3.21.87 \(\int \frac {(A+B x) (a+b x+c x^2)^2}{(d+e x)^5} \, dx\)

Optimal. Leaf size=289 \[ \frac {2 A c e (2 c d-b e)-B \left (-2 c e (4 b d-a e)+b^2 e^2+10 c^2 d^2\right )}{e^6 (d+e x)}+\frac {B \left (-6 c d e (2 b d-a e)+b e^2 (3 b d-2 a e)+10 c^2 d^3\right )-A e \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{2 e^6 (d+e x)^2}+\frac {\left (a e^2-b d e+c d^2\right ) \left (2 A e (2 c d-b e)-B \left (5 c d^2-e (3 b d-a e)\right )\right )}{3 e^6 (d+e x)^3}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )^2}{4 e^6 (d+e x)^4}-\frac {c \log (d+e x) (-A c e-2 b B e+5 B c d)}{e^6}+\frac {B c^2 x}{e^5} \]

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Rubi [A]  time = 0.36, antiderivative size = 287, 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 = {771} \begin {gather*} \frac {2 A c e (2 c d-b e)-B \left (-2 c e (4 b d-a e)+b^2 e^2+10 c^2 d^2\right )}{e^6 (d+e x)}+\frac {B \left (-6 c d e (2 b d-a e)+b e^2 (3 b d-2 a e)+10 c^2 d^3\right )-A e \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{2 e^6 (d+e x)^2}-\frac {\left (a e^2-b d e+c d^2\right ) \left (-B e (3 b d-a e)-2 A e (2 c d-b e)+5 B c d^2\right )}{3 e^6 (d+e x)^3}+\frac {(B d-A e) \left (a e^2-b d e+c d^2\right )^2}{4 e^6 (d+e x)^4}-\frac {c \log (d+e x) (-A c e-2 b B e+5 B c d)}{e^6}+\frac {B c^2 x}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/(d + e*x)^5,x]

[Out]

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

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 391, normalized size = 1.35 \begin {gather*} -\frac {A e \left (e^2 \left (3 a^2 e^2+2 a b e (d+4 e x)+b^2 \left (d^2+4 d e x+6 e^2 x^2\right )\right )+2 c e \left (a e \left (d^2+4 d e x+6 e^2 x^2\right )+3 b \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+c^2 (-d) \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )+B \left (e^2 \left (a^2 e^2 (d+4 e x)+2 a b e \left (d^2+4 d e x+6 e^2 x^2\right )+3 b^2 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+2 c e \left (3 a e \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )-b d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )+c^2 \left (77 d^5+248 d^4 e x+252 d^3 e^2 x^2+48 d^2 e^3 x^3-48 d e^4 x^4-12 e^5 x^5\right )\right )+12 c (d+e x)^4 \log (d+e x) (-A c e-2 b B e+5 B c d)}{12 e^6 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/(d + e*x)^5,x]

[Out]

-1/12*(A*e*(-(c^2*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3*x^3)) + e^2*(3*a^2*e^2 + 2*a*b*e*(d + 4*e*x)
+ b^2*(d^2 + 4*d*e*x + 6*e^2*x^2)) + 2*c*e*(a*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3*b*(d^3 + 4*d^2*e*x + 6*d*e^2*
x^2 + 4*e^3*x^3))) + B*(c^2*(77*d^5 + 248*d^4*e*x + 252*d^3*e^2*x^2 + 48*d^2*e^3*x^3 - 48*d*e^4*x^4 - 12*e^5*x
^5) + e^2*(a^2*e^2*(d + 4*e*x) + 2*a*b*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + 3*b^2*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 +
4*e^3*x^3)) + 2*c*e*(3*a*e*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) - b*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*
x^2 + 48*e^3*x^3))) + 12*c*(5*B*c*d - 2*b*B*e - A*c*e)*(d + e*x)^4*Log[d + e*x])/(e^6*(d + e*x)^4)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{(d+e x)^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^2)/(d + e*x)^5,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^2)/(d + e*x)^5, x]

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fricas [B]  time = 0.38, size = 572, normalized size = 1.98 \begin {gather*} \frac {12 \, B c^{2} e^{5} x^{5} + 48 \, B c^{2} d e^{4} x^{4} - 77 \, B c^{2} d^{5} - 3 \, A a^{2} e^{5} + 25 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e - 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{3} e^{2} - {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} - {\left (B a^{2} + 2 \, A a b\right )} d e^{4} - 12 \, {\left (4 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} e^{5}\right )} x^{3} - 6 \, {\left (42 \, B c^{2} d^{3} e^{2} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d e^{4} + {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{5}\right )} x^{2} - 4 \, {\left (62 \, B c^{2} d^{4} e - 22 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{2} e^{3} + {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{4} + {\left (B a^{2} + 2 \, A a b\right )} e^{5}\right )} x - 12 \, {\left (5 \, B c^{2} d^{5} - {\left (2 \, B b c + A c^{2}\right )} d^{4} e + {\left (5 \, B c^{2} d e^{4} - {\left (2 \, B b c + A c^{2}\right )} e^{5}\right )} x^{4} + 4 \, {\left (5 \, B c^{2} d^{2} e^{3} - {\left (2 \, B b c + A c^{2}\right )} d e^{4}\right )} x^{3} + 6 \, {\left (5 \, B c^{2} d^{3} e^{2} - {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3}\right )} x^{2} + 4 \, {\left (5 \, B c^{2} d^{4} e - {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="fricas")

[Out]

1/12*(12*B*c^2*e^5*x^5 + 48*B*c^2*d*e^4*x^4 - 77*B*c^2*d^5 - 3*A*a^2*e^5 + 25*(2*B*b*c + A*c^2)*d^4*e - 3*(B*b
^2 + 2*(B*a + A*b)*c)*d^3*e^2 - (2*B*a*b + A*b^2 + 2*A*a*c)*d^2*e^3 - (B*a^2 + 2*A*a*b)*d*e^4 - 12*(4*B*c^2*d^
2*e^3 - 4*(2*B*b*c + A*c^2)*d*e^4 + (B*b^2 + 2*(B*a + A*b)*c)*e^5)*x^3 - 6*(42*B*c^2*d^3*e^2 - 18*(2*B*b*c + A
*c^2)*d^2*e^3 + 3*(B*b^2 + 2*(B*a + A*b)*c)*d*e^4 + (2*B*a*b + A*b^2 + 2*A*a*c)*e^5)*x^2 - 4*(62*B*c^2*d^4*e -
22*(2*B*b*c + A*c^2)*d^3*e^2 + 3*(B*b^2 + 2*(B*a + A*b)*c)*d^2*e^3 + (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^4 + (B*a
^2 + 2*A*a*b)*e^5)*x - 12*(5*B*c^2*d^5 - (2*B*b*c + A*c^2)*d^4*e + (5*B*c^2*d*e^4 - (2*B*b*c + A*c^2)*e^5)*x^4
+ 4*(5*B*c^2*d^2*e^3 - (2*B*b*c + A*c^2)*d*e^4)*x^3 + 6*(5*B*c^2*d^3*e^2 - (2*B*b*c + A*c^2)*d^2*e^3)*x^2 + 4
*(5*B*c^2*d^4*e - (2*B*b*c + A*c^2)*d^3*e^2)*x)*log(e*x + d))/(e^10*x^4 + 4*d*e^9*x^3 + 6*d^2*e^8*x^2 + 4*d^3*
e^7*x + d^4*e^6)

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giac [B]  time = 0.21, size = 719, normalized size = 2.49 \begin {gather*} {\left (x e + d\right )} B c^{2} e^{\left (-6\right )} + {\left (5 \, B c^{2} d - 2 \, B b c e - A c^{2} e\right )} e^{\left (-6\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - \frac {1}{12} \, {\left (\frac {120 \, B c^{2} d^{2} e^{22}}{x e + d} - \frac {60 \, B c^{2} d^{3} e^{22}}{{\left (x e + d\right )}^{2}} + \frac {20 \, B c^{2} d^{4} e^{22}}{{\left (x e + d\right )}^{3}} - \frac {3 \, B c^{2} d^{5} e^{22}}{{\left (x e + d\right )}^{4}} - \frac {96 \, B b c d e^{23}}{x e + d} - \frac {48 \, A c^{2} d e^{23}}{x e + d} + \frac {72 \, B b c d^{2} e^{23}}{{\left (x e + d\right )}^{2}} + \frac {36 \, A c^{2} d^{2} e^{23}}{{\left (x e + d\right )}^{2}} - \frac {32 \, B b c d^{3} e^{23}}{{\left (x e + d\right )}^{3}} - \frac {16 \, A c^{2} d^{3} e^{23}}{{\left (x e + d\right )}^{3}} + \frac {6 \, B b c d^{4} e^{23}}{{\left (x e + d\right )}^{4}} + \frac {3 \, A c^{2} d^{4} e^{23}}{{\left (x e + d\right )}^{4}} + \frac {12 \, B b^{2} e^{24}}{x e + d} + \frac {24 \, B a c e^{24}}{x e + d} + \frac {24 \, A b c e^{24}}{x e + d} - \frac {18 \, B b^{2} d e^{24}}{{\left (x e + d\right )}^{2}} - \frac {36 \, B a c d e^{24}}{{\left (x e + d\right )}^{2}} - \frac {36 \, A b c d e^{24}}{{\left (x e + d\right )}^{2}} + \frac {12 \, B b^{2} d^{2} e^{24}}{{\left (x e + d\right )}^{3}} + \frac {24 \, B a c d^{2} e^{24}}{{\left (x e + d\right )}^{3}} + \frac {24 \, A b c d^{2} e^{24}}{{\left (x e + d\right )}^{3}} - \frac {3 \, B b^{2} d^{3} e^{24}}{{\left (x e + d\right )}^{4}} - \frac {6 \, B a c d^{3} e^{24}}{{\left (x e + d\right )}^{4}} - \frac {6 \, A b c d^{3} e^{24}}{{\left (x e + d\right )}^{4}} + \frac {12 \, B a b e^{25}}{{\left (x e + d\right )}^{2}} + \frac {6 \, A b^{2} e^{25}}{{\left (x e + d\right )}^{2}} + \frac {12 \, A a c e^{25}}{{\left (x e + d\right )}^{2}} - \frac {16 \, B a b d e^{25}}{{\left (x e + d\right )}^{3}} - \frac {8 \, A b^{2} d e^{25}}{{\left (x e + d\right )}^{3}} - \frac {16 \, A a c d e^{25}}{{\left (x e + d\right )}^{3}} + \frac {6 \, B a b d^{2} e^{25}}{{\left (x e + d\right )}^{4}} + \frac {3 \, A b^{2} d^{2} e^{25}}{{\left (x e + d\right )}^{4}} + \frac {6 \, A a c d^{2} e^{25}}{{\left (x e + d\right )}^{4}} + \frac {4 \, B a^{2} e^{26}}{{\left (x e + d\right )}^{3}} + \frac {8 \, A a b e^{26}}{{\left (x e + d\right )}^{3}} - \frac {3 \, B a^{2} d e^{26}}{{\left (x e + d\right )}^{4}} - \frac {6 \, A a b d e^{26}}{{\left (x e + d\right )}^{4}} + \frac {3 \, A a^{2} e^{27}}{{\left (x e + d\right )}^{4}}\right )} e^{\left (-28\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="giac")

[Out]

(x*e + d)*B*c^2*e^(-6) + (5*B*c^2*d - 2*B*b*c*e - A*c^2*e)*e^(-6)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - 1/12*
(120*B*c^2*d^2*e^22/(x*e + d) - 60*B*c^2*d^3*e^22/(x*e + d)^2 + 20*B*c^2*d^4*e^22/(x*e + d)^3 - 3*B*c^2*d^5*e^
22/(x*e + d)^4 - 96*B*b*c*d*e^23/(x*e + d) - 48*A*c^2*d*e^23/(x*e + d) + 72*B*b*c*d^2*e^23/(x*e + d)^2 + 36*A*
c^2*d^2*e^23/(x*e + d)^2 - 32*B*b*c*d^3*e^23/(x*e + d)^3 - 16*A*c^2*d^3*e^23/(x*e + d)^3 + 6*B*b*c*d^4*e^23/(x
*e + d)^4 + 3*A*c^2*d^4*e^23/(x*e + d)^4 + 12*B*b^2*e^24/(x*e + d) + 24*B*a*c*e^24/(x*e + d) + 24*A*b*c*e^24/(
x*e + d) - 18*B*b^2*d*e^24/(x*e + d)^2 - 36*B*a*c*d*e^24/(x*e + d)^2 - 36*A*b*c*d*e^24/(x*e + d)^2 + 12*B*b^2*
d^2*e^24/(x*e + d)^3 + 24*B*a*c*d^2*e^24/(x*e + d)^3 + 24*A*b*c*d^2*e^24/(x*e + d)^3 - 3*B*b^2*d^3*e^24/(x*e +
d)^4 - 6*B*a*c*d^3*e^24/(x*e + d)^4 - 6*A*b*c*d^3*e^24/(x*e + d)^4 + 12*B*a*b*e^25/(x*e + d)^2 + 6*A*b^2*e^25
/(x*e + d)^2 + 12*A*a*c*e^25/(x*e + d)^2 - 16*B*a*b*d*e^25/(x*e + d)^3 - 8*A*b^2*d*e^25/(x*e + d)^3 - 16*A*a*c
*d*e^25/(x*e + d)^3 + 6*B*a*b*d^2*e^25/(x*e + d)^4 + 3*A*b^2*d^2*e^25/(x*e + d)^4 + 6*A*a*c*d^2*e^25/(x*e + d)
^4 + 4*B*a^2*e^26/(x*e + d)^3 + 8*A*a*b*e^26/(x*e + d)^3 - 3*B*a^2*d*e^26/(x*e + d)^4 - 6*A*a*b*d*e^26/(x*e +
d)^4 + 3*A*a^2*e^27/(x*e + d)^4)*e^(-28)

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maple [B]  time = 0.05, size = 710, normalized size = 2.46 \begin {gather*} -\frac {A \,a^{2}}{4 \left (e x +d \right )^{4} e}+\frac {A a b d}{2 \left (e x +d \right )^{4} e^{2}}-\frac {A a c \,d^{2}}{2 \left (e x +d \right )^{4} e^{3}}-\frac {A \,b^{2} d^{2}}{4 \left (e x +d \right )^{4} e^{3}}+\frac {A b c \,d^{3}}{2 \left (e x +d \right )^{4} e^{4}}-\frac {A \,c^{2} d^{4}}{4 \left (e x +d \right )^{4} e^{5}}+\frac {B \,a^{2} d}{4 \left (e x +d \right )^{4} e^{2}}-\frac {B a b \,d^{2}}{2 \left (e x +d \right )^{4} e^{3}}+\frac {B a c \,d^{3}}{2 \left (e x +d \right )^{4} e^{4}}+\frac {B \,b^{2} d^{3}}{4 \left (e x +d \right )^{4} e^{4}}-\frac {B b c \,d^{4}}{2 \left (e x +d \right )^{4} e^{5}}+\frac {B \,c^{2} d^{5}}{4 \left (e x +d \right )^{4} e^{6}}-\frac {2 A a b}{3 \left (e x +d \right )^{3} e^{2}}+\frac {4 A a c d}{3 \left (e x +d \right )^{3} e^{3}}+\frac {2 A \,b^{2} d}{3 \left (e x +d \right )^{3} e^{3}}-\frac {2 A b c \,d^{2}}{\left (e x +d \right )^{3} e^{4}}+\frac {4 A \,c^{2} d^{3}}{3 \left (e x +d \right )^{3} e^{5}}-\frac {B \,a^{2}}{3 \left (e x +d \right )^{3} e^{2}}+\frac {4 B a b d}{3 \left (e x +d \right )^{3} e^{3}}-\frac {2 B a c \,d^{2}}{\left (e x +d \right )^{3} e^{4}}-\frac {B \,b^{2} d^{2}}{\left (e x +d \right )^{3} e^{4}}+\frac {8 B b c \,d^{3}}{3 \left (e x +d \right )^{3} e^{5}}-\frac {5 B \,c^{2} d^{4}}{3 \left (e x +d \right )^{3} e^{6}}-\frac {A a c}{\left (e x +d \right )^{2} e^{3}}-\frac {A \,b^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {3 A b c d}{\left (e x +d \right )^{2} e^{4}}-\frac {3 A \,c^{2} d^{2}}{\left (e x +d \right )^{2} e^{5}}-\frac {B a b}{\left (e x +d \right )^{2} e^{3}}+\frac {3 B a c d}{\left (e x +d \right )^{2} e^{4}}+\frac {3 B \,b^{2} d}{2 \left (e x +d \right )^{2} e^{4}}-\frac {6 B b c \,d^{2}}{\left (e x +d \right )^{2} e^{5}}+\frac {5 B \,c^{2} d^{3}}{\left (e x +d \right )^{2} e^{6}}-\frac {2 A b c}{\left (e x +d \right ) e^{4}}+\frac {4 A \,c^{2} d}{\left (e x +d \right ) e^{5}}+\frac {A \,c^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {2 B a c}{\left (e x +d \right ) e^{4}}-\frac {B \,b^{2}}{\left (e x +d \right ) e^{4}}+\frac {8 B b c d}{\left (e x +d \right ) e^{5}}+\frac {2 B b c \ln \left (e x +d \right )}{e^{5}}-\frac {10 B \,c^{2} d^{2}}{\left (e x +d \right ) e^{6}}-\frac {5 B \,c^{2} d \ln \left (e x +d \right )}{e^{6}}+\frac {B \,c^{2} x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^2/(e*x+d)^5,x)

[Out]

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

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maxima [A]  time = 0.78, size = 417, normalized size = 1.44 \begin {gather*} -\frac {77 \, B c^{2} d^{5} + 3 \, A a^{2} e^{5} - 25 \, {\left (2 \, B b c + A c^{2}\right )} d^{4} e + 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{3} e^{2} + {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d^{2} e^{3} + {\left (B a^{2} + 2 \, A a b\right )} d e^{4} + 12 \, {\left (10 \, B c^{2} d^{2} e^{3} - 4 \, {\left (2 \, B b c + A c^{2}\right )} d e^{4} + {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} e^{5}\right )} x^{3} + 6 \, {\left (50 \, B c^{2} d^{3} e^{2} - 18 \, {\left (2 \, B b c + A c^{2}\right )} d^{2} e^{3} + 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d e^{4} + {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} e^{5}\right )} x^{2} + 4 \, {\left (65 \, B c^{2} d^{4} e - 22 \, {\left (2 \, B b c + A c^{2}\right )} d^{3} e^{2} + 3 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} d^{2} e^{3} + {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} d e^{4} + {\left (B a^{2} + 2 \, A a b\right )} e^{5}\right )} x}{12 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )}} + \frac {B c^{2} x}{e^{5}} - \frac {{\left (5 \, B c^{2} d - {\left (2 \, B b c + A c^{2}\right )} e\right )} \log \left (e x + d\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/(e*x+d)^5,x, algorithm="maxima")

[Out]

-1/12*(77*B*c^2*d^5 + 3*A*a^2*e^5 - 25*(2*B*b*c + A*c^2)*d^4*e + 3*(B*b^2 + 2*(B*a + A*b)*c)*d^3*e^2 + (2*B*a*
b + A*b^2 + 2*A*a*c)*d^2*e^3 + (B*a^2 + 2*A*a*b)*d*e^4 + 12*(10*B*c^2*d^2*e^3 - 4*(2*B*b*c + A*c^2)*d*e^4 + (B
*b^2 + 2*(B*a + A*b)*c)*e^5)*x^3 + 6*(50*B*c^2*d^3*e^2 - 18*(2*B*b*c + A*c^2)*d^2*e^3 + 3*(B*b^2 + 2*(B*a + A*
b)*c)*d*e^4 + (2*B*a*b + A*b^2 + 2*A*a*c)*e^5)*x^2 + 4*(65*B*c^2*d^4*e - 22*(2*B*b*c + A*c^2)*d^3*e^2 + 3*(B*b
^2 + 2*(B*a + A*b)*c)*d^2*e^3 + (2*B*a*b + A*b^2 + 2*A*a*c)*d*e^4 + (B*a^2 + 2*A*a*b)*e^5)*x)/(e^10*x^4 + 4*d*
e^9*x^3 + 6*d^2*e^8*x^2 + 4*d^3*e^7*x + d^4*e^6) + B*c^2*x/e^5 - (5*B*c^2*d - (2*B*b*c + A*c^2)*e)*log(e*x + d
)/e^6

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mupad [B]  time = 2.42, size = 475, normalized size = 1.64 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,c^2\,e-5\,B\,c^2\,d+2\,B\,b\,c\,e\right )}{e^6}-\frac {x^3\,\left (B\,b^2\,e^4-8\,B\,b\,c\,d\,e^3+2\,A\,b\,c\,e^4+10\,B\,c^2\,d^2\,e^2-4\,A\,c^2\,d\,e^3+2\,B\,a\,c\,e^4\right )+x^2\,\left (\frac {3\,B\,b^2\,d\,e^3}{2}+\frac {A\,b^2\,e^4}{2}-18\,B\,b\,c\,d^2\,e^2+3\,A\,b\,c\,d\,e^3+B\,a\,b\,e^4+25\,B\,c^2\,d^3\,e-9\,A\,c^2\,d^2\,e^2+3\,B\,a\,c\,d\,e^3+A\,a\,c\,e^4\right )+x\,\left (\frac {B\,a^2\,e^4}{3}+\frac {2\,B\,a\,b\,d\,e^3}{3}+\frac {2\,A\,a\,b\,e^4}{3}+2\,B\,a\,c\,d^2\,e^2+\frac {2\,A\,a\,c\,d\,e^3}{3}+B\,b^2\,d^2\,e^2+\frac {A\,b^2\,d\,e^3}{3}-\frac {44\,B\,b\,c\,d^3\,e}{3}+2\,A\,b\,c\,d^2\,e^2+\frac {65\,B\,c^2\,d^4}{3}-\frac {22\,A\,c^2\,d^3\,e}{3}\right )+\frac {B\,a^2\,d\,e^4+3\,A\,a^2\,e^5+2\,B\,a\,b\,d^2\,e^3+2\,A\,a\,b\,d\,e^4+6\,B\,a\,c\,d^3\,e^2+2\,A\,a\,c\,d^2\,e^3+3\,B\,b^2\,d^3\,e^2+A\,b^2\,d^2\,e^3-50\,B\,b\,c\,d^4\,e+6\,A\,b\,c\,d^3\,e^2+77\,B\,c^2\,d^5-25\,A\,c^2\,d^4\,e}{12\,e}}{d^4\,e^5+4\,d^3\,e^6\,x+6\,d^2\,e^7\,x^2+4\,d\,e^8\,x^3+e^9\,x^4}+\frac {B\,c^2\,x}{e^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^2)/(d + e*x)^5,x)

[Out]

(log(d + e*x)*(A*c^2*e - 5*B*c^2*d + 2*B*b*c*e))/e^6 - (x^3*(B*b^2*e^4 + 2*A*b*c*e^4 + 2*B*a*c*e^4 - 4*A*c^2*d
*e^3 + 10*B*c^2*d^2*e^2 - 8*B*b*c*d*e^3) + x^2*((A*b^2*e^4)/2 + A*a*c*e^4 + B*a*b*e^4 + (3*B*b^2*d*e^3)/2 + 25
*B*c^2*d^3*e - 9*A*c^2*d^2*e^2 + 3*A*b*c*d*e^3 + 3*B*a*c*d*e^3 - 18*B*b*c*d^2*e^2) + x*((B*a^2*e^4)/3 + (65*B*
c^2*d^4)/3 + (2*A*a*b*e^4)/3 + (A*b^2*d*e^3)/3 - (22*A*c^2*d^3*e)/3 + B*b^2*d^2*e^2 + (2*A*a*c*d*e^3)/3 + (2*B
*a*b*d*e^3)/3 - (44*B*b*c*d^3*e)/3 + 2*A*b*c*d^2*e^2 + 2*B*a*c*d^2*e^2) + (3*A*a^2*e^5 + 77*B*c^2*d^5 + B*a^2*
d*e^4 - 25*A*c^2*d^4*e + A*b^2*d^2*e^3 + 3*B*b^2*d^3*e^2 + 2*A*a*b*d*e^4 - 50*B*b*c*d^4*e + 2*A*a*c*d^2*e^3 +
2*B*a*b*d^2*e^3 + 6*A*b*c*d^3*e^2 + 6*B*a*c*d^3*e^2)/(12*e))/(d^4*e^5 + e^9*x^4 + 4*d^3*e^6*x + 4*d*e^8*x^3 +
6*d^2*e^7*x^2) + (B*c^2*x)/e^5

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**2/(e*x+d)**5,x)

[Out]

Timed out

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