3.16.12 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^7} \, dx\)

Optimal. Leaf size=193 \[ \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (B d-A e)}{6 e (d+e x)^6 (b d-a e)} \]

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Rubi [A]  time = 0.15, antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {770, 78, 45, 37} \begin {gather*} \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{60 e (d+e x)^4 (b d-a e)^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (-3 a B e+A b e+2 b B d)}{15 e (d+e x)^5 (b d-a e)^2}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (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)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^7,x]

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 229, normalized size = 1.19 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (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 )\right )}{60 e^5 (a+b x) (d+e x)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^7,x]

[Out]

-1/60*(Sqrt[(a + b*x)^2]*(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*(a + b*x)*(d + e*x)^6)

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IntegrateAlgebraic [F]  time = 180.03, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^7,x]

[Out]

$Aborted

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fricas [B]  time = 0.42, size = 317, normalized size = 1.64 \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)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(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 = 0.18, size = 426, normalized size = 2.21 \begin {gather*} -\frac {{\left (30 \, B b^{3} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 40 \, B b^{3} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 30 \, B b^{3} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B b^{3} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + 2 \, B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 60 \, B a b^{2} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 20 \, A b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 45 \, B a b^{2} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, A b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, B a b^{2} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, A b^{3} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a b^{2} d^{3} e \mathrm {sgn}\left (b x + a\right ) + A b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 45 \, B a^{2} b x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 45 \, A a b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 18 \, B a^{2} b d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 18 \, A a b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 12 \, B a^{3} x e^{4} \mathrm {sgn}\left (b x + a\right ) + 36 \, A a^{2} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 2 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 6 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\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)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^7,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 316, normalized size = 1.64 \begin {gather*} -\frac {\left (30 B \,b^{3} e^{4} x^{4}+20 A \,b^{3} e^{4} x^{3}+60 B a \,b^{2} e^{4} x^{3}+40 B \,b^{3} d \,e^{3} x^{3}+45 A a \,b^{2} e^{4} x^{2}+15 A \,b^{3} d \,e^{3} x^{2}+45 B \,a^{2} b \,e^{4} x^{2}+45 B a \,b^{2} d \,e^{3} x^{2}+30 B \,b^{3} d^{2} e^{2} x^{2}+36 A \,a^{2} b \,e^{4} x +18 A a \,b^{2} d \,e^{3} x +6 A \,b^{3} d^{2} e^{2} x +12 B \,a^{3} e^{4} x +18 B \,a^{2} b d \,e^{3} x +18 B a \,b^{2} d^{2} e^{2} x +12 B \,b^{3} d^{3} e x +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}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (e x +d \right )^{6} \left (b x +a \right )^{3} 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)^(3/2)/(e*x+d)^7,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \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)^(3/2)/(e*x+d)^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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mupad [B]  time = 2.17, size = 577, normalized size = 2.99 \begin {gather*} -\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{3\,e^5}-\frac {B\,b^3\,d}{3\,e^5}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}-\frac {\left (\frac {A\,a^3}{6\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{6\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{6\,e}-\frac {B\,b^3\,d}{6\,e^2}\right )}{e}-\frac {a\,b\,\left (A\,b+B\,a\right )}{2\,e}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^6}-\frac {\left (\frac {B\,a^3\,e^3-3\,B\,a^2\,b\,d\,e^2+3\,A\,a^2\,b\,e^3+3\,B\,a\,b^2\,d^2\,e-3\,A\,a\,b^2\,d\,e^2-B\,b^3\,d^3+A\,b^3\,d^2\,e}{5\,e^5}-\frac {d\,\left (\frac {3\,B\,a^2\,b\,e^3-3\,B\,a\,b^2\,d\,e^2+3\,A\,a\,b^2\,e^3+B\,b^3\,d^2\,e-A\,b^3\,d\,e^2}{5\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{5\,e^3}-\frac {B\,b^3\,d}{5\,e^3}\right )}{e}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^5}-\frac {\left (\frac {3\,B\,a^2\,b\,e^2-6\,B\,a\,b^2\,d\,e+3\,A\,a\,b^2\,e^2+3\,B\,b^3\,d^2-2\,A\,b^3\,d\,e}{4\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{4\,e^4}-\frac {B\,b^3\,d}{4\,e^4}\right )}{e}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^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)^(3/2))/(d + e*x)^7,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{7}}\, dx \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)**(3/2)/(e*x+d)**7,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/(d + e*x)**7, x)

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