3.1135 \(\int \frac {(A+B x) (d+e x)^3}{(a+b x)^3} \, dx\)

Optimal. Leaf size=141 \[ -\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 (a+b x)}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x)^2}+\frac {3 e (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5}+\frac {e^2 x (-3 a B e+A b e+3 b B d)}{b^4}+\frac {B e^3 x^2}{2 b^3} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \[ \frac {e^2 x (-3 a B e+A b e+3 b B d)}{b^4}-\frac {(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 (a+b x)}-\frac {(A b-a B) (b d-a e)^3}{2 b^5 (a+b x)^2}+\frac {3 e (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5}+\frac {B e^3 x^2}{2 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.16, size = 245, normalized size = 1.74 \[ \frac {-A b \left (5 a^3 e^3+a^2 b e^2 (4 e x-9 d)+a b^2 e \left (3 d^2-12 d e x-4 e^2 x^2\right )+b^3 \left (d^3+6 d^2 e x-2 e^3 x^3\right )\right )+B \left (7 a^4 e^3+a^3 b e^2 (2 e x-15 d)+a^2 b^2 e \left (9 d^2-12 d e x-11 e^2 x^2\right )-a b^3 \left (d^3-12 d^2 e x-12 d e^2 x^2+4 e^3 x^3\right )+b^4 x \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )\right )+6 e (a+b x)^2 (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{2 b^5 (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(5*a^3*e^3 + a^2*b*e^2*(-9*d + 4*e*x) + a*b^2*e*(3*d^2 - 12*d*e*x - 4*e^2*x^2) + b^3*(d^3 + 6*d^2*e*x -
2*e^3*x^3))) + B*(7*a^4*e^3 + a^3*b*e^2*(-15*d + 2*e*x) + a^2*b^2*e*(9*d^2 - 12*d*e*x - 11*e^2*x^2) + b^4*x*(
-2*d^3 + 6*d*e^2*x^2 + e^3*x^3) - a*b^3*(d^3 - 12*d^2*e*x - 12*d*e^2*x^2 + 4*e^3*x^3)) + 6*e*(b*d - a*e)*(b*B*
d + A*b*e - 2*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)^2)

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fricas [B]  time = 0.90, size = 442, normalized size = 3.13 \[ \frac {B b^{4} e^{3} x^{4} - {\left (B a b^{3} + A b^{4}\right )} d^{3} + 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e - 3 \, {\left (5 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} d e^{2} + {\left (7 \, B a^{4} - 5 \, A a^{3} b\right )} e^{3} + 2 \, {\left (3 \, B b^{4} d e^{2} - {\left (2 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + {\left (12 \, B a b^{3} d e^{2} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} e^{3}\right )} x^{2} - 2 \, {\left (B b^{4} d^{3} - 3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} d^{2} e + 6 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} - {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (B a^{2} b^{2} d^{2} e - {\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} d e^{2} + {\left (2 \, B a^{4} - A a^{3} b\right )} e^{3} + {\left (B b^{4} d^{2} e - {\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 2 \, {\left (B a b^{3} d^{2} e - {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d e^{2} + {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^4*e^3*x^4 - (B*a*b^3 + A*b^4)*d^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*d^2*e - 3*(5*B*a^3*b - 3*A*a^2*b^2)*d*e
^2 + (7*B*a^4 - 5*A*a^3*b)*e^3 + 2*(3*B*b^4*d*e^2 - (2*B*a*b^3 - A*b^4)*e^3)*x^3 + (12*B*a*b^3*d*e^2 - (11*B*a
^2*b^2 - 4*A*a*b^3)*e^3)*x^2 - 2*(B*b^4*d^3 - 3*(2*B*a*b^3 - A*b^4)*d^2*e + 6*(B*a^2*b^2 - A*a*b^3)*d*e^2 - (B
*a^3*b - 2*A*a^2*b^2)*e^3)*x + 6*(B*a^2*b^2*d^2*e - (3*B*a^3*b - A*a^2*b^2)*d*e^2 + (2*B*a^4 - A*a^3*b)*e^3 +
(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 2*(B*a*b^3*d^2*e - (3*B*a^2*b^2
- A*a*b^3)*d*e^2 + (2*B*a^3*b - A*a^2*b^2)*e^3)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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giac [A]  time = 1.22, size = 272, normalized size = 1.93 \[ \frac {3 \, {\left (B b^{2} d^{2} e - 3 \, B a b d e^{2} + A b^{2} d e^{2} + 2 \, B a^{2} e^{3} - A a b e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {B b^{3} x^{2} e^{3} + 6 \, B b^{3} d x e^{2} - 6 \, B a b^{2} x e^{3} + 2 \, A b^{3} x e^{3}}{2 \, b^{6}} - \frac {B a b^{3} d^{3} + A b^{4} d^{3} - 9 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 15 \, B a^{3} b d e^{2} - 9 \, A a^{2} b^{2} d e^{2} - 7 \, B a^{4} e^{3} + 5 \, A a^{3} b e^{3} + 2 \, {\left (B b^{4} d^{3} - 6 \, B a b^{3} d^{2} e + 3 \, A b^{4} d^{2} e + 9 \, B a^{2} b^{2} d e^{2} - 6 \, A a b^{3} d e^{2} - 4 \, B a^{3} b e^{3} + 3 \, A a^{2} b^{2} e^{3}\right )} x}{2 \, {\left (b x + a\right )}^{2} b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(B*b^2*d^2*e - 3*B*a*b*d*e^2 + A*b^2*d*e^2 + 2*B*a^2*e^3 - A*a*b*e^3)*log(abs(b*x + a))/b^5 + 1/2*(B*b^3*x^2
*e^3 + 6*B*b^3*d*x*e^2 - 6*B*a*b^2*x*e^3 + 2*A*b^3*x*e^3)/b^6 - 1/2*(B*a*b^3*d^3 + A*b^4*d^3 - 9*B*a^2*b^2*d^2
*e + 3*A*a*b^3*d^2*e + 15*B*a^3*b*d*e^2 - 9*A*a^2*b^2*d*e^2 - 7*B*a^4*e^3 + 5*A*a^3*b*e^3 + 2*(B*b^4*d^3 - 6*B
*a*b^3*d^2*e + 3*A*b^4*d^2*e + 9*B*a^2*b^2*d*e^2 - 6*A*a*b^3*d*e^2 - 4*B*a^3*b*e^3 + 3*A*a^2*b^2*e^3)*x)/((b*x
+ a)^2*b^5)

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maple [B]  time = 0.01, size = 404, normalized size = 2.87 \[ \frac {A \,a^{3} e^{3}}{2 \left (b x +a \right )^{2} b^{4}}-\frac {3 A \,a^{2} d \,e^{2}}{2 \left (b x +a \right )^{2} b^{3}}+\frac {3 A a \,d^{2} e}{2 \left (b x +a \right )^{2} b^{2}}-\frac {A \,d^{3}}{2 \left (b x +a \right )^{2} b}-\frac {B \,a^{4} e^{3}}{2 \left (b x +a \right )^{2} b^{5}}+\frac {3 B \,a^{3} d \,e^{2}}{2 \left (b x +a \right )^{2} b^{4}}-\frac {3 B \,a^{2} d^{2} e}{2 \left (b x +a \right )^{2} b^{3}}+\frac {B a \,d^{3}}{2 \left (b x +a \right )^{2} b^{2}}+\frac {B \,e^{3} x^{2}}{2 b^{3}}-\frac {3 A \,a^{2} e^{3}}{\left (b x +a \right ) b^{4}}+\frac {6 A a d \,e^{2}}{\left (b x +a \right ) b^{3}}-\frac {3 A a \,e^{3} \ln \left (b x +a \right )}{b^{4}}-\frac {3 A \,d^{2} e}{\left (b x +a \right ) b^{2}}+\frac {3 A d \,e^{2} \ln \left (b x +a \right )}{b^{3}}+\frac {A \,e^{3} x}{b^{3}}+\frac {4 B \,a^{3} e^{3}}{\left (b x +a \right ) b^{5}}-\frac {9 B \,a^{2} d \,e^{2}}{\left (b x +a \right ) b^{4}}+\frac {6 B \,a^{2} e^{3} \ln \left (b x +a \right )}{b^{5}}+\frac {6 B a \,d^{2} e}{\left (b x +a \right ) b^{3}}-\frac {9 B a d \,e^{2} \ln \left (b x +a \right )}{b^{4}}-\frac {3 B a \,e^{3} x}{b^{4}}-\frac {B \,d^{3}}{\left (b x +a \right ) b^{2}}+\frac {3 B \,d^{2} e \ln \left (b x +a \right )}{b^{3}}+\frac {3 B d \,e^{2} x}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.66, size = 282, normalized size = 2.00 \[ -\frac {{\left (B a b^{3} + A b^{4}\right )} d^{3} - 3 \, {\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 3 \, {\left (5 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} d e^{2} - {\left (7 \, B a^{4} - 5 \, A a^{3} b\right )} e^{3} + 2 \, {\left (B b^{4} d^{3} - 3 \, {\left (2 \, B a b^{3} - A b^{4}\right )} d^{2} e + 3 \, {\left (3 \, B a^{2} b^{2} - 2 \, A a b^{3}\right )} d e^{2} - {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} e^{3}\right )} x}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} + \frac {B b e^{3} x^{2} + 2 \, {\left (3 \, B b d e^{2} - {\left (3 \, B a - A b\right )} e^{3}\right )} x}{2 \, b^{4}} + \frac {3 \, {\left (B b^{2} d^{2} e - {\left (3 \, B a b - A b^{2}\right )} d e^{2} + {\left (2 \, B a^{2} - A a b\right )} e^{3}\right )} \log \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.17, size = 290, normalized size = 2.06 \[ x\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{b^3}-\frac {3\,B\,a\,e^3}{b^4}\right )-\frac {\frac {-7\,B\,a^4\,e^3+15\,B\,a^3\,b\,d\,e^2+5\,A\,a^3\,b\,e^3-9\,B\,a^2\,b^2\,d^2\,e-9\,A\,a^2\,b^2\,d\,e^2+B\,a\,b^3\,d^3+3\,A\,a\,b^3\,d^2\,e+A\,b^4\,d^3}{2\,b}+x\,\left (-4\,B\,a^3\,e^3+9\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3-6\,B\,a\,b^2\,d^2\,e-6\,A\,a\,b^2\,d\,e^2+B\,b^3\,d^3+3\,A\,b^3\,d^2\,e\right )}{a^2\,b^4+2\,a\,b^5\,x+b^6\,x^2}+\frac {\ln \left (a+b\,x\right )\,\left (6\,B\,a^2\,e^3-9\,B\,a\,b\,d\,e^2-3\,A\,a\,b\,e^3+3\,B\,b^2\,d^2\,e+3\,A\,b^2\,d\,e^2\right )}{b^5}+\frac {B\,e^3\,x^2}{2\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((A*e^3 + 3*B*d*e^2)/b^3 - (3*B*a*e^3)/b^4) - ((A*b^4*d^3 - 7*B*a^4*e^3 + 5*A*a^3*b*e^3 + B*a*b^3*d^3 - 9*A*
a^2*b^2*d*e^2 - 9*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e + 15*B*a^3*b*d*e^2)/(2*b) + x*(B*b^3*d^3 - 4*B*a^3*e^3 + 3
*A*a^2*b*e^3 + 3*A*b^3*d^2*e - 6*A*a*b^2*d*e^2 - 6*B*a*b^2*d^2*e + 9*B*a^2*b*d*e^2))/(a^2*b^4 + b^6*x^2 + 2*a*
b^5*x) + (log(a + b*x)*(6*B*a^2*e^3 - 3*A*a*b*e^3 + 3*A*b^2*d*e^2 + 3*B*b^2*d^2*e - 9*B*a*b*d*e^2))/b^5 + (B*e
^3*x^2)/(2*b^3)

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sympy [B]  time = 5.09, size = 299, normalized size = 2.12 \[ \frac {B e^{3} x^{2}}{2 b^{3}} + x \left (\frac {A e^{3}}{b^{3}} - \frac {3 B a e^{3}}{b^{4}} + \frac {3 B d e^{2}}{b^{3}}\right ) + \frac {- 5 A a^{3} b e^{3} + 9 A a^{2} b^{2} d e^{2} - 3 A a b^{3} d^{2} e - A b^{4} d^{3} + 7 B a^{4} e^{3} - 15 B a^{3} b d e^{2} + 9 B a^{2} b^{2} d^{2} e - B a b^{3} d^{3} + x \left (- 6 A a^{2} b^{2} e^{3} + 12 A a b^{3} d e^{2} - 6 A b^{4} d^{2} e + 8 B a^{3} b e^{3} - 18 B a^{2} b^{2} d e^{2} + 12 B a b^{3} d^{2} e - 2 B b^{4} d^{3}\right )}{2 a^{2} b^{5} + 4 a b^{6} x + 2 b^{7} x^{2}} + \frac {3 e \left (a e - b d\right ) \left (- A b e + 2 B a e - B b d\right ) \log {\left (a + b x \right )}}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*e**3*x**2/(2*b**3) + x*(A*e**3/b**3 - 3*B*a*e**3/b**4 + 3*B*d*e**2/b**3) + (-5*A*a**3*b*e**3 + 9*A*a**2*b**2
*d*e**2 - 3*A*a*b**3*d**2*e - A*b**4*d**3 + 7*B*a**4*e**3 - 15*B*a**3*b*d*e**2 + 9*B*a**2*b**2*d**2*e - B*a*b*
*3*d**3 + x*(-6*A*a**2*b**2*e**3 + 12*A*a*b**3*d*e**2 - 6*A*b**4*d**2*e + 8*B*a**3*b*e**3 - 18*B*a**2*b**2*d*e
**2 + 12*B*a*b**3*d**2*e - 2*B*b**4*d**3))/(2*a**2*b**5 + 4*a*b**6*x + 2*b**7*x**2) + 3*e*(a*e - b*d)*(-A*b*e
+ 2*B*a*e - B*b*d)*log(a + b*x)/b**5

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