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

Optimal. Leaf size=295 \[ \frac {\log (d+e x) \left (a^2 C e^4+2 a c e^2 \left (6 C d^2-e (3 B d-A e)\right )+c^2 d^2 \left (15 C d^2-2 e (5 B d-3 A e)\right )\right )}{e^7}+\frac {\left (a e^2+c d^2\right ) \left (a e^2 (2 C d-B e)+c d \left (6 C d^2-e (5 B d-4 A e)\right )\right )}{e^7 (d+e x)}-\frac {\left (a e^2+c d^2\right )^2 \left (A e^2-B d e+C d^2\right )}{2 e^7 (d+e x)^2}-\frac {c x \left (2 a e^2 (3 C d-B e)+c d \left (10 C d^2-3 e (2 B d-A e)\right )\right )}{e^6}+\frac {c x^2 \left (2 a C e^2+c \left (6 C d^2-e (3 B d-A e)\right )\right )}{2 e^5}-\frac {c^2 x^3 (3 C d-B e)}{3 e^4}+\frac {c^2 C x^4}{4 e^3} \]

[Out]

-c*(2*a*e^2*(-B*e+3*C*d)+c*d*(10*C*d^2-3*e*(-A*e+2*B*d)))*x/e^6+1/2*c*(2*a*C*e^2+c*(6*C*d^2-e*(-A*e+3*B*d)))*x
^2/e^5-1/3*c^2*(-B*e+3*C*d)*x^3/e^4+1/4*c^2*C*x^4/e^3-1/2*(a*e^2+c*d^2)^2*(A*e^2-B*d*e+C*d^2)/e^7/(e*x+d)^2+(a
*e^2+c*d^2)*(a*e^2*(-B*e+2*C*d)+c*d*(6*C*d^2-e*(-4*A*e+5*B*d)))/e^7/(e*x+d)+(a^2*C*e^4+c^2*d^2*(15*C*d^2-2*e*(
-3*A*e+5*B*d))+2*a*c*e^2*(6*C*d^2-e*(-A*e+3*B*d)))*ln(e*x+d)/e^7

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Rubi [A]  time = 0.49, antiderivative size = 292, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1628} \[ \frac {\log (d+e x) \left (a^2 C e^4+2 a c e^2 \left (6 C d^2-e (3 B d-A e)\right )+c^2 \left (15 C d^4-2 d^2 e (5 B d-3 A e)\right )\right )}{e^7}+\frac {c x^2 \left (2 a C e^2-c e (3 B d-A e)+6 c C d^2\right )}{2 e^5}-\frac {c x \left (2 a e^2 (3 C d-B e)-3 c d e (2 B d-A e)+10 c C d^3\right )}{e^6}+\frac {\left (a e^2+c d^2\right ) \left (a e^2 (2 C d-B e)-c d e (5 B d-4 A e)+6 c C d^3\right )}{e^7 (d+e x)}-\frac {\left (a e^2+c d^2\right )^2 \left (A e^2-B d e+C d^2\right )}{2 e^7 (d+e x)^2}-\frac {c^2 x^3 (3 C d-B e)}{3 e^4}+\frac {c^2 C x^4}{4 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(10*c*C*d^3 - 3*c*d*e*(2*B*d - A*e) + 2*a*e^2*(3*C*d - B*e))*x)/e^6) + (c*(6*c*C*d^2 + 2*a*C*e^2 - c*e*(3
*B*d - A*e))*x^2)/(2*e^5) - (c^2*(3*C*d - B*e)*x^3)/(3*e^4) + (c^2*C*x^4)/(4*e^3) - ((c*d^2 + a*e^2)^2*(C*d^2
- B*d*e + A*e^2))/(2*e^7*(d + e*x)^2) + ((c*d^2 + a*e^2)*(6*c*C*d^3 - c*d*e*(5*B*d - 4*A*e) + a*e^2*(2*C*d - B
*e)))/(e^7*(d + e*x)) + ((a^2*C*e^4 + c^2*(15*C*d^4 - 2*d^2*e*(5*B*d - 3*A*e)) + 2*a*c*e^2*(6*C*d^2 - e*(3*B*d
- A*e)))*Log[d + e*x])/e^7

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 )^2 \left (A+B x+C x^2\right )}{(d+e x)^3} \, dx &=\int \left (\frac {c \left (-10 c C d^3+3 c d e (2 B d-A e)-2 a e^2 (3 C d-B e)\right )}{e^6}+\frac {c \left (6 c C d^2+2 a C e^2-c e (3 B d-A e)\right ) x}{e^5}+\frac {c^2 (-3 C d+B e) x^2}{e^4}+\frac {c^2 C x^3}{e^3}+\frac {\left (c d^2+a e^2\right )^2 \left (C d^2-B d e+A e^2\right )}{e^6 (d+e x)^3}+\frac {\left (c d^2+a e^2\right ) \left (-6 c C d^3+c d e (5 B d-4 A e)-a e^2 (2 C d-B e)\right )}{e^6 (d+e x)^2}+\frac {a^2 C e^4+c^2 \left (15 C d^4-2 d^2 e (5 B d-3 A e)\right )+2 a c e^2 \left (6 C d^2-e (3 B d-A e)\right )}{e^6 (d+e x)}\right ) \, dx\\ &=-\frac {c \left (10 c C d^3-3 c d e (2 B d-A e)+2 a e^2 (3 C d-B e)\right ) x}{e^6}+\frac {c \left (6 c C d^2+2 a C e^2-c e (3 B d-A e)\right ) x^2}{2 e^5}-\frac {c^2 (3 C d-B e) x^3}{3 e^4}+\frac {c^2 C x^4}{4 e^3}-\frac {\left (c d^2+a e^2\right )^2 \left (C d^2-B d e+A e^2\right )}{2 e^7 (d+e x)^2}+\frac {\left (c d^2+a e^2\right ) \left (6 c C d^3-c d e (5 B d-4 A e)+a e^2 (2 C d-B e)\right )}{e^7 (d+e x)}+\frac {\left (a^2 C e^4+c^2 \left (15 C d^4-2 d^2 e (5 B d-3 A e)\right )+2 a c e^2 \left (6 C d^2-e (3 B d-A e)\right )\right ) \log (d+e x)}{e^7}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 274, normalized size = 0.93 \[ \frac {12 \log (d+e x) \left (a^2 C e^4+2 a c e^2 \left (e (A e-3 B d)+6 C d^2\right )+c^2 \left (2 d^2 e (3 A e-5 B d)+15 C d^4\right )\right )-12 c e x \left (-2 a e^2 (B e-3 C d)+3 c d e (A e-2 B d)+10 c C d^3\right )+6 c e^2 x^2 \left (2 a C e^2+c e (A e-3 B d)+6 c C d^2\right )-\frac {6 \left (a e^2+c d^2\right )^2 \left (e (A e-B d)+C d^2\right )}{(d+e x)^2}+\frac {12 \left (a e^2+c d^2\right ) \left (a e^2 (2 C d-B e)+c d e (4 A e-5 B d)+6 c C d^3\right )}{d+e x}+4 c^2 e^3 x^3 (B e-3 C d)+3 c^2 C e^4 x^4}{12 e^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12*c*e*(10*c*C*d^3 + 3*c*d*e*(-2*B*d + A*e) - 2*a*e^2*(-3*C*d + B*e))*x + 6*c*e^2*(6*c*C*d^2 + 2*a*C*e^2 + c
*e*(-3*B*d + A*e))*x^2 + 4*c^2*e^3*(-3*C*d + B*e)*x^3 + 3*c^2*C*e^4*x^4 - (6*(c*d^2 + a*e^2)^2*(C*d^2 + e*(-(B
*d) + A*e)))/(d + e*x)^2 + (12*(c*d^2 + a*e^2)*(6*c*C*d^3 + c*d*e*(-5*B*d + 4*A*e) + a*e^2*(2*C*d - B*e)))/(d
+ e*x) + 12*(a^2*C*e^4 + 2*a*c*e^2*(6*C*d^2 + e*(-3*B*d + A*e)) + c^2*(15*C*d^4 + 2*d^2*e*(-5*B*d + 3*A*e)))*L
og[d + e*x])/(12*e^7)

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fricas [B]  time = 0.91, size = 608, normalized size = 2.06 \[ \frac {3 \, C c^{2} e^{6} x^{6} + 66 \, C c^{2} d^{6} - 54 \, B c^{2} d^{5} e - 60 \, B a c d^{3} e^{3} - 6 \, B a^{2} d e^{5} - 6 \, A a^{2} e^{6} + 42 \, {\left (2 \, C a c + A c^{2}\right )} d^{4} e^{2} + 18 \, {\left (C a^{2} + 2 \, A a c\right )} d^{2} e^{4} - 2 \, {\left (3 \, C c^{2} d e^{5} - 2 \, B c^{2} e^{6}\right )} x^{5} + {\left (15 \, C c^{2} d^{2} e^{4} - 10 \, B c^{2} d e^{5} + 6 \, {\left (2 \, C a c + A c^{2}\right )} e^{6}\right )} x^{4} - 4 \, {\left (15 \, C c^{2} d^{3} e^{3} - 10 \, B c^{2} d^{2} e^{4} - 6 \, B a c e^{6} + 6 \, {\left (2 \, C a c + A c^{2}\right )} d e^{5}\right )} x^{3} - 6 \, {\left (34 \, C c^{2} d^{4} e^{2} - 21 \, B c^{2} d^{3} e^{3} - 8 \, B a c d e^{5} + 11 \, {\left (2 \, C a c + A c^{2}\right )} d^{2} e^{4}\right )} x^{2} - 12 \, {\left (4 \, C c^{2} d^{5} e - B c^{2} d^{4} e^{2} + 4 \, B a c d^{2} e^{4} + B a^{2} e^{6} - {\left (2 \, C a c + A c^{2}\right )} d^{3} e^{3} - 2 \, {\left (C a^{2} + 2 \, A a c\right )} d e^{5}\right )} x + 12 \, {\left (15 \, C c^{2} d^{6} - 10 \, B c^{2} d^{5} e - 6 \, B a c d^{3} e^{3} + 6 \, {\left (2 \, C a c + A c^{2}\right )} d^{4} e^{2} + {\left (C a^{2} + 2 \, A a c\right )} d^{2} e^{4} + {\left (15 \, C c^{2} d^{4} e^{2} - 10 \, B c^{2} d^{3} e^{3} - 6 \, B a c d e^{5} + 6 \, {\left (2 \, C a c + A c^{2}\right )} d^{2} e^{4} + {\left (C a^{2} + 2 \, A a c\right )} e^{6}\right )} x^{2} + 2 \, {\left (15 \, C c^{2} d^{5} e - 10 \, B c^{2} d^{4} e^{2} - 6 \, B a c d^{2} e^{4} + 6 \, {\left (2 \, C a c + A c^{2}\right )} d^{3} e^{3} + {\left (C a^{2} + 2 \, A a c\right )} d e^{5}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{9} x^{2} + 2 \, d e^{8} x + d^{2} e^{7}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*C*c^2*e^6*x^6 + 66*C*c^2*d^6 - 54*B*c^2*d^5*e - 60*B*a*c*d^3*e^3 - 6*B*a^2*d*e^5 - 6*A*a^2*e^6 + 42*(2
*C*a*c + A*c^2)*d^4*e^2 + 18*(C*a^2 + 2*A*a*c)*d^2*e^4 - 2*(3*C*c^2*d*e^5 - 2*B*c^2*e^6)*x^5 + (15*C*c^2*d^2*e
^4 - 10*B*c^2*d*e^5 + 6*(2*C*a*c + A*c^2)*e^6)*x^4 - 4*(15*C*c^2*d^3*e^3 - 10*B*c^2*d^2*e^4 - 6*B*a*c*e^6 + 6*
(2*C*a*c + A*c^2)*d*e^5)*x^3 - 6*(34*C*c^2*d^4*e^2 - 21*B*c^2*d^3*e^3 - 8*B*a*c*d*e^5 + 11*(2*C*a*c + A*c^2)*d
^2*e^4)*x^2 - 12*(4*C*c^2*d^5*e - B*c^2*d^4*e^2 + 4*B*a*c*d^2*e^4 + B*a^2*e^6 - (2*C*a*c + A*c^2)*d^3*e^3 - 2*
(C*a^2 + 2*A*a*c)*d*e^5)*x + 12*(15*C*c^2*d^6 - 10*B*c^2*d^5*e - 6*B*a*c*d^3*e^3 + 6*(2*C*a*c + A*c^2)*d^4*e^2
+ (C*a^2 + 2*A*a*c)*d^2*e^4 + (15*C*c^2*d^4*e^2 - 10*B*c^2*d^3*e^3 - 6*B*a*c*d*e^5 + 6*(2*C*a*c + A*c^2)*d^2*
e^4 + (C*a^2 + 2*A*a*c)*e^6)*x^2 + 2*(15*C*c^2*d^5*e - 10*B*c^2*d^4*e^2 - 6*B*a*c*d^2*e^4 + 6*(2*C*a*c + A*c^2
)*d^3*e^3 + (C*a^2 + 2*A*a*c)*d*e^5)*x)*log(e*x + d))/(e^9*x^2 + 2*d*e^8*x + d^2*e^7)

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giac [A]  time = 0.16, size = 397, normalized size = 1.35 \[ {\left (15 \, C c^{2} d^{4} - 10 \, B c^{2} d^{3} e + 12 \, C a c d^{2} e^{2} + 6 \, A c^{2} d^{2} e^{2} - 6 \, B a c d e^{3} + C a^{2} e^{4} + 2 \, A a c e^{4}\right )} e^{\left (-7\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{12} \, {\left (3 \, C c^{2} x^{4} e^{9} - 12 \, C c^{2} d x^{3} e^{8} + 36 \, C c^{2} d^{2} x^{2} e^{7} - 120 \, C c^{2} d^{3} x e^{6} + 4 \, B c^{2} x^{3} e^{9} - 18 \, B c^{2} d x^{2} e^{8} + 72 \, B c^{2} d^{2} x e^{7} + 12 \, C a c x^{2} e^{9} + 6 \, A c^{2} x^{2} e^{9} - 72 \, C a c d x e^{8} - 36 \, A c^{2} d x e^{8} + 24 \, B a c x e^{9}\right )} e^{\left (-12\right )} + \frac {{\left (11 \, C c^{2} d^{6} - 9 \, B c^{2} d^{5} e + 14 \, C a c d^{4} e^{2} + 7 \, A c^{2} d^{4} e^{2} - 10 \, B a c d^{3} e^{3} + 3 \, C a^{2} d^{2} e^{4} + 6 \, A a c d^{2} e^{4} - B a^{2} d e^{5} - A a^{2} e^{6} + 2 \, {\left (6 \, C c^{2} d^{5} e - 5 \, B c^{2} d^{4} e^{2} + 8 \, C a c d^{3} e^{3} + 4 \, A c^{2} d^{3} e^{3} - 6 \, B a c d^{2} e^{4} + 2 \, C a^{2} d e^{5} + 4 \, A a c d e^{5} - B a^{2} e^{6}\right )} x\right )} e^{\left (-7\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*C*c^2*d^4 - 10*B*c^2*d^3*e + 12*C*a*c*d^2*e^2 + 6*A*c^2*d^2*e^2 - 6*B*a*c*d*e^3 + C*a^2*e^4 + 2*A*a*c*e^4)
*e^(-7)*log(abs(x*e + d)) + 1/12*(3*C*c^2*x^4*e^9 - 12*C*c^2*d*x^3*e^8 + 36*C*c^2*d^2*x^2*e^7 - 120*C*c^2*d^3*
x*e^6 + 4*B*c^2*x^3*e^9 - 18*B*c^2*d*x^2*e^8 + 72*B*c^2*d^2*x*e^7 + 12*C*a*c*x^2*e^9 + 6*A*c^2*x^2*e^9 - 72*C*
a*c*d*x*e^8 - 36*A*c^2*d*x*e^8 + 24*B*a*c*x*e^9)*e^(-12) + 1/2*(11*C*c^2*d^6 - 9*B*c^2*d^5*e + 14*C*a*c*d^4*e^
2 + 7*A*c^2*d^4*e^2 - 10*B*a*c*d^3*e^3 + 3*C*a^2*d^2*e^4 + 6*A*a*c*d^2*e^4 - B*a^2*d*e^5 - A*a^2*e^6 + 2*(6*C*
c^2*d^5*e - 5*B*c^2*d^4*e^2 + 8*C*a*c*d^3*e^3 + 4*A*c^2*d^3*e^3 - 6*B*a*c*d^2*e^4 + 2*C*a^2*d*e^5 + 4*A*a*c*d*
e^5 - B*a^2*e^6)*x)*e^(-7)/(x*e + d)^2

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maple [A]  time = 0.01, size = 563, normalized size = 1.91 \[ \frac {C \,c^{2} x^{4}}{4 e^{3}}+\frac {B \,c^{2} x^{3}}{3 e^{3}}-\frac {C \,c^{2} d \,x^{3}}{e^{4}}-\frac {A \,a^{2}}{2 \left (e x +d \right )^{2} e}-\frac {A a c \,d^{2}}{\left (e x +d \right )^{2} e^{3}}-\frac {A \,c^{2} d^{4}}{2 \left (e x +d \right )^{2} e^{5}}+\frac {A \,c^{2} x^{2}}{2 e^{3}}+\frac {B \,a^{2} d}{2 \left (e x +d \right )^{2} e^{2}}+\frac {B a c \,d^{3}}{\left (e x +d \right )^{2} e^{4}}+\frac {B \,c^{2} d^{5}}{2 \left (e x +d \right )^{2} e^{6}}-\frac {3 B \,c^{2} d \,x^{2}}{2 e^{4}}-\frac {C \,a^{2} d^{2}}{2 \left (e x +d \right )^{2} e^{3}}-\frac {C a c \,d^{4}}{\left (e x +d \right )^{2} e^{5}}+\frac {C a c \,x^{2}}{e^{3}}-\frac {C \,c^{2} d^{6}}{2 \left (e x +d \right )^{2} e^{7}}+\frac {3 C \,c^{2} d^{2} x^{2}}{e^{5}}+\frac {4 A a c d}{\left (e x +d \right ) e^{3}}+\frac {2 A a c \ln \left (e x +d \right )}{e^{3}}+\frac {4 A \,c^{2} d^{3}}{\left (e x +d \right ) e^{5}}+\frac {6 A \,c^{2} d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {3 A \,c^{2} d x}{e^{4}}-\frac {B \,a^{2}}{\left (e x +d \right ) e^{2}}-\frac {6 B a c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {6 B a c d \ln \left (e x +d \right )}{e^{4}}+\frac {2 B a c x}{e^{3}}-\frac {5 B \,c^{2} d^{4}}{\left (e x +d \right ) e^{6}}-\frac {10 B \,c^{2} d^{3} \ln \left (e x +d \right )}{e^{6}}+\frac {6 B \,c^{2} d^{2} x}{e^{5}}+\frac {2 C \,a^{2} d}{\left (e x +d \right ) e^{3}}+\frac {C \,a^{2} \ln \left (e x +d \right )}{e^{3}}+\frac {8 C a c \,d^{3}}{\left (e x +d \right ) e^{5}}+\frac {12 C a c \,d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {6 C a c d x}{e^{4}}+\frac {6 C \,c^{2} d^{5}}{\left (e x +d \right ) e^{7}}+\frac {15 C \,c^{2} d^{4} \ln \left (e x +d \right )}{e^{7}}-\frac {10 C \,c^{2} d^{3} x}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*c/e^4*C*x*a*d+4/e^3/(e*x+d)*A*a*c*d-6/e^4/(e*x+d)*B*a*c*d^2+8/e^5/(e*x+d)*C*a*c*d^3-6/e^4*ln(e*x+d)*B*a*c*d
+1/4*c^2*C*x^4/e^3+12/e^5*ln(e*x+d)*C*a*c*d^2-1/e^3/(e*x+d)^2*A*d^2*a*c+1/e^4/(e*x+d)^2*B*a*c*d^3-1/e^5/(e*x+d
)^2*C*a*c*d^4+1/e^3*ln(e*x+d)*a^2*C-1/2/e/(e*x+d)^2*A*a^2+1/3*c^2/e^3*B*x^3+1/2*c^2/e^3*A*x^2-1/e^2/(e*x+d)*B*
a^2-c^2/e^4*C*x^3*d-3/2*c^2/e^4*B*x^2*d-10/e^6*ln(e*x+d)*B*c^2*d^3+15/e^7*ln(e*x+d)*C*c^2*d^4-1/2/e^5/(e*x+d)^
2*A*c^2*d^4+1/2/e^2/(e*x+d)^2*B*d*a^2+1/2/e^6/(e*x+d)^2*B*c^2*d^5-1/2/e^3/(e*x+d)^2*C*d^2*a^2-1/2/e^7/(e*x+d)^
2*C*c^2*d^6+c/e^3*C*x^2*a+3*c^2/e^5*C*x^2*d^2-3*c^2/e^4*A*x*d+2*c/e^3*B*x*a+6*c^2/e^5*B*x*d^2-10*c^2/e^6*C*x*d
^3+4/e^5/(e*x+d)*A*c^2*d^3-5/e^6/(e*x+d)*B*c^2*d^4+2/e^3/(e*x+d)*C*a^2*d+6/e^7/(e*x+d)*C*c^2*d^5+2/e^3*ln(e*x+
d)*A*a*c+6/e^5*ln(e*x+d)*A*c^2*d^2

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maxima [A]  time = 0.49, size = 402, normalized size = 1.36 \[ \frac {11 \, C c^{2} d^{6} - 9 \, B c^{2} d^{5} e - 10 \, B a c d^{3} e^{3} - B a^{2} d e^{5} - A a^{2} e^{6} + 7 \, {\left (2 \, C a c + A c^{2}\right )} d^{4} e^{2} + 3 \, {\left (C a^{2} + 2 \, A a c\right )} d^{2} e^{4} + 2 \, {\left (6 \, C c^{2} d^{5} e - 5 \, B c^{2} d^{4} e^{2} - 6 \, B a c d^{2} e^{4} - B a^{2} e^{6} + 4 \, {\left (2 \, C a c + A c^{2}\right )} d^{3} e^{3} + 2 \, {\left (C a^{2} + 2 \, A a c\right )} d e^{5}\right )} x}{2 \, {\left (e^{9} x^{2} + 2 \, d e^{8} x + d^{2} e^{7}\right )}} + \frac {3 \, C c^{2} e^{3} x^{4} - 4 \, {\left (3 \, C c^{2} d e^{2} - B c^{2} e^{3}\right )} x^{3} + 6 \, {\left (6 \, C c^{2} d^{2} e - 3 \, B c^{2} d e^{2} + {\left (2 \, C a c + A c^{2}\right )} e^{3}\right )} x^{2} - 12 \, {\left (10 \, C c^{2} d^{3} - 6 \, B c^{2} d^{2} e - 2 \, B a c e^{3} + 3 \, {\left (2 \, C a c + A c^{2}\right )} d e^{2}\right )} x}{12 \, e^{6}} + \frac {{\left (15 \, C c^{2} d^{4} - 10 \, B c^{2} d^{3} e - 6 \, B a c d e^{3} + 6 \, {\left (2 \, C a c + A c^{2}\right )} d^{2} e^{2} + {\left (C a^{2} + 2 \, A a c\right )} e^{4}\right )} \log \left (e x + d\right )}{e^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(11*C*c^2*d^6 - 9*B*c^2*d^5*e - 10*B*a*c*d^3*e^3 - B*a^2*d*e^5 - A*a^2*e^6 + 7*(2*C*a*c + A*c^2)*d^4*e^2 +
3*(C*a^2 + 2*A*a*c)*d^2*e^4 + 2*(6*C*c^2*d^5*e - 5*B*c^2*d^4*e^2 - 6*B*a*c*d^2*e^4 - B*a^2*e^6 + 4*(2*C*a*c +
A*c^2)*d^3*e^3 + 2*(C*a^2 + 2*A*a*c)*d*e^5)*x)/(e^9*x^2 + 2*d*e^8*x + d^2*e^7) + 1/12*(3*C*c^2*e^3*x^4 - 4*(3
*C*c^2*d*e^2 - B*c^2*e^3)*x^3 + 6*(6*C*c^2*d^2*e - 3*B*c^2*d*e^2 + (2*C*a*c + A*c^2)*e^3)*x^2 - 12*(10*C*c^2*d
^3 - 6*B*c^2*d^2*e - 2*B*a*c*e^3 + 3*(2*C*a*c + A*c^2)*d*e^2)*x)/e^6 + (15*C*c^2*d^4 - 10*B*c^2*d^3*e - 6*B*a*
c*d*e^3 + 6*(2*C*a*c + A*c^2)*d^2*e^2 + (C*a^2 + 2*A*a*c)*e^4)*log(e*x + d)/e^7

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mupad [B]  time = 3.82, size = 495, normalized size = 1.68 \[ x\,\left (\frac {3\,d\,\left (\frac {3\,d\,\left (\frac {B\,c^2}{e^3}-\frac {3\,C\,c^2\,d}{e^4}\right )}{e}-\frac {A\,c^2+2\,C\,a\,c}{e^3}+\frac {3\,C\,c^2\,d^2}{e^5}\right )}{e}-\frac {3\,d^2\,\left (\frac {B\,c^2}{e^3}-\frac {3\,C\,c^2\,d}{e^4}\right )}{e^2}+\frac {2\,B\,a\,c}{e^3}-\frac {C\,c^2\,d^3}{e^6}\right )+x^3\,\left (\frac {B\,c^2}{3\,e^3}-\frac {C\,c^2\,d}{e^4}\right )-x^2\,\left (\frac {3\,d\,\left (\frac {B\,c^2}{e^3}-\frac {3\,C\,c^2\,d}{e^4}\right )}{2\,e}-\frac {A\,c^2+2\,C\,a\,c}{2\,e^3}+\frac {3\,C\,c^2\,d^2}{2\,e^5}\right )+\frac {\frac {3\,C\,a^2\,d^2\,e^4-B\,a^2\,d\,e^5-A\,a^2\,e^6+14\,C\,a\,c\,d^4\,e^2-10\,B\,a\,c\,d^3\,e^3+6\,A\,a\,c\,d^2\,e^4+11\,C\,c^2\,d^6-9\,B\,c^2\,d^5\,e+7\,A\,c^2\,d^4\,e^2}{2\,e}+x\,\left (2\,C\,a^2\,d\,e^4-B\,a^2\,e^5+8\,C\,a\,c\,d^3\,e^2-6\,B\,a\,c\,d^2\,e^3+4\,A\,a\,c\,d\,e^4+6\,C\,c^2\,d^5-5\,B\,c^2\,d^4\,e+4\,A\,c^2\,d^3\,e^2\right )}{d^2\,e^6+2\,d\,e^7\,x+e^8\,x^2}+\frac {\ln \left (d+e\,x\right )\,\left (C\,a^2\,e^4+12\,C\,a\,c\,d^2\,e^2-6\,B\,a\,c\,d\,e^3+2\,A\,a\,c\,e^4+15\,C\,c^2\,d^4-10\,B\,c^2\,d^3\,e+6\,A\,c^2\,d^2\,e^2\right )}{e^7}+\frac {C\,c^2\,x^4}{4\,e^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((3*d*((3*d*((B*c^2)/e^3 - (3*C*c^2*d)/e^4))/e - (A*c^2 + 2*C*a*c)/e^3 + (3*C*c^2*d^2)/e^5))/e - (3*d^2*((B*
c^2)/e^3 - (3*C*c^2*d)/e^4))/e^2 + (2*B*a*c)/e^3 - (C*c^2*d^3)/e^6) + x^3*((B*c^2)/(3*e^3) - (C*c^2*d)/e^4) -
x^2*((3*d*((B*c^2)/e^3 - (3*C*c^2*d)/e^4))/(2*e) - (A*c^2 + 2*C*a*c)/(2*e^3) + (3*C*c^2*d^2)/(2*e^5)) + ((11*C
*c^2*d^6 - A*a^2*e^6 - B*a^2*d*e^5 - 9*B*c^2*d^5*e + 7*A*c^2*d^4*e^2 + 3*C*a^2*d^2*e^4 + 6*A*a*c*d^2*e^4 - 10*
B*a*c*d^3*e^3 + 14*C*a*c*d^4*e^2)/(2*e) + x*(6*C*c^2*d^5 - B*a^2*e^5 + 2*C*a^2*d*e^4 - 5*B*c^2*d^4*e + 4*A*c^2
*d^3*e^2 + 4*A*a*c*d*e^4 - 6*B*a*c*d^2*e^3 + 8*C*a*c*d^3*e^2))/(d^2*e^6 + e^8*x^2 + 2*d*e^7*x) + (log(d + e*x)
*(C*a^2*e^4 + 15*C*c^2*d^4 + 2*A*a*c*e^4 - 10*B*c^2*d^3*e + 6*A*c^2*d^2*e^2 - 6*B*a*c*d*e^3 + 12*C*a*c*d^2*e^2
))/e^7 + (C*c^2*x^4)/(4*e^3)

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sympy [A]  time = 14.20, size = 474, normalized size = 1.61 \[ \frac {C c^{2} x^{4}}{4 e^{3}} + x^{3} \left (\frac {B c^{2}}{3 e^{3}} - \frac {C c^{2} d}{e^{4}}\right ) + x^{2} \left (\frac {A c^{2}}{2 e^{3}} - \frac {3 B c^{2} d}{2 e^{4}} + \frac {C a c}{e^{3}} + \frac {3 C c^{2} d^{2}}{e^{5}}\right ) + x \left (- \frac {3 A c^{2} d}{e^{4}} + \frac {2 B a c}{e^{3}} + \frac {6 B c^{2} d^{2}}{e^{5}} - \frac {6 C a c d}{e^{4}} - \frac {10 C c^{2} d^{3}}{e^{6}}\right ) + \frac {- A a^{2} e^{6} + 6 A a c d^{2} e^{4} + 7 A c^{2} d^{4} e^{2} - B a^{2} d e^{5} - 10 B a c d^{3} e^{3} - 9 B c^{2} d^{5} e + 3 C a^{2} d^{2} e^{4} + 14 C a c d^{4} e^{2} + 11 C c^{2} d^{6} + x \left (8 A a c d e^{5} + 8 A c^{2} d^{3} e^{3} - 2 B a^{2} e^{6} - 12 B a c d^{2} e^{4} - 10 B c^{2} d^{4} e^{2} + 4 C a^{2} d e^{5} + 16 C a c d^{3} e^{3} + 12 C c^{2} d^{5} e\right )}{2 d^{2} e^{7} + 4 d e^{8} x + 2 e^{9} x^{2}} + \frac {\left (2 A a c e^{4} + 6 A c^{2} d^{2} e^{2} - 6 B a c d e^{3} - 10 B c^{2} d^{3} e + C a^{2} e^{4} + 12 C a c d^{2} e^{2} + 15 C c^{2} d^{4}\right ) \log {\left (d + e x \right )}}{e^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C*c**2*x**4/(4*e**3) + x**3*(B*c**2/(3*e**3) - C*c**2*d/e**4) + x**2*(A*c**2/(2*e**3) - 3*B*c**2*d/(2*e**4) +
C*a*c/e**3 + 3*C*c**2*d**2/e**5) + x*(-3*A*c**2*d/e**4 + 2*B*a*c/e**3 + 6*B*c**2*d**2/e**5 - 6*C*a*c*d/e**4 -
10*C*c**2*d**3/e**6) + (-A*a**2*e**6 + 6*A*a*c*d**2*e**4 + 7*A*c**2*d**4*e**2 - B*a**2*d*e**5 - 10*B*a*c*d**3*
e**3 - 9*B*c**2*d**5*e + 3*C*a**2*d**2*e**4 + 14*C*a*c*d**4*e**2 + 11*C*c**2*d**6 + x*(8*A*a*c*d*e**5 + 8*A*c*
*2*d**3*e**3 - 2*B*a**2*e**6 - 12*B*a*c*d**2*e**4 - 10*B*c**2*d**4*e**2 + 4*C*a**2*d*e**5 + 16*C*a*c*d**3*e**3
+ 12*C*c**2*d**5*e))/(2*d**2*e**7 + 4*d*e**8*x + 2*e**9*x**2) + (2*A*a*c*e**4 + 6*A*c**2*d**2*e**2 - 6*B*a*c*
d*e**3 - 10*B*c**2*d**3*e + C*a**2*e**4 + 12*C*a*c*d**2*e**2 + 15*C*c**2*d**4)*log(d + e*x)/e**7

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