c d^{2} - 3 \, B c d e + C a e^{2} + A c e^{2}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (C c x^{2} e^{3} - 6 \, C c d x e^{2} + 2 \, B c x e^{3}\right )} e^{\left (-6\right )} + \frac {{\left (7 \, C c d^{4} - 5 \, B c d^{3} e + 3 \, C a d^{2} e^{2} + 3 \, A c d^{2} e^{2} - B a d e^{3} - A a e^{4} + 2 \, {\left (4 \, C c d^{3} e - 3 \, B c d^{2} e^{2} + 2 \, C a d e^{3} + 2 \, A c d e^{3} - B a e^{4}\right )} x\right )} e^{\left (-5\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)*(C*x^2+B*x+A)/(e*x+d)^3,x, algorithm="giac")

[Out]

(6*C*c*d^2 - 3*B*c*d*e + C*a*e^2 + A*c*e^2)*e^(-5)*log(abs(x*e + d)) + 1/2*(C*c*x^2*e^3 - 6*C*c*d*x*e^2 + 2*B*
c*x*e^3)*e^(-6) + 1/2*(7*C*c*d^4 - 5*B*c*d^3*e + 3*C*a*d^2*e^2 + 3*A*c*d^2*e^2 - B*a*d*e^3 - A*a*e^4 + 2*(4*C*
c*d^3*e - 3*B*c*d^2*e^2 + 2*C*a*d*e^3 + 2*A*c*d*e^3 - B*a*e^4)*x)*e^(-5)/(x*e + d)^2

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maple [A]  time = 0.01, size = 257, normalized size = 1.65 \[ -\frac {A a}{2 \left (e x +d \right )^{2} e}-\frac {A c \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {B a d}{2 \left (e x +d \right )^{2} e^{2}}+\frac {B c \,d^{3}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {C a \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {C c \,d^{4}}{2 \left (e x +d \right )^{2} e^{5}}+\frac {C c \,x^{2}}{2 e^{3}}+\frac {2 A c d}{\left (e x +d \right ) e^{3}}+\frac {A c \ln \left (e x +d \right )}{e^{3}}-\frac {B a}{\left (e x +d \right ) e^{2}}-\frac {3 B c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 B c d \ln \left (e x +d \right )}{e^{4}}+\frac {B c x}{e^{3}}+\frac {2 C a d}{\left (e x +d \right ) e^{3}}+\frac {C a \ln \left (e x +d \right )}{e^{3}}+\frac {4 C c \,d^{3}}{\left (e x +d \right ) e^{5}}+\frac {6 C c \,d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {3 C c d x}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)*(C*x^2+B*x+A)/(e*x+d)^3,x)

[Out]

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

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maxima [A]  time = 0.47, size = 177, normalized size = 1.13 \[ \frac {7 \, C c d^{4} - 5 \, B c d^{3} e - B a d e^{3} - A a e^{4} + 3 \, {\left (C a + A c\right )} d^{2} e^{2} + 2 \, {\left (4 \, C c d^{3} e - 3 \, B c d^{2} e^{2} - B a e^{4} + 2 \, {\left (C a + A c\right )} d e^{3}\right )} x}{2 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} + \frac {C c e x^{2} - 2 \, {\left (3 \, C c d - B c e\right )} x}{2 \, e^{4}} + \frac {{\left (6 \, C c d^{2} - 3 \, B c d e + {\left (C a + A c\right )} e^{2}\right )} \log \left (e x + d\right )}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)*(C*x^2+B*x+A)/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*(7*C*c*d^4 - 5*B*c*d^3*e - B*a*d*e^3 - A*a*e^4 + 3*(C*a + A*c)*d^2*e^2 + 2*(4*C*c*d^3*e - 3*B*c*d^2*e^2 -
B*a*e^4 + 2*(C*a + A*c)*d*e^3)*x)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5) + 1/2*(C*c*e*x^2 - 2*(3*C*c*d - B*c*e)*x)/e^
4 + (6*C*c*d^2 - 3*B*c*d*e + (C*a + A*c)*e^2)*log(e*x + d)/e^5

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mupad [B]  time = 0.09, size = 185, normalized size = 1.19 \[ \frac {x\,\left (4\,C\,c\,d^3-B\,a\,e^3+2\,A\,c\,d\,e^2+2\,C\,a\,d\,e^2-3\,B\,c\,d^2\,e\right )-\frac {A\,a\,e^4-7\,C\,c\,d^4+B\,a\,d\,e^3+5\,B\,c\,d^3\,e-3\,A\,c\,d^2\,e^2-3\,C\,a\,d^2\,e^2}{2\,e}}{d^2\,e^4+2\,d\,e^5\,x+e^6\,x^2}+x\,\left (\frac {B\,c}{e^3}-\frac {3\,C\,c\,d}{e^4}\right )+\frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e^2+C\,a\,e^2+6\,C\,c\,d^2-3\,B\,c\,d\,e\right )}{e^5}+\frac {C\,c\,x^2}{2\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)*(A + B*x + C*x^2))/(d + e*x)^3,x)

[Out]

(x*(4*C*c*d^3 - B*a*e^3 + 2*A*c*d*e^2 + 2*C*a*d*e^2 - 3*B*c*d^2*e) - (A*a*e^4 - 7*C*c*d^4 + B*a*d*e^3 + 5*B*c*
d^3*e - 3*A*c*d^2*e^2 - 3*C*a*d^2*e^2)/(2*e))/(d^2*e^4 + e^6*x^2 + 2*d*e^5*x) + x*((B*c)/e^3 - (3*C*c*d)/e^4)
+ (log(d + e*x)*(A*c*e^2 + C*a*e^2 + 6*C*c*d^2 - 3*B*c*d*e))/e^5 + (C*c*x^2)/(2*e^3)

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sympy [A]  time = 5.29, size = 206, normalized size = 1.32 \[ \frac {C c x^{2}}{2 e^{3}} + x \left (\frac {B c}{e^{3}} - \frac {3 C c d}{e^{4}}\right ) + \frac {- A a e^{4} + 3 A c d^{2} e^{2} - B a d e^{3} - 5 B c d^{3} e + 3 C a d^{2} e^{2} + 7 C c d^{4} + x \left (4 A c d e^{3} - 2 B a e^{4} - 6 B c d^{2} e^{2} + 4 C a d e^{3} + 8 C c d^{3} e\right )}{2 d^{2} e^{5} + 4 d e^{6} x + 2 e^{7} x^{2}} + \frac {\left (A c e^{2} - 3 B c d e + C a e^{2} + 6 C c d^{2}\right ) \log {\left (d + e x \right )}}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)*(C*x**2+B*x+A)/(e*x+d)**3,x)

[Out]

C*c*x**2/(2*e**3) + x*(B*c/e**3 - 3*C*c*d/e**4) + (-A*a*e**4 + 3*A*c*d**2*e**2 - B*a*d*e**3 - 5*B*c*d**3*e + 3
*C*a*d**2*e**2 + 7*C*c*d**4 + x*(4*A*c*d*e**3 - 2*B*a*e**4 - 6*B*c*d**2*e**2 + 4*C*a*d*e**3 + 8*C*c*d**3*e))/(
2*d**2*e**5 + 4*d*e**6*x + 2*e**7*x**2) + (A*c*e**2 - 3*B*c*d*e + C*a*e**2 + 6*C*c*d**2)*log(d + e*x)/e**5

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