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

Optimal. Leaf size=188 \[ -\frac {b^3 (d+e x)^3 (-4 a B e-A b e+5 b B d)}{3 e^6}+\frac {b^2 (d+e x)^2 (b d-a e) (-3 a B e-2 A b e+5 b B d)}{e^6}+\frac {(b d-a e)^4 (B d-A e)}{e^6 (d+e x)}+\frac {(b d-a e)^3 \log (d+e x) (-a B e-4 A b e+5 b B d)}{e^6}-\frac {2 b x (b d-a e)^2 (-2 a B e-3 A b e+5 b B d)}{e^5}+\frac {b^4 B (d+e x)^4}{4 e^6} \]

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Rubi [A]  time = 0.32, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {27, 77} \begin {gather*} \frac {b^2 (d+e x)^2 (b d-a e) (-3 a B e-2 A b e+5 b B d)}{e^6}-\frac {b^3 (d+e x)^3 (-4 a B e-A b e+5 b B d)}{3 e^6}+\frac {(b d-a e)^4 (B d-A e)}{e^6 (d+e x)}-\frac {2 b x (b d-a e)^2 (-2 a B e-3 A b e+5 b B d)}{e^5}+\frac {(b d-a e)^3 \log (d+e x) (-a B e-4 A b e+5 b B d)}{e^6}+\frac {b^4 B (d+e x)^4}{4 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 354, normalized size = 1.88 \begin {gather*} \frac {12 a^4 e^4 (B d-A e)+48 a^3 b e^3 \left (A d e+B \left (-d^2+d e x+e^2 x^2\right )\right )+36 a^2 b^2 e^2 \left (2 A e \left (-d^2+d e x+e^2 x^2\right )+B \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )\right )+8 a b^3 e \left (3 A e \left (2 d^3-4 d^2 e x-3 d e^2 x^2+e^3 x^3\right )+2 B \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )\right )+12 (d+e x) (b d-a e)^3 \log (d+e x) (-a B e-4 A b e+5 b B d)+b^4 \left (4 A e \left (-3 d^4+9 d^3 e x+6 d^2 e^2 x^2-2 d e^3 x^3+e^4 x^4\right )+B \left (12 d^5-48 d^4 e x-30 d^3 e^2 x^2+10 d^2 e^3 x^3-5 d e^4 x^4+3 e^5 x^5\right )\right )}{12 e^6 (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^4*e^4*(B*d - A*e) + 48*a^3*b*e^3*(A*d*e + B*(-d^2 + d*e*x + e^2*x^2)) + 36*a^2*b^2*e^2*(2*A*e*(-d^2 + d*
e*x + e^2*x^2) + B*(2*d^3 - 4*d^2*e*x - 3*d*e^2*x^2 + e^3*x^3)) + 8*a*b^3*e*(3*A*e*(2*d^3 - 4*d^2*e*x - 3*d*e^
2*x^2 + e^3*x^3) + 2*B*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4)) + b^4*(4*A*e*(-3*d^4 + 9*
d^3*e*x + 6*d^2*e^2*x^2 - 2*d*e^3*x^3 + e^4*x^4) + B*(12*d^5 - 48*d^4*e*x - 30*d^3*e^2*x^2 + 10*d^2*e^3*x^3 -
5*d*e^4*x^4 + 3*e^5*x^5)) + 12*(b*d - a*e)^3*(5*b*B*d - 4*A*b*e - a*B*e)*(d + e*x)*Log[d + e*x])/(12*e^6*(d +
e*x))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.42, size = 598, normalized size = 3.18 \begin {gather*} \frac {3 \, B b^{4} e^{5} x^{5} + 12 \, B b^{4} d^{5} - 12 \, A a^{4} e^{5} - 12 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 24 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 24 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} + 12 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} - {\left (5 \, B b^{4} d e^{4} - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} e^{5}\right )} x^{4} + 2 \, {\left (5 \, B b^{4} d^{2} e^{3} - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d e^{4} + 6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{5}\right )} x^{3} - 6 \, {\left (5 \, B b^{4} d^{3} e^{2} - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{3} + 6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{4} - 4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{5}\right )} x^{2} - 12 \, {\left (4 \, B b^{4} d^{4} e - 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} + 4 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4}\right )} x + 12 \, {\left (5 \, B b^{4} d^{5} - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} + {\left (5 \, B b^{4} d^{4} e - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} + 6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4} + {\left (B a^{4} + 4 \, A a^{3} b\right )} e^{5}\right )} x\right )} \log \left (e x + d\right )}{12 \, {\left (e^{7} x + d e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*b^4*e^5*x^5 + 12*B*b^4*d^5 - 12*A*a^4*e^5 - 12*(4*B*a*b^3 + A*b^4)*d^4*e + 24*(3*B*a^2*b^2 + 2*A*a*b
^3)*d^3*e^2 - 24*(2*B*a^3*b + 3*A*a^2*b^2)*d^2*e^3 + 12*(B*a^4 + 4*A*a^3*b)*d*e^4 - (5*B*b^4*d*e^4 - 4*(4*B*a*
b^3 + A*b^4)*e^5)*x^4 + 2*(5*B*b^4*d^2*e^3 - 4*(4*B*a*b^3 + A*b^4)*d*e^4 + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*e^5)*x^
3 - 6*(5*B*b^4*d^3*e^2 - 4*(4*B*a*b^3 + A*b^4)*d^2*e^3 + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d*e^4 - 4*(2*B*a^3*b + 3*
A*a^2*b^2)*e^5)*x^2 - 12*(4*B*b^4*d^4*e - 3*(4*B*a*b^3 + A*b^4)*d^3*e^2 + 4*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^3
- 2*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^4)*x + 12*(5*B*b^4*d^5 - 4*(4*B*a*b^3 + A*b^4)*d^4*e + 6*(3*B*a^2*b^2 + 2*A*
a*b^3)*d^3*e^2 - 4*(2*B*a^3*b + 3*A*a^2*b^2)*d^2*e^3 + (B*a^4 + 4*A*a^3*b)*d*e^4 + (5*B*b^4*d^4*e - 4*(4*B*a*b
^3 + A*b^4)*d^3*e^2 + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^3 - 4*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^4 + (B*a^4 + 4*A*a
^3*b)*e^5)*x)*log(e*x + d))/(e^7*x + d*e^6)

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giac [B]  time = 0.19, size = 526, normalized size = 2.80 \begin {gather*} \frac {1}{12} \, {\left (3 \, B b^{4} - \frac {4 \, {\left (5 \, B b^{4} d e - 4 \, B a b^{3} e^{2} - A b^{4} e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac {12 \, {\left (5 \, B b^{4} d^{2} e^{2} - 8 \, B a b^{3} d e^{3} - 2 \, A b^{4} d e^{3} + 3 \, B a^{2} b^{2} e^{4} + 2 \, A a b^{3} e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac {24 \, {\left (5 \, B b^{4} d^{3} e^{3} - 12 \, B a b^{3} d^{2} e^{4} - 3 \, A b^{4} d^{2} e^{4} + 9 \, B a^{2} b^{2} d e^{5} + 6 \, A a b^{3} d e^{5} - 2 \, B a^{3} b e^{6} - 3 \, A a^{2} b^{2} e^{6}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}}\right )} {\left (x e + d\right )}^{4} e^{\left (-6\right )} - {\left (5 \, B b^{4} d^{4} - 16 \, B a b^{3} d^{3} e - 4 \, A b^{4} d^{3} e + 18 \, B a^{2} b^{2} d^{2} e^{2} + 12 \, A a b^{3} d^{2} e^{2} - 8 \, B a^{3} b d e^{3} - 12 \, A a^{2} b^{2} d e^{3} + B a^{4} e^{4} + 4 \, A a^{3} b e^{4}\right )} e^{\left (-6\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {B b^{4} d^{5} e^{4}}{x e + d} - \frac {4 \, B a b^{3} d^{4} e^{5}}{x e + d} - \frac {A b^{4} d^{4} e^{5}}{x e + d} + \frac {6 \, B a^{2} b^{2} d^{3} e^{6}}{x e + d} + \frac {4 \, A a b^{3} d^{3} e^{6}}{x e + d} - \frac {4 \, B a^{3} b d^{2} e^{7}}{x e + d} - \frac {6 \, A a^{2} b^{2} d^{2} e^{7}}{x e + d} + \frac {B a^{4} d e^{8}}{x e + d} + \frac {4 \, A a^{3} b d e^{8}}{x e + d} - \frac {A a^{4} e^{9}}{x e + d}\right )} e^{\left (-10\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*B*b^4 - 4*(5*B*b^4*d*e - 4*B*a*b^3*e^2 - A*b^4*e^2)*e^(-1)/(x*e + d) + 12*(5*B*b^4*d^2*e^2 - 8*B*a*b^3
*d*e^3 - 2*A*b^4*d*e^3 + 3*B*a^2*b^2*e^4 + 2*A*a*b^3*e^4)*e^(-2)/(x*e + d)^2 - 24*(5*B*b^4*d^3*e^3 - 12*B*a*b^
3*d^2*e^4 - 3*A*b^4*d^2*e^4 + 9*B*a^2*b^2*d*e^5 + 6*A*a*b^3*d*e^5 - 2*B*a^3*b*e^6 - 3*A*a^2*b^2*e^6)*e^(-3)/(x
*e + d)^3)*(x*e + d)^4*e^(-6) - (5*B*b^4*d^4 - 16*B*a*b^3*d^3*e - 4*A*b^4*d^3*e + 18*B*a^2*b^2*d^2*e^2 + 12*A*
a*b^3*d^2*e^2 - 8*B*a^3*b*d*e^3 - 12*A*a^2*b^2*d*e^3 + B*a^4*e^4 + 4*A*a^3*b*e^4)*e^(-6)*log(abs(x*e + d)*e^(-
1)/(x*e + d)^2) + (B*b^4*d^5*e^4/(x*e + d) - 4*B*a*b^3*d^4*e^5/(x*e + d) - A*b^4*d^4*e^5/(x*e + d) + 6*B*a^2*b
^2*d^3*e^6/(x*e + d) + 4*A*a*b^3*d^3*e^6/(x*e + d) - 4*B*a^3*b*d^2*e^7/(x*e + d) - 6*A*a^2*b^2*d^2*e^7/(x*e +
d) + B*a^4*d*e^8/(x*e + d) + 4*A*a^3*b*d*e^8/(x*e + d) - A*a^4*e^9/(x*e + d))*e^(-10)

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maple [B]  time = 0.06, size = 564, normalized size = 3.00 \begin {gather*} \frac {B \,b^{4} x^{4}}{4 e^{2}}+\frac {A \,b^{4} x^{3}}{3 e^{2}}+\frac {4 B a \,b^{3} x^{3}}{3 e^{2}}-\frac {2 B \,b^{4} d \,x^{3}}{3 e^{3}}+\frac {2 A a \,b^{3} x^{2}}{e^{2}}-\frac {A \,b^{4} d \,x^{2}}{e^{3}}+\frac {3 B \,a^{2} b^{2} x^{2}}{e^{2}}-\frac {4 B a \,b^{3} d \,x^{2}}{e^{3}}+\frac {3 B \,b^{4} d^{2} x^{2}}{2 e^{4}}-\frac {A \,a^{4}}{\left (e x +d \right ) e}+\frac {4 A \,a^{3} b d}{\left (e x +d \right ) e^{2}}+\frac {4 A \,a^{3} b \ln \left (e x +d \right )}{e^{2}}-\frac {6 A \,a^{2} b^{2} d^{2}}{\left (e x +d \right ) e^{3}}-\frac {12 A \,a^{2} b^{2} d \ln \left (e x +d \right )}{e^{3}}+\frac {6 A \,a^{2} b^{2} x}{e^{2}}+\frac {4 A a \,b^{3} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {12 A a \,b^{3} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {8 A a \,b^{3} d x}{e^{3}}-\frac {A \,b^{4} d^{4}}{\left (e x +d \right ) e^{5}}-\frac {4 A \,b^{4} d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {3 A \,b^{4} d^{2} x}{e^{4}}+\frac {B \,a^{4} d}{\left (e x +d \right ) e^{2}}+\frac {B \,a^{4} \ln \left (e x +d \right )}{e^{2}}-\frac {4 B \,a^{3} b \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {8 B \,a^{3} b d \ln \left (e x +d \right )}{e^{3}}+\frac {4 B \,a^{3} b x}{e^{2}}+\frac {6 B \,a^{2} b^{2} d^{3}}{\left (e x +d \right ) e^{4}}+\frac {18 B \,a^{2} b^{2} d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {12 B \,a^{2} b^{2} d x}{e^{3}}-\frac {4 B a \,b^{3} d^{4}}{\left (e x +d \right ) e^{5}}-\frac {16 B a \,b^{3} d^{3} \ln \left (e x +d \right )}{e^{5}}+\frac {12 B a \,b^{3} d^{2} x}{e^{4}}+\frac {B \,b^{4} d^{5}}{\left (e x +d \right ) e^{6}}+\frac {5 B \,b^{4} d^{4} \ln \left (e x +d \right )}{e^{6}}-\frac {4 B \,b^{4} d^{3} x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/e/(e*x+d)*A*a^4+1/4*b^4/e^2*B*x^4+1/3*b^4/e^2*A*x^3+1/e^2*ln(e*x+d)*B*a^4-4*b^3/e^3*B*x^2*a*d+18/e^4*ln(e*x
+d)*B*a^2*b^2*d^2-16/e^5*ln(e*x+d)*B*a*b^3*d^3+4/e^2/(e*x+d)*A*d*a^3*b-6/e^3/(e*x+d)*A*a^2*b^2*d^2+4/e^4/(e*x+
d)*A*a*b^3*d^3-8/e^3*ln(e*x+d)*B*a^3*b*d-4/e^3/(e*x+d)*B*a^3*b*d^2+6/e^4/(e*x+d)*B*a^2*b^2*d^3-4/e^5/(e*x+d)*B
*a*b^3*d^4-12/e^3*ln(e*x+d)*A*a^2*b^2*d+12/e^4*ln(e*x+d)*A*a*b^3*d^2+2*b^3/e^2*A*x^2*a-b^4/e^3*A*x^2*d+1/e^6/(
e*x+d)*B*b^4*d^5+4/3*b^3/e^2*B*x^3*a-2/3*b^4/e^3*B*x^3*d+4/e^2*ln(e*x+d)*A*a^3*b-4/e^5*ln(e*x+d)*A*b^4*d^3+5/e
^6*ln(e*x+d)*B*b^4*d^4-8*b^3/e^3*A*a*d*x-12*b^2/e^3*B*a^2*d*x+12*b^3/e^4*B*a*d^2*x+6*b^2/e^2*A*a^2*x+3*b^4/e^4
*A*d^2*x+4*b/e^2*B*a^3*x-4*b^4/e^5*B*d^3*x-1/e^5/(e*x+d)*A*b^4*d^4+1/e^2/(e*x+d)*B*d*a^4+3*b^2/e^2*B*x^2*a^2+3
/2*b^4/e^4*B*x^2*d^2

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maxima [B]  time = 0.51, size = 410, normalized size = 2.18 \begin {gather*} \frac {B b^{4} d^{5} - A a^{4} e^{5} - {\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4}}{e^{7} x + d e^{6}} + \frac {3 \, B b^{4} e^{3} x^{4} - 4 \, {\left (2 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} + A b^{4}\right )} e^{3}\right )} x^{3} + 6 \, {\left (3 \, B b^{4} d^{2} e - 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d e^{2} + 2 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{3}\right )} x^{2} - 12 \, {\left (4 \, B b^{4} d^{3} - 3 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e + 4 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{2} - 2 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{3}\right )} x}{12 \, e^{5}} + \frac {{\left (5 \, B b^{4} d^{4} - 4 \, {\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e + 6 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{2} - 4 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{3} + {\left (B a^{4} + 4 \, A a^{3} b\right )} e^{4}\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)*(b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^2,x, algorithm="maxima")

[Out]

(B*b^4*d^5 - A*a^4*e^5 - (4*B*a*b^3 + A*b^4)*d^4*e + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 2*(2*B*a^3*b + 3*A*
a^2*b^2)*d^2*e^3 + (B*a^4 + 4*A*a^3*b)*d*e^4)/(e^7*x + d*e^6) + 1/12*(3*B*b^4*e^3*x^4 - 4*(2*B*b^4*d*e^2 - (4*
B*a*b^3 + A*b^4)*e^3)*x^3 + 6*(3*B*b^4*d^2*e - 2*(4*B*a*b^3 + A*b^4)*d*e^2 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*e^3)*
x^2 - 12*(4*B*b^4*d^3 - 3*(4*B*a*b^3 + A*b^4)*d^2*e + 4*(3*B*a^2*b^2 + 2*A*a*b^3)*d*e^2 - 2*(2*B*a^3*b + 3*A*a
^2*b^2)*e^3)*x)/e^5 + (5*B*b^4*d^4 - 4*(4*B*a*b^3 + A*b^4)*d^3*e + 6*(3*B*a^2*b^2 + 2*A*a*b^3)*d^2*e^2 - 4*(2*
B*a^3*b + 3*A*a^2*b^2)*d*e^3 + (B*a^4 + 4*A*a^3*b)*e^4)*log(e*x + d)/e^6

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mupad [B]  time = 0.10, size = 486, normalized size = 2.59 \begin {gather*} x^3\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{3\,e^2}-\frac {2\,B\,b^4\,d}{3\,e^3}\right )-x^2\,\left (\frac {d\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e^2}-\frac {2\,B\,b^4\,d}{e^3}\right )}{e}-\frac {a\,b^2\,\left (2\,A\,b+3\,B\,a\right )}{e^2}+\frac {B\,b^4\,d^2}{2\,e^4}\right )+x\,\left (\frac {2\,d\,\left (\frac {2\,d\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e^2}-\frac {2\,B\,b^4\,d}{e^3}\right )}{e}-\frac {2\,a\,b^2\,\left (2\,A\,b+3\,B\,a\right )}{e^2}+\frac {B\,b^4\,d^2}{e^4}\right )}{e}-\frac {d^2\,\left (\frac {A\,b^4+4\,B\,a\,b^3}{e^2}-\frac {2\,B\,b^4\,d}{e^3}\right )}{e^2}+\frac {2\,a^2\,b\,\left (3\,A\,b+2\,B\,a\right )}{e^2}\right )+\frac {\ln \left (d+e\,x\right )\,\left (B\,a^4\,e^4-8\,B\,a^3\,b\,d\,e^3+4\,A\,a^3\,b\,e^4+18\,B\,a^2\,b^2\,d^2\,e^2-12\,A\,a^2\,b^2\,d\,e^3-16\,B\,a\,b^3\,d^3\,e+12\,A\,a\,b^3\,d^2\,e^2+5\,B\,b^4\,d^4-4\,A\,b^4\,d^3\,e\right )}{e^6}-\frac {-B\,a^4\,d\,e^4+A\,a^4\,e^5+4\,B\,a^3\,b\,d^2\,e^3-4\,A\,a^3\,b\,d\,e^4-6\,B\,a^2\,b^2\,d^3\,e^2+6\,A\,a^2\,b^2\,d^2\,e^3+4\,B\,a\,b^3\,d^4\,e-4\,A\,a\,b^3\,d^3\,e^2-B\,b^4\,d^5+A\,b^4\,d^4\,e}{e\,\left (x\,e^6+d\,e^5\right )}+\frac {B\,b^4\,x^4}{4\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*((A*b^4 + 4*B*a*b^3)/(3*e^2) - (2*B*b^4*d)/(3*e^3)) - x^2*((d*((A*b^4 + 4*B*a*b^3)/e^2 - (2*B*b^4*d)/e^3))
/e - (a*b^2*(2*A*b + 3*B*a))/e^2 + (B*b^4*d^2)/(2*e^4)) + x*((2*d*((2*d*((A*b^4 + 4*B*a*b^3)/e^2 - (2*B*b^4*d)
/e^3))/e - (2*a*b^2*(2*A*b + 3*B*a))/e^2 + (B*b^4*d^2)/e^4))/e - (d^2*((A*b^4 + 4*B*a*b^3)/e^2 - (2*B*b^4*d)/e
^3))/e^2 + (2*a^2*b*(3*A*b + 2*B*a))/e^2) + (log(d + e*x)*(B*a^4*e^4 + 5*B*b^4*d^4 + 4*A*a^3*b*e^4 - 4*A*b^4*d
^3*e + 12*A*a*b^3*d^2*e^2 - 12*A*a^2*b^2*d*e^3 + 18*B*a^2*b^2*d^2*e^2 - 16*B*a*b^3*d^3*e - 8*B*a^3*b*d*e^3))/e
^6 - (A*a^4*e^5 - B*b^4*d^5 + A*b^4*d^4*e - B*a^4*d*e^4 - 4*A*a*b^3*d^3*e^2 + 4*B*a^3*b*d^2*e^3 + 6*A*a^2*b^2*
d^2*e^3 - 6*B*a^2*b^2*d^3*e^2 - 4*A*a^3*b*d*e^4 + 4*B*a*b^3*d^4*e)/(e*(d*e^5 + e^6*x)) + (B*b^4*x^4)/(4*e^2)

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sympy [B]  time = 2.09, size = 396, normalized size = 2.11 \begin {gather*} \frac {B b^{4} x^{4}}{4 e^{2}} + x^{3} \left (\frac {A b^{4}}{3 e^{2}} + \frac {4 B a b^{3}}{3 e^{2}} - \frac {2 B b^{4} d}{3 e^{3}}\right ) + x^{2} \left (\frac {2 A a b^{3}}{e^{2}} - \frac {A b^{4} d}{e^{3}} + \frac {3 B a^{2} b^{2}}{e^{2}} - \frac {4 B a b^{3} d}{e^{3}} + \frac {3 B b^{4} d^{2}}{2 e^{4}}\right ) + x \left (\frac {6 A a^{2} b^{2}}{e^{2}} - \frac {8 A a b^{3} d}{e^{3}} + \frac {3 A b^{4} d^{2}}{e^{4}} + \frac {4 B a^{3} b}{e^{2}} - \frac {12 B a^{2} b^{2} d}{e^{3}} + \frac {12 B a b^{3} d^{2}}{e^{4}} - \frac {4 B b^{4} d^{3}}{e^{5}}\right ) + \frac {- A a^{4} e^{5} + 4 A a^{3} b d e^{4} - 6 A a^{2} b^{2} d^{2} e^{3} + 4 A a b^{3} d^{3} e^{2} - A b^{4} d^{4} e + B a^{4} d e^{4} - 4 B a^{3} b d^{2} e^{3} + 6 B a^{2} b^{2} d^{3} e^{2} - 4 B a b^{3} d^{4} e + B b^{4} d^{5}}{d e^{6} + e^{7} x} + \frac {\left (a e - b d\right )^{3} \left (4 A b e + B a e - 5 B b d\right ) \log {\left (d + e x \right )}}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*b**4*x**4/(4*e**2) + x**3*(A*b**4/(3*e**2) + 4*B*a*b**3/(3*e**2) - 2*B*b**4*d/(3*e**3)) + x**2*(2*A*a*b**3/e
**2 - A*b**4*d/e**3 + 3*B*a**2*b**2/e**2 - 4*B*a*b**3*d/e**3 + 3*B*b**4*d**2/(2*e**4)) + x*(6*A*a**2*b**2/e**2
- 8*A*a*b**3*d/e**3 + 3*A*b**4*d**2/e**4 + 4*B*a**3*b/e**2 - 12*B*a**2*b**2*d/e**3 + 12*B*a*b**3*d**2/e**4 -
4*B*b**4*d**3/e**5) + (-A*a**4*e**5 + 4*A*a**3*b*d*e**4 - 6*A*a**2*b**2*d**2*e**3 + 4*A*a*b**3*d**3*e**2 - A*b
**4*d**4*e + B*a**4*d*e**4 - 4*B*a**3*b*d**2*e**3 + 6*B*a**2*b**2*d**3*e**2 - 4*B*a*b**3*d**4*e + B*b**4*d**5)
/(d*e**6 + e**7*x) + (a*e - b*d)**3*(4*A*b*e + B*a*e - 5*B*b*d)*log(d + e*x)/e**6

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