3.15.80 \(\int \frac {A+B x}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=157 \[ -\frac {b (A b-a B)}{(a+b x) (b d-a e)^3}+\frac {a B e-2 A b e+b B d}{(d+e x) (b d-a e)^3}+\frac {B d-A e}{2 (d+e x)^2 (b d-a e)^2}+\frac {b \log (a+b x) (2 a B e-3 A b e+b B d)}{(b d-a e)^4}-\frac {b \log (d+e x) (2 a B e-3 A b e+b B d)}{(b d-a e)^4} \]

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Rubi [A]  time = 0.15, antiderivative size = 157, 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 (A b-a B)}{(a+b x) (b d-a e)^3}+\frac {a B e-2 A b e+b B d}{(d+e x) (b d-a e)^3}+\frac {B d-A e}{2 (d+e x)^2 (b d-a e)^2}+\frac {b \log (a+b x) (2 a B e-3 A b e+b B d)}{(b d-a e)^4}-\frac {b \log (d+e x) (2 a B e-3 A b e+b B d)}{(b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 146, normalized size = 0.93 \begin {gather*} \frac {\frac {(b d-a e)^2 (B d-A e)}{(d+e x)^2}-\frac {2 b (A b-a B) (b d-a e)}{a+b x}+\frac {2 (b d-a e) (a B e-2 A b e+b B d)}{d+e x}+2 b \log (a+b x) (2 a B e-3 A b e+b B d)-2 b \log (d+e x) (2 a B e-3 A b e+b B d)}{2 (b d-a e)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x}{(d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

IntegrateAlgebraic[(A + B*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)), x]

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fricas [B]  time = 0.44, size = 801, normalized size = 5.10 \begin {gather*} -\frac {A a^{3} e^{3} - {\left (5 \, B a b^{2} - 2 \, A b^{3}\right )} d^{3} + {\left (4 \, B a^{2} b + 3 \, A a b^{2}\right )} d^{2} e + {\left (B a^{3} - 6 \, A a^{2} b\right )} d e^{2} - 2 \, {\left (B b^{3} d^{2} e + {\left (B a b^{2} - 3 \, A b^{3}\right )} d e^{2} - {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} - {\left (3 \, B b^{3} d^{3} + {\left (4 \, B a b^{2} - 9 \, A b^{3}\right )} d^{2} e - {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} d e^{2} - {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} e^{3}\right )} x - 2 \, {\left (B a b^{2} d^{3} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e + {\left (B b^{3} d e^{2} + {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} e^{3}\right )} x^{3} + {\left (2 \, B b^{3} d^{2} e + {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} d e^{2} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} + {\left (B b^{3} d^{3} + {\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} d^{2} e + 2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d e^{2}\right )} x\right )} \log \left (b x + a\right ) + 2 \, {\left (B a b^{2} d^{3} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e + {\left (B b^{3} d e^{2} + {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} e^{3}\right )} x^{3} + {\left (2 \, B b^{3} d^{2} e + {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} d e^{2} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} + {\left (B b^{3} d^{3} + {\left (4 \, B a b^{2} - 3 \, A b^{3}\right )} d^{2} e + 2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a b^{4} d^{6} - 4 \, a^{2} b^{3} d^{5} e + 6 \, a^{3} b^{2} d^{4} e^{2} - 4 \, a^{4} b d^{3} e^{3} + a^{5} d^{2} e^{4} + {\left (b^{5} d^{4} e^{2} - 4 \, a b^{4} d^{3} e^{3} + 6 \, a^{2} b^{3} d^{2} e^{4} - 4 \, a^{3} b^{2} d e^{5} + a^{4} b e^{6}\right )} x^{3} + {\left (2 \, b^{5} d^{5} e - 7 \, a b^{4} d^{4} e^{2} + 8 \, a^{2} b^{3} d^{3} e^{3} - 2 \, a^{3} b^{2} d^{2} e^{4} - 2 \, a^{4} b d e^{5} + a^{5} e^{6}\right )} x^{2} + {\left (b^{5} d^{6} - 2 \, a b^{4} d^{5} e - 2 \, a^{2} b^{3} d^{4} e^{2} + 8 \, a^{3} b^{2} d^{3} e^{3} - 7 \, a^{4} b d^{2} e^{4} + 2 \, a^{5} d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.16, size = 393, normalized size = 2.50 \begin {gather*} \frac {{\left (B b^{3} d + 2 \, B a b^{2} e - 3 \, A b^{3} e\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5} d^{4} - 4 \, a b^{4} d^{3} e + 6 \, a^{2} b^{3} d^{2} e^{2} - 4 \, a^{3} b^{2} d e^{3} + a^{4} b e^{4}} - \frac {{\left (B b^{2} d e + 2 \, B a b e^{2} - 3 \, A b^{2} e^{2}\right )} \log \left ({\left | x e + d \right |}\right )}{b^{4} d^{4} e - 4 \, a b^{3} d^{3} e^{2} + 6 \, a^{2} b^{2} d^{2} e^{3} - 4 \, a^{3} b d e^{4} + a^{4} e^{5}} + \frac {5 \, B a b^{2} d^{3} - 2 \, A b^{3} d^{3} - 4 \, B a^{2} b d^{2} e - 3 \, A a b^{2} d^{2} e - B a^{3} d e^{2} + 6 \, A a^{2} b d e^{2} - A a^{3} e^{3} + 2 \, {\left (B b^{3} d^{2} e + B a b^{2} d e^{2} - 3 \, A b^{3} d e^{2} - 2 \, B a^{2} b e^{3} + 3 \, A a b^{2} e^{3}\right )} x^{2} + {\left (3 \, B b^{3} d^{3} + 4 \, B a b^{2} d^{2} e - 9 \, A b^{3} d^{2} e - 5 \, B a^{2} b d e^{2} + 6 \, A a b^{2} d e^{2} - 2 \, B a^{3} e^{3} + 3 \, A a^{2} b e^{3}\right )} x}{2 \, {\left (b d - a e\right )}^{4} {\left (b x + a\right )} {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b^3*d + 2*B*a*b^2*e - 3*A*b^3*e)*log(abs(b*x + a))/(b^5*d^4 - 4*a*b^4*d^3*e + 6*a^2*b^3*d^2*e^2 - 4*a^3*b^2
*d*e^3 + a^4*b*e^4) - (B*b^2*d*e + 2*B*a*b*e^2 - 3*A*b^2*e^2)*log(abs(x*e + d))/(b^4*d^4*e - 4*a*b^3*d^3*e^2 +
6*a^2*b^2*d^2*e^3 - 4*a^3*b*d*e^4 + a^4*e^5) + 1/2*(5*B*a*b^2*d^3 - 2*A*b^3*d^3 - 4*B*a^2*b*d^2*e - 3*A*a*b^2
*d^2*e - B*a^3*d*e^2 + 6*A*a^2*b*d*e^2 - A*a^3*e^3 + 2*(B*b^3*d^2*e + B*a*b^2*d*e^2 - 3*A*b^3*d*e^2 - 2*B*a^2*
b*e^3 + 3*A*a*b^2*e^3)*x^2 + (3*B*b^3*d^3 + 4*B*a*b^2*d^2*e - 9*A*b^3*d^2*e - 5*B*a^2*b*d*e^2 + 6*A*a*b^2*d*e^
2 - 2*B*a^3*e^3 + 3*A*a^2*b*e^3)*x)/((b*d - a*e)^4*(b*x + a)*(x*e + d)^2)

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maple [A]  time = 0.06, size = 289, normalized size = 1.84 \begin {gather*} -\frac {3 A \,b^{2} e \ln \left (b x +a \right )}{\left (a e -b d \right )^{4}}+\frac {3 A \,b^{2} e \ln \left (e x +d \right )}{\left (a e -b d \right )^{4}}+\frac {2 B a b e \ln \left (b x +a \right )}{\left (a e -b d \right )^{4}}-\frac {2 B a b e \ln \left (e x +d \right )}{\left (a e -b d \right )^{4}}+\frac {B \,b^{2} d \ln \left (b x +a \right )}{\left (a e -b d \right )^{4}}-\frac {B \,b^{2} d \ln \left (e x +d \right )}{\left (a e -b d \right )^{4}}+\frac {A \,b^{2}}{\left (a e -b d \right )^{3} \left (b x +a \right )}+\frac {2 A b e}{\left (a e -b d \right )^{3} \left (e x +d \right )}-\frac {B a b}{\left (a e -b d \right )^{3} \left (b x +a \right )}-\frac {B a e}{\left (a e -b d \right )^{3} \left (e x +d \right )}-\frac {B b d}{\left (a e -b d \right )^{3} \left (e x +d \right )}-\frac {A e}{2 \left (a e -b d \right )^{2} \left (e x +d \right )^{2}}+\frac {B d}{2 \left (a e -b d \right )^{2} \left (e x +d \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2/(a*e-b*d)^3/(b*x+a)*A-b/(a*e-b*d)^3/(b*x+a)*B*a-3*b^2/(a*e-b*d)^4*ln(b*x+a)*A*e+2*b/(a*e-b*d)^4*ln(b*x+a)*
a*B*e+b^2/(a*e-b*d)^4*ln(b*x+a)*B*d-1/2/(a*e-b*d)^2/(e*x+d)^2*A*e+1/2/(a*e-b*d)^2/(e*x+d)^2*B*d+3*b^2/(a*e-b*d
)^4*ln(e*x+d)*A*e-2*b/(a*e-b*d)^4*ln(e*x+d)*a*B*e-b^2/(a*e-b*d)^4*ln(e*x+d)*B*d+2/(a*e-b*d)^3/(e*x+d)*A*b*e-1/
(a*e-b*d)^3/(e*x+d)*a*B*e-1/(a*e-b*d)^3/(e*x+d)*B*b*d

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maxima [B]  time = 0.80, size = 479, normalized size = 3.05 \begin {gather*} \frac {{\left (B b^{2} d + {\left (2 \, B a b - 3 \, A b^{2}\right )} e\right )} \log \left (b x + a\right )}{b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}} - \frac {{\left (B b^{2} d + {\left (2 \, B a b - 3 \, A b^{2}\right )} e\right )} \log \left (e x + d\right )}{b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}} + \frac {A a^{2} e^{2} + {\left (5 \, B a b - 2 \, A b^{2}\right )} d^{2} + {\left (B a^{2} - 5 \, A a b\right )} d e + 2 \, {\left (B b^{2} d e + {\left (2 \, B a b - 3 \, A b^{2}\right )} e^{2}\right )} x^{2} + {\left (3 \, B b^{2} d^{2} + {\left (7 \, B a b - 9 \, A b^{2}\right )} d e + {\left (2 \, B a^{2} - 3 \, A a b\right )} e^{2}\right )} x}{2 \, {\left (a b^{3} d^{5} - 3 \, a^{2} b^{2} d^{4} e + 3 \, a^{3} b d^{3} e^{2} - a^{4} d^{2} e^{3} + {\left (b^{4} d^{3} e^{2} - 3 \, a b^{3} d^{2} e^{3} + 3 \, a^{2} b^{2} d e^{4} - a^{3} b e^{5}\right )} x^{3} + {\left (2 \, b^{4} d^{4} e - 5 \, a b^{3} d^{3} e^{2} + 3 \, a^{2} b^{2} d^{2} e^{3} + a^{3} b d e^{4} - a^{4} e^{5}\right )} x^{2} + {\left (b^{4} d^{5} - a b^{3} d^{4} e - 3 \, a^{2} b^{2} d^{3} e^{2} + 5 \, a^{3} b d^{2} e^{3} - 2 \, a^{4} d e^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*b^2*d + (2*B*a*b - 3*A*b^2)*e)*log(b*x + a)/(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 +
a^4*e^4) - (B*b^2*d + (2*B*a*b - 3*A*b^2)*e)*log(e*x + d)/(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3
*b*d*e^3 + a^4*e^4) + 1/2*(A*a^2*e^2 + (5*B*a*b - 2*A*b^2)*d^2 + (B*a^2 - 5*A*a*b)*d*e + 2*(B*b^2*d*e + (2*B*a
*b - 3*A*b^2)*e^2)*x^2 + (3*B*b^2*d^2 + (7*B*a*b - 9*A*b^2)*d*e + (2*B*a^2 - 3*A*a*b)*e^2)*x)/(a*b^3*d^5 - 3*a
^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4 - a^3*b*e^5)*x
^3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e
- 3*a^2*b^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x)

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mupad [B]  time = 2.66, size = 454, normalized size = 2.89 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {\left (b^2\,\left (3\,A\,e-B\,d\right )-2\,B\,a\,b\,e\right )\,\left (\frac {a^4\,e^4-2\,a^3\,b\,d\,e^3+2\,a\,b^3\,d^3\,e-b^4\,d^4}{a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3}+2\,b\,e\,x\right )\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^4\,\left (B\,b^2\,d-3\,A\,b^2\,e+2\,B\,a\,b\,e\right )}\right )\,\left (b^2\,\left (3\,A\,e-B\,d\right )-2\,B\,a\,b\,e\right )}{{\left (a\,e-b\,d\right )}^4}-\frac {\frac {B\,a^2\,d\,e+A\,a^2\,e^2+5\,B\,a\,b\,d^2-5\,A\,a\,b\,d\,e-2\,A\,b^2\,d^2}{2\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}+\frac {x\,\left (a\,e+3\,b\,d\right )\,\left (2\,B\,a\,e-3\,A\,b\,e+B\,b\,d\right )}{2\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}+\frac {b\,e\,x^2\,\left (2\,B\,a\,e-3\,A\,b\,e+B\,b\,d\right )}{a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3}}{x\,\left (b\,d^2+2\,a\,e\,d\right )+a\,d^2+x^2\,\left (a\,e^2+2\,b\,d\,e\right )+b\,e^2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(((b^2*(3*A*e - B*d) - 2*B*a*b*e)*((a^4*e^4 - b^4*d^4 + 2*a*b^3*d^3*e - 2*a^3*b*d*e^3)/(a^3*e^3 - b^3*
d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + 2*b*e*x)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/((a*e - b
*d)^4*(B*b^2*d - 3*A*b^2*e + 2*B*a*b*e)))*(b^2*(3*A*e - B*d) - 2*B*a*b*e))/(a*e - b*d)^4 - ((A*a^2*e^2 - 2*A*b
^2*d^2 + 5*B*a*b*d^2 + B*a^2*d*e - 5*A*a*b*d*e)/(2*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2)) + (x*(
a*e + 3*b*d)*(2*B*a*e - 3*A*b*e + B*b*d))/(2*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2)) + (b*e*x^2*(
2*B*a*e - 3*A*b*e + B*b*d))/(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2))/(x*(b*d^2 + 2*a*d*e) + a*d^2
+ x^2*(a*e^2 + 2*b*d*e) + b*e^2*x^3)

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sympy [B]  time = 3.58, size = 1066, normalized size = 6.79 \begin {gather*} - \frac {b \left (- 3 A b e + 2 B a e + B b d\right ) \log {\left (x + \frac {- 3 A a b^{2} e^{2} - 3 A b^{3} d e + 2 B a^{2} b e^{2} + 3 B a b^{2} d e + B b^{3} d^{2} - \frac {a^{5} b e^{5} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} + \frac {5 a^{4} b^{2} d e^{4} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} - \frac {10 a^{3} b^{3} d^{2} e^{3} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} + \frac {10 a^{2} b^{4} d^{3} e^{2} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} - \frac {5 a b^{5} d^{4} e \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} + \frac {b^{6} d^{5} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}}}{- 6 A b^{3} e^{2} + 4 B a b^{2} e^{2} + 2 B b^{3} d e} \right )}}{\left (a e - b d\right )^{4}} + \frac {b \left (- 3 A b e + 2 B a e + B b d\right ) \log {\left (x + \frac {- 3 A a b^{2} e^{2} - 3 A b^{3} d e + 2 B a^{2} b e^{2} + 3 B a b^{2} d e + B b^{3} d^{2} + \frac {a^{5} b e^{5} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} - \frac {5 a^{4} b^{2} d e^{4} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} + \frac {10 a^{3} b^{3} d^{2} e^{3} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} - \frac {10 a^{2} b^{4} d^{3} e^{2} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} + \frac {5 a b^{5} d^{4} e \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}} - \frac {b^{6} d^{5} \left (- 3 A b e + 2 B a e + B b d\right )}{\left (a e - b d\right )^{4}}}{- 6 A b^{3} e^{2} + 4 B a b^{2} e^{2} + 2 B b^{3} d e} \right )}}{\left (a e - b d\right )^{4}} + \frac {- A a^{2} e^{2} + 5 A a b d e + 2 A b^{2} d^{2} - B a^{2} d e - 5 B a b d^{2} + x^{2} \left (6 A b^{2} e^{2} - 4 B a b e^{2} - 2 B b^{2} d e\right ) + x \left (3 A a b e^{2} + 9 A b^{2} d e - 2 B a^{2} e^{2} - 7 B a b d e - 3 B b^{2} d^{2}\right )}{2 a^{4} d^{2} e^{3} - 6 a^{3} b d^{3} e^{2} + 6 a^{2} b^{2} d^{4} e - 2 a b^{3} d^{5} + x^{3} \left (2 a^{3} b e^{5} - 6 a^{2} b^{2} d e^{4} + 6 a b^{3} d^{2} e^{3} - 2 b^{4} d^{3} e^{2}\right ) + x^{2} \left (2 a^{4} e^{5} - 2 a^{3} b d e^{4} - 6 a^{2} b^{2} d^{2} e^{3} + 10 a b^{3} d^{3} e^{2} - 4 b^{4} d^{4} e\right ) + x \left (4 a^{4} d e^{4} - 10 a^{3} b d^{2} e^{3} + 6 a^{2} b^{2} d^{3} e^{2} + 2 a b^{3} d^{4} e - 2 b^{4} d^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*(-3*A*b*e + 2*B*a*e + B*b*d)*log(x + (-3*A*a*b**2*e**2 - 3*A*b**3*d*e + 2*B*a**2*b*e**2 + 3*B*a*b**2*d*e +
B*b**3*d**2 - a**5*b*e**5*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 + 5*a**4*b**2*d*e**4*(-3*A*b*e + 2*B*a*e
+ B*b*d)/(a*e - b*d)**4 - 10*a**3*b**3*d**2*e**3*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 + 10*a**2*b**4*d
**3*e**2*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 - 5*a*b**5*d**4*e*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d
)**4 + b**6*d**5*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4)/(-6*A*b**3*e**2 + 4*B*a*b**2*e**2 + 2*B*b**3*d*e
))/(a*e - b*d)**4 + b*(-3*A*b*e + 2*B*a*e + B*b*d)*log(x + (-3*A*a*b**2*e**2 - 3*A*b**3*d*e + 2*B*a**2*b*e**2
+ 3*B*a*b**2*d*e + B*b**3*d**2 + a**5*b*e**5*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 - 5*a**4*b**2*d*e**4*
(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 + 10*a**3*b**3*d**2*e**3*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)*
*4 - 10*a**2*b**4*d**3*e**2*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4 + 5*a*b**5*d**4*e*(-3*A*b*e + 2*B*a*e
+ B*b*d)/(a*e - b*d)**4 - b**6*d**5*(-3*A*b*e + 2*B*a*e + B*b*d)/(a*e - b*d)**4)/(-6*A*b**3*e**2 + 4*B*a*b**2*
e**2 + 2*B*b**3*d*e))/(a*e - b*d)**4 + (-A*a**2*e**2 + 5*A*a*b*d*e + 2*A*b**2*d**2 - B*a**2*d*e - 5*B*a*b*d**2
+ x**2*(6*A*b**2*e**2 - 4*B*a*b*e**2 - 2*B*b**2*d*e) + x*(3*A*a*b*e**2 + 9*A*b**2*d*e - 2*B*a**2*e**2 - 7*B*a
*b*d*e - 3*B*b**2*d**2))/(2*a**4*d**2*e**3 - 6*a**3*b*d**3*e**2 + 6*a**2*b**2*d**4*e - 2*a*b**3*d**5 + x**3*(2
*a**3*b*e**5 - 6*a**2*b**2*d*e**4 + 6*a*b**3*d**2*e**3 - 2*b**4*d**3*e**2) + x**2*(2*a**4*e**5 - 2*a**3*b*d*e*
*4 - 6*a**2*b**2*d**2*e**3 + 10*a*b**3*d**3*e**2 - 4*b**4*d**4*e) + x*(4*a**4*d*e**4 - 10*a**3*b*d**2*e**3 + 6
*a**2*b**2*d**3*e**2 + 2*a*b**3*d**4*e - 2*b**4*d**5))

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