3.10.8 \(\int \frac {(a+b x)^3 (A+B x)}{(d+e x)^7} \, dx\)

Optimal. Leaf size=133 \[ \frac {b (a+b x)^4 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac {(a+b x)^4 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac {(a+b x)^4 (B d-A e)}{6 e (d+e x)^6 (b d-a e)} \]

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Rubi [A]  time = 0.08, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {78, 45, 37} \begin {gather*} \frac {b (a+b x)^4 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac {(a+b x)^4 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac {(a+b x)^4 (B d-A e)}{6 e (d+e x)^6 (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^7,x]

[Out]

-((B*d - A*e)*(a + b*x)^4)/(6*e*(b*d - a*e)*(d + e*x)^6) + ((2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^4)/(15*e*(b*
d - a*e)^2*(d + e*x)^5) + (b*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^4)/(60*e*(b*d - a*e)^3*(d + e*x)^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

\begin {align*} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^7} \, dx &=-\frac {(B d-A e) (a+b x)^4}{6 e (b d-a e) (d+e x)^6}+\frac {(2 b B d+A b e-3 a B e) \int \frac {(a+b x)^3}{(d+e x)^6} \, dx}{3 e (b d-a e)}\\ &=-\frac {(B d-A e) (a+b x)^4}{6 e (b d-a e) (d+e x)^6}+\frac {(2 b B d+A b e-3 a B e) (a+b x)^4}{15 e (b d-a e)^2 (d+e x)^5}+\frac {(b (2 b B d+A b e-3 a B e)) \int \frac {(a+b x)^3}{(d+e x)^5} \, dx}{15 e (b d-a e)^2}\\ &=-\frac {(B d-A e) (a+b x)^4}{6 e (b d-a e) (d+e x)^6}+\frac {(2 b B d+A b e-3 a B e) (a+b x)^4}{15 e (b d-a e)^2 (d+e x)^5}+\frac {b (2 b B d+A b e-3 a B e) (a+b x)^4}{60 e (b d-a e)^3 (d+e x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 211, normalized size = 1.59 \begin {gather*} -\frac {2 a^3 e^3 (5 A e+B (d+6 e x))+3 a^2 b e^2 \left (2 A e (d+6 e x)+B \left (d^2+6 d e x+15 e^2 x^2\right )\right )+3 a b^2 e \left (A e \left (d^2+6 d e x+15 e^2 x^2\right )+B \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )\right )+b^3 \left (A e \left (d^3+6 d^2 e x+15 d e^2 x^2+20 e^3 x^3\right )+2 B \left (d^4+6 d^3 e x+15 d^2 e^2 x^2+20 d e^3 x^3+15 e^4 x^4\right )\right )}{60 e^5 (d+e x)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^7,x]

[Out]

-1/60*(2*a^3*e^3*(5*A*e + B*(d + 6*e*x)) + 3*a^2*b*e^2*(2*A*e*(d + 6*e*x) + B*(d^2 + 6*d*e*x + 15*e^2*x^2)) +
3*a*b^2*e*(A*e*(d^2 + 6*d*e*x + 15*e^2*x^2) + B*(d^3 + 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3)) + b^3*(A*e*(d^3
+ 6*d^2*e*x + 15*d*e^2*x^2 + 20*e^3*x^3) + 2*B*(d^4 + 6*d^3*e*x + 15*d^2*e^2*x^2 + 20*d*e^3*x^3 + 15*e^4*x^4)
))/(e^5*(d + e*x)^6)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x)^3 (A+B x)}{(d+e x)^7} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x)^3*(A + B*x))/(d + e*x)^7,x]

[Out]

IntegrateAlgebraic[((a + b*x)^3*(A + B*x))/(d + e*x)^7, x]

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fricas [B]  time = 0.71, size = 317, normalized size = 2.38 \begin {gather*} -\frac {30 \, B b^{3} e^{4} x^{4} + 2 \, B b^{3} d^{4} + 10 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 20 \, {\left (2 \, B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 15 \, {\left (2 \, B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^7,x, algorithm="fricas")

[Out]

-1/60*(30*B*b^3*e^4*x^4 + 2*B*b^3*d^4 + 10*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^2)*d^2*e
^2 + 2*(B*a^3 + 3*A*a^2*b)*d*e^3 + 20*(2*B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 15*(2*B*b^3*d^2*e^2 + (3
*B*a*b^2 + A*b^3)*d*e^3 + 3*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 6*(2*B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 3*
(B*a^2*b + A*a*b^2)*d*e^3 + 2*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e
^8*x^3 + 15*d^4*e^7*x^2 + 6*d^5*e^6*x + d^6*e^5)

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giac [B]  time = 1.17, size = 282, normalized size = 2.12 \begin {gather*} -\frac {{\left (30 \, B b^{3} x^{4} e^{4} + 40 \, B b^{3} d x^{3} e^{3} + 30 \, B b^{3} d^{2} x^{2} e^{2} + 12 \, B b^{3} d^{3} x e + 2 \, B b^{3} d^{4} + 60 \, B a b^{2} x^{3} e^{4} + 20 \, A b^{3} x^{3} e^{4} + 45 \, B a b^{2} d x^{2} e^{3} + 15 \, A b^{3} d x^{2} e^{3} + 18 \, B a b^{2} d^{2} x e^{2} + 6 \, A b^{3} d^{2} x e^{2} + 3 \, B a b^{2} d^{3} e + A b^{3} d^{3} e + 45 \, B a^{2} b x^{2} e^{4} + 45 \, A a b^{2} x^{2} e^{4} + 18 \, B a^{2} b d x e^{3} + 18 \, A a b^{2} d x e^{3} + 3 \, B a^{2} b d^{2} e^{2} + 3 \, A a b^{2} d^{2} e^{2} + 12 \, B a^{3} x e^{4} + 36 \, A a^{2} b x e^{4} + 2 \, B a^{3} d e^{3} + 6 \, A a^{2} b d e^{3} + 10 \, A a^{3} e^{4}\right )} e^{\left (-5\right )}}{60 \, {\left (x e + d\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^7,x, algorithm="giac")

[Out]

-1/60*(30*B*b^3*x^4*e^4 + 40*B*b^3*d*x^3*e^3 + 30*B*b^3*d^2*x^2*e^2 + 12*B*b^3*d^3*x*e + 2*B*b^3*d^4 + 60*B*a*
b^2*x^3*e^4 + 20*A*b^3*x^3*e^4 + 45*B*a*b^2*d*x^2*e^3 + 15*A*b^3*d*x^2*e^3 + 18*B*a*b^2*d^2*x*e^2 + 6*A*b^3*d^
2*x*e^2 + 3*B*a*b^2*d^3*e + A*b^3*d^3*e + 45*B*a^2*b*x^2*e^4 + 45*A*a*b^2*x^2*e^4 + 18*B*a^2*b*d*x*e^3 + 18*A*
a*b^2*d*x*e^3 + 3*B*a^2*b*d^2*e^2 + 3*A*a*b^2*d^2*e^2 + 12*B*a^3*x*e^4 + 36*A*a^2*b*x*e^4 + 2*B*a^3*d*e^3 + 6*
A*a^2*b*d*e^3 + 10*A*a^3*e^4)*e^(-5)/(x*e + d)^6

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maple [B]  time = 0.01, size = 281, normalized size = 2.11 \begin {gather*} -\frac {B \,b^{3}}{2 \left (e x +d \right )^{2} e^{5}}-\frac {\left (A b e +3 B a e -4 B b d \right ) b^{2}}{3 \left (e x +d \right )^{3} e^{5}}-\frac {3 \left (A a b \,e^{2}-A d \,b^{2} e +B \,a^{2} e^{2}-3 B d a b e +2 B \,b^{2} d^{2}\right ) b}{4 \left (e x +d \right )^{4} e^{5}}-\frac {3 A \,a^{2} b \,e^{3}-6 A d a \,b^{2} e^{2}+3 A \,d^{2} b^{3} e +B \,a^{3} e^{3}-6 B d \,a^{2} b \,e^{2}+9 B \,d^{2} a \,b^{2} e -4 B \,b^{3} d^{3}}{5 \left (e x +d \right )^{5} e^{5}}-\frac {A \,a^{3} e^{4}-3 A d \,a^{2} b \,e^{3}+3 A \,d^{2} a \,b^{2} e^{2}-A \,d^{3} b^{3} e -B d \,a^{3} e^{3}+3 B \,d^{2} a^{2} b \,e^{2}-3 B \,d^{3} a \,b^{2} e +B \,b^{3} d^{4}}{6 \left (e x +d \right )^{6} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^7,x)

[Out]

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

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maxima [B]  time = 0.58, size = 317, normalized size = 2.38 \begin {gather*} -\frac {30 \, B b^{3} e^{4} x^{4} + 2 \, B b^{3} d^{4} + 10 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 20 \, {\left (2 \, B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 15 \, {\left (2 \, B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 6 \, {\left (2 \, B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{60 \, {\left (e^{11} x^{6} + 6 \, d e^{10} x^{5} + 15 \, d^{2} e^{9} x^{4} + 20 \, d^{3} e^{8} x^{3} + 15 \, d^{4} e^{7} x^{2} + 6 \, d^{5} e^{6} x + d^{6} e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^7,x, algorithm="maxima")

[Out]

-1/60*(30*B*b^3*e^4*x^4 + 2*B*b^3*d^4 + 10*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^2)*d^2*e
^2 + 2*(B*a^3 + 3*A*a^2*b)*d*e^3 + 20*(2*B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 15*(2*B*b^3*d^2*e^2 + (3
*B*a*b^2 + A*b^3)*d*e^3 + 3*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 6*(2*B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 3*
(B*a^2*b + A*a*b^2)*d*e^3 + 2*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^11*x^6 + 6*d*e^10*x^5 + 15*d^2*e^9*x^4 + 20*d^3*e
^8*x^3 + 15*d^4*e^7*x^2 + 6*d^5*e^6*x + d^6*e^5)

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mupad [B]  time = 1.15, size = 321, normalized size = 2.41 \begin {gather*} -\frac {\frac {2\,B\,a^3\,d\,e^3+10\,A\,a^3\,e^4+3\,B\,a^2\,b\,d^2\,e^2+6\,A\,a^2\,b\,d\,e^3+3\,B\,a\,b^2\,d^3\,e+3\,A\,a\,b^2\,d^2\,e^2+2\,B\,b^3\,d^4+A\,b^3\,d^3\,e}{60\,e^5}+\frac {x\,\left (2\,B\,a^3\,e^3+3\,B\,a^2\,b\,d\,e^2+6\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e+3\,A\,a\,b^2\,d\,e^2+2\,B\,b^3\,d^3+A\,b^3\,d^2\,e\right )}{10\,e^4}+\frac {b^2\,x^3\,\left (A\,b\,e+3\,B\,a\,e+2\,B\,b\,d\right )}{3\,e^2}+\frac {b\,x^2\,\left (3\,B\,a^2\,e^2+3\,B\,a\,b\,d\,e+3\,A\,a\,b\,e^2+2\,B\,b^2\,d^2+A\,b^2\,d\,e\right )}{4\,e^3}+\frac {B\,b^3\,x^4}{2\,e}}{d^6+6\,d^5\,e\,x+15\,d^4\,e^2\,x^2+20\,d^3\,e^3\,x^3+15\,d^2\,e^4\,x^4+6\,d\,e^5\,x^5+e^6\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x)^3)/(d + e*x)^7,x)

[Out]

-((10*A*a^3*e^4 + 2*B*b^3*d^4 + A*b^3*d^3*e + 2*B*a^3*d*e^3 + 3*A*a*b^2*d^2*e^2 + 3*B*a^2*b*d^2*e^2 + 6*A*a^2*
b*d*e^3 + 3*B*a*b^2*d^3*e)/(60*e^5) + (x*(2*B*a^3*e^3 + 2*B*b^3*d^3 + 6*A*a^2*b*e^3 + A*b^3*d^2*e + 3*A*a*b^2*
d*e^2 + 3*B*a*b^2*d^2*e + 3*B*a^2*b*d*e^2))/(10*e^4) + (b^2*x^3*(A*b*e + 3*B*a*e + 2*B*b*d))/(3*e^2) + (b*x^2*
(3*B*a^2*e^2 + 2*B*b^2*d^2 + 3*A*a*b*e^2 + A*b^2*d*e + 3*B*a*b*d*e))/(4*e^3) + (B*b^3*x^4)/(2*e))/(d^6 + e^6*x
^6 + 6*d*e^5*x^5 + 15*d^4*e^2*x^2 + 20*d^3*e^3*x^3 + 15*d^2*e^4*x^4 + 6*d^5*e*x)

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sympy [B]  time = 157.68, size = 386, normalized size = 2.90 \begin {gather*} \frac {- 10 A a^{3} e^{4} - 6 A a^{2} b d e^{3} - 3 A a b^{2} d^{2} e^{2} - A b^{3} d^{3} e - 2 B a^{3} d e^{3} - 3 B a^{2} b d^{2} e^{2} - 3 B a b^{2} d^{3} e - 2 B b^{3} d^{4} - 30 B b^{3} e^{4} x^{4} + x^{3} \left (- 20 A b^{3} e^{4} - 60 B a b^{2} e^{4} - 40 B b^{3} d e^{3}\right ) + x^{2} \left (- 45 A a b^{2} e^{4} - 15 A b^{3} d e^{3} - 45 B a^{2} b e^{4} - 45 B a b^{2} d e^{3} - 30 B b^{3} d^{2} e^{2}\right ) + x \left (- 36 A a^{2} b e^{4} - 18 A a b^{2} d e^{3} - 6 A b^{3} d^{2} e^{2} - 12 B a^{3} e^{4} - 18 B a^{2} b d e^{3} - 18 B a b^{2} d^{2} e^{2} - 12 B b^{3} d^{3} e\right )}{60 d^{6} e^{5} + 360 d^{5} e^{6} x + 900 d^{4} e^{7} x^{2} + 1200 d^{3} e^{8} x^{3} + 900 d^{2} e^{9} x^{4} + 360 d e^{10} x^{5} + 60 e^{11} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**7,x)

[Out]

(-10*A*a**3*e**4 - 6*A*a**2*b*d*e**3 - 3*A*a*b**2*d**2*e**2 - A*b**3*d**3*e - 2*B*a**3*d*e**3 - 3*B*a**2*b*d**
2*e**2 - 3*B*a*b**2*d**3*e - 2*B*b**3*d**4 - 30*B*b**3*e**4*x**4 + x**3*(-20*A*b**3*e**4 - 60*B*a*b**2*e**4 -
40*B*b**3*d*e**3) + x**2*(-45*A*a*b**2*e**4 - 15*A*b**3*d*e**3 - 45*B*a**2*b*e**4 - 45*B*a*b**2*d*e**3 - 30*B*
b**3*d**2*e**2) + x*(-36*A*a**2*b*e**4 - 18*A*a*b**2*d*e**3 - 6*A*b**3*d**2*e**2 - 12*B*a**3*e**4 - 18*B*a**2*
b*d*e**3 - 18*B*a*b**2*d**2*e**2 - 12*B*b**3*d**3*e))/(60*d**6*e**5 + 360*d**5*e**6*x + 900*d**4*e**7*x**2 + 1
200*d**3*e**8*x**3 + 900*d**2*e**9*x**4 + 360*d*e**10*x**5 + 60*e**11*x**6)

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