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

Optimal. Leaf size=145 \[ \frac {e^2 (a+b x)^2 (-4 a B e+A b e+3 b B d)}{2 b^5}-\frac {(A b-a B) (b d-a e)^3}{b^5 (a+b x)}+\frac {(b d-a e)^2 \log (a+b x) (-4 a B e+3 A b e+b B d)}{b^5}+\frac {3 e x (b d-a e) (-2 a B e+A b e+b B d)}{b^4}+\frac {B e^3 (a+b x)^3}{3 b^5} \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 145, 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 {e^2 (a+b x)^2 (-4 a B e+A b e+3 b B d)}{2 b^5}-\frac {(A b-a B) (b d-a e)^3}{b^5 (a+b x)}+\frac {3 e x (b d-a e) (-2 a B e+A b e+b B d)}{b^4}+\frac {(b d-a e)^2 \log (a+b x) (-4 a B e+3 A b e+b B d)}{b^5}+\frac {B e^3 (a+b x)^3}{3 b^5} \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]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 250, normalized size = 1.72 \begin {gather*} \frac {-3 A b \left (-2 a^3 e^3+2 a^2 b e^2 (3 d+2 e x)+3 a b^2 e \left (-2 d^2-2 d e x+e^2 x^2\right )+b^3 \left (2 d^3-6 d e^2 x^2-e^3 x^3\right )\right )+B \left (-6 a^4 e^3+18 a^3 b e^2 (d+e x)+6 a^2 b^2 e \left (-3 d^2-6 d e x+2 e^2 x^2\right )+a b^3 \left (6 d^3+18 d^2 e x-27 d e^2 x^2-4 e^3 x^3\right )+b^4 e x^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+6 (a+b x) (b d-a e)^2 \log (a+b x) (-4 a B e+3 A b e+b B d)}{6 b^5 (a+b x)} \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]

(B*(-6*a^4*e^3 + 18*a^3*b*e^2*(d + e*x) + 6*a^2*b^2*e*(-3*d^2 - 6*d*e*x + 2*e^2*x^2) + b^4*e*x^2*(18*d^2 + 9*d
*e*x + 2*e^2*x^2) + a*b^3*(6*d^3 + 18*d^2*e*x - 27*d*e^2*x^2 - 4*e^3*x^3)) - 3*A*b*(-2*a^3*e^3 + 2*a^2*b*e^2*(
3*d + 2*e*x) + 3*a*b^2*e*(-2*d^2 - 2*d*e*x + e^2*x^2) + b^3*(2*d^3 - 6*d*e^2*x^2 - e^3*x^3)) + 6*(b*d - a*e)^2
*(b*B*d + 3*A*b*e - 4*a*B*e)*(a + b*x)*Log[a + b*x])/(6*b^5*(a + b*x))

________________________________________________________________________________________

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

________________________________________________________________________________________

fricas [B]  time = 0.43, size = 417, normalized size = 2.88 \begin {gather*} \frac {2 \, B b^{4} e^{3} x^{4} + 6 \, {\left (B a b^{3} - A b^{4}\right )} d^{3} - 18 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 18 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} d e^{2} - 6 \, {\left (B a^{4} - A a^{3} b\right )} e^{3} + {\left (9 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, B b^{4} d^{2} e - 3 \, {\left (3 \, B a b^{3} - 2 \, A b^{4}\right )} d e^{2} + {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B a b^{3} d^{2} e - 3 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} d e^{2} + {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \, {\left (B a b^{3} d^{3} - 3 \, {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 3 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} d e^{2} - {\left (4 \, B a^{4} - 3 \, A a^{3} b\right )} e^{3} + {\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\right )} \log \left (b x + a\right )}{6 \, {\left (b^{6} x + a b^{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, algorithm="fricas")

[Out]

1/6*(2*B*b^4*e^3*x^4 + 6*(B*a*b^3 - A*b^4)*d^3 - 18*(B*a^2*b^2 - A*a*b^3)*d^2*e + 18*(B*a^3*b - A*a^2*b^2)*d*e
^2 - 6*(B*a^4 - A*a^3*b)*e^3 + (9*B*b^4*d*e^2 - (4*B*a*b^3 - 3*A*b^4)*e^3)*x^3 + 3*(6*B*b^4*d^2*e - 3*(3*B*a*b
^3 - 2*A*b^4)*d*e^2 + (4*B*a^2*b^2 - 3*A*a*b^3)*e^3)*x^2 + 6*(3*B*a*b^3*d^2*e - 3*(2*B*a^2*b^2 - A*a*b^3)*d*e^
2 + (3*B*a^3*b - 2*A*a^2*b^2)*e^3)*x + 6*(B*a*b^3*d^3 - 3*(2*B*a^2*b^2 - A*a*b^3)*d^2*e + 3*(3*B*a^3*b - 2*A*a
^2*b^2)*d*e^2 - (4*B*a^4 - 3*A*a^3*b)*e^3 + (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)*log(b*x + a))/(b^6*x + a*b^5)

________________________________________________________________________________________

giac [B]  time = 0.16, size = 282, normalized size = 1.94 \begin {gather*} \frac {{\left (B b^{3} d^{3} - 6 \, B a b^{2} d^{2} e + 3 \, A b^{3} d^{2} e + 9 \, B a^{2} b d e^{2} - 6 \, A a b^{2} d e^{2} - 4 \, B a^{3} e^{3} + 3 \, A a^{2} b e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} + \frac {2 \, B b^{4} x^{3} e^{3} + 9 \, B b^{4} d x^{2} e^{2} + 18 \, B b^{4} d^{2} x e - 6 \, B a b^{3} x^{2} e^{3} + 3 \, A b^{4} x^{2} e^{3} - 36 \, B a b^{3} d x e^{2} + 18 \, A b^{4} d x e^{2} + 18 \, B a^{2} b^{2} x e^{3} - 12 \, A a b^{3} x e^{3}}{6 \, b^{6}} + \frac {B a b^{3} d^{3} - A b^{4} d^{3} - 3 \, B a^{2} b^{2} d^{2} e + 3 \, A a b^{3} d^{2} e + 3 \, B a^{3} b d e^{2} - 3 \, A a^{2} b^{2} d e^{2} - B a^{4} e^{3} + A a^{3} b e^{3}}{{\left (b x + a\right )} b^{5}} \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^3 - 6*B*a*b^2*d^2*e + 3*A*b^3*d^2*e + 9*B*a^2*b*d*e^2 - 6*A*a*b^2*d*e^2 - 4*B*a^3*e^3 + 3*A*a^2*b*e^3
)*log(abs(b*x + a))/b^5 + 1/6*(2*B*b^4*x^3*e^3 + 9*B*b^4*d*x^2*e^2 + 18*B*b^4*d^2*x*e - 6*B*a*b^3*x^2*e^3 + 3*
A*b^4*x^2*e^3 - 36*B*a*b^3*d*x*e^2 + 18*A*b^4*d*x*e^2 + 18*B*a^2*b^2*x*e^3 - 12*A*a*b^3*x*e^3)/b^6 + (B*a*b^3*
d^3 - A*b^4*d^3 - 3*B*a^2*b^2*d^2*e + 3*A*a*b^3*d^2*e + 3*B*a^3*b*d*e^2 - 3*A*a^2*b^2*d*e^2 - B*a^4*e^3 + A*a^
3*b*e^3)/((b*x + a)*b^5)

________________________________________________________________________________________

maple [B]  time = 0.06, size = 376, normalized size = 2.59 \begin {gather*} \frac {B \,e^{3} x^{3}}{3 b^{2}}+\frac {A \,e^{3} x^{2}}{2 b^{2}}-\frac {B a \,e^{3} x^{2}}{b^{3}}+\frac {3 B d \,e^{2} x^{2}}{2 b^{2}}+\frac {A \,a^{3} e^{3}}{\left (b x +a \right ) b^{4}}-\frac {3 A \,a^{2} d \,e^{2}}{\left (b x +a \right ) b^{3}}+\frac {3 A \,a^{2} e^{3} \ln \left (b x +a \right )}{b^{4}}+\frac {3 A a \,d^{2} e}{\left (b x +a \right ) b^{2}}-\frac {6 A a d \,e^{2} \ln \left (b x +a \right )}{b^{3}}-\frac {2 A a \,e^{3} x}{b^{3}}-\frac {A \,d^{3}}{\left (b x +a \right ) b}+\frac {3 A \,d^{2} e \ln \left (b x +a \right )}{b^{2}}+\frac {3 A d \,e^{2} x}{b^{2}}-\frac {B \,a^{4} e^{3}}{\left (b x +a \right ) b^{5}}+\frac {3 B \,a^{3} d \,e^{2}}{\left (b x +a \right ) b^{4}}-\frac {4 B \,a^{3} e^{3} \ln \left (b x +a \right )}{b^{5}}-\frac {3 B \,a^{2} d^{2} e}{\left (b x +a \right ) b^{3}}+\frac {9 B \,a^{2} d \,e^{2} \ln \left (b x +a \right )}{b^{4}}+\frac {3 B \,a^{2} e^{3} x}{b^{4}}+\frac {B a \,d^{3}}{\left (b x +a \right ) b^{2}}-\frac {6 B a \,d^{2} e \ln \left (b x +a \right )}{b^{3}}-\frac {6 B a d \,e^{2} x}{b^{3}}+\frac {B \,d^{3} \ln \left (b x +a \right )}{b^{2}}+\frac {3 B \,d^{2} e x}{b^{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]

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

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 273, normalized size = 1.88 \begin {gather*} \frac {{\left (B a b^{3} - A b^{4}\right )} d^{3} - 3 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} d^{2} e + 3 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} d e^{2} - {\left (B a^{4} - A a^{3} b\right )} e^{3}}{b^{6} x + a b^{5}} + \frac {2 \, B b^{2} e^{3} x^{3} + 3 \, {\left (3 \, B b^{2} d e^{2} - {\left (2 \, B a b - A b^{2}\right )} e^{3}\right )} x^{2} + 6 \, {\left (3 \, B b^{2} d^{2} e - 3 \, {\left (2 \, B a b - A b^{2}\right )} d e^{2} + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{3}\right )} x}{6 \, b^{4}} + \frac {{\left (B b^{3} d^{3} - 3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} d^{2} e + 3 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} d e^{2} - {\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} e^{3}\right )} \log \left (b x + a\right )}{b^{5}} \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*a*b^3 - A*b^4)*d^3 - 3*(B*a^2*b^2 - A*a*b^3)*d^2*e + 3*(B*a^3*b - A*a^2*b^2)*d*e^2 - (B*a^4 - A*a^3*b)*e^3
)/(b^6*x + a*b^5) + 1/6*(2*B*b^2*e^3*x^3 + 3*(3*B*b^2*d*e^2 - (2*B*a*b - A*b^2)*e^3)*x^2 + 6*(3*B*b^2*d^2*e -
3*(2*B*a*b - A*b^2)*d*e^2 + (3*B*a^2 - 2*A*a*b)*e^3)*x)/b^4 + (B*b^3*d^3 - 3*(2*B*a*b^2 - A*b^3)*d^2*e + 3*(3*
B*a^2*b - 2*A*a*b^2)*d*e^2 - (4*B*a^3 - 3*A*a^2*b)*e^3)*log(b*x + a)/b^5

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 293, normalized size = 2.02 \begin {gather*} x^2\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{2\,b^2}-\frac {B\,a\,e^3}{b^3}\right )-x\,\left (\frac {2\,a\,\left (\frac {A\,e^3+3\,B\,d\,e^2}{b^2}-\frac {2\,B\,a\,e^3}{b^3}\right )}{b}-\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{b^2}+\frac {B\,a^2\,e^3}{b^4}\right )+\frac {\ln \left (a+b\,x\right )\,\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 )}{b^5}-\frac {B\,a^4\,e^3-3\,B\,a^3\,b\,d\,e^2-A\,a^3\,b\,e^3+3\,B\,a^2\,b^2\,d^2\,e+3\,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}{b\,\left (x\,b^5+a\,b^4\right )}+\frac {B\,e^3\,x^3}{3\,b^2} \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]

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

________________________________________________________________________________________

sympy [A]  time = 1.43, size = 257, normalized size = 1.77 \begin {gather*} \frac {B e^{3} x^{3}}{3 b^{2}} + x^{2} \left (\frac {A e^{3}}{2 b^{2}} - \frac {B a e^{3}}{b^{3}} + \frac {3 B d e^{2}}{2 b^{2}}\right ) + x \left (- \frac {2 A a e^{3}}{b^{3}} + \frac {3 A d e^{2}}{b^{2}} + \frac {3 B a^{2} e^{3}}{b^{4}} - \frac {6 B a d e^{2}}{b^{3}} + \frac {3 B d^{2} e}{b^{2}}\right ) + \frac {A a^{3} b e^{3} - 3 A a^{2} b^{2} d e^{2} + 3 A a b^{3} d^{2} e - A b^{4} d^{3} - B a^{4} e^{3} + 3 B a^{3} b d e^{2} - 3 B a^{2} b^{2} d^{2} e + B a b^{3} d^{3}}{a b^{5} + b^{6} x} - \frac {\left (a e - b d\right )^{2} \left (- 3 A b e + 4 B a e - B b d\right ) \log {\left (a + b x \right )}}{b^{5}} \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*e**3*x**3/(3*b**2) + x**2*(A*e**3/(2*b**2) - B*a*e**3/b**3 + 3*B*d*e**2/(2*b**2)) + x*(-2*A*a*e**3/b**3 + 3*
A*d*e**2/b**2 + 3*B*a**2*e**3/b**4 - 6*B*a*d*e**2/b**3 + 3*B*d**2*e/b**2) + (A*a**3*b*e**3 - 3*A*a**2*b**2*d*e
**2 + 3*A*a*b**3*d**2*e - A*b**4*d**3 - B*a**4*e**3 + 3*B*a**3*b*d*e**2 - 3*B*a**2*b**2*d**2*e + B*a*b**3*d**3
)/(a*b**5 + b**6*x) - (a*e - b*d)**2*(-3*A*b*e + 4*B*a*e - B*b*d)*log(a + b*x)/b**5

________________________________________________________________________________________