3.16.14 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^9} \, dx\)

Optimal. Leaf size=298 \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+4 b B d)}{5 e^5 (a+b x) (d+e x)^5}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{2 e^5 (a+b x) (d+e x)^6}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (a+b x) (d+e x)^7}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{8 e^5 (a+b x) (d+e x)^8}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4} \]

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Rubi [A]  time = 0.18, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 77} \begin {gather*} \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (-3 a B e-A b e+4 b B d)}{5 e^5 (a+b x) (d+e x)^5}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (-a B e-A b e+2 b B d)}{2 e^5 (a+b x) (d+e x)^6}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (a+b x) (d+e x)^7}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{8 e^5 (a+b x) (d+e x)^8}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4} \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)^9,x]

[Out]

-((b*d - a*e)^3*(B*d - A*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*e^5*(a + b*x)*(d + e*x)^8) + ((b*d - a*e)^2*(4*b
*B*d - 3*A*b*e - a*B*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)*(d + e*x)^7) - (b*(b*d - a*e)*(2*b*B*d
- A*b*e - a*B*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*e^5*(a + b*x)*(d + e*x)^6) + (b^2*(4*b*B*d - A*b*e - 3*a*B
*e)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x)*(d + e*x)^5) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*e^
5*(a + b*x)*(d + e*x)^4)

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]))))

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)^9} \, 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)^9} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (-\frac {b^3 (b d-a e)^3 (-B d+A e)}{e^4 (d+e x)^9}+\frac {b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e)}{e^4 (d+e x)^8}-\frac {3 b^4 (b d-a e) (-2 b B d+A b e+a B e)}{e^4 (d+e x)^7}+\frac {b^5 (-4 b B d+A b e+3 a B e)}{e^4 (d+e x)^6}+\frac {b^6 B}{e^4 (d+e x)^5}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac {(b d-a e)^3 (B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{8 e^5 (a+b x) (d+e x)^8}+\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x) (d+e x)^7}-\frac {b (b d-a e) (2 b B d-A b e-a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^5 (a+b x) (d+e x)^6}+\frac {b^2 (4 b B d-A b e-3 a B e) \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x) (d+e x)^5}-\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2}}{4 e^5 (a+b x) (d+e x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 229, normalized size = 0.77 \begin {gather*} -\frac {\sqrt {(a+b x)^2} \left (5 a^3 e^3 (7 A e+B (d+8 e x))+5 a^2 b e^2 \left (3 A e (d+8 e x)+B \left (d^2+8 d e x+28 e^2 x^2\right )\right )+a b^2 e \left (5 A e \left (d^2+8 d e x+28 e^2 x^2\right )+3 B \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )\right )+b^3 \left (A e \left (d^3+8 d^2 e x+28 d e^2 x^2+56 e^3 x^3\right )+B \left (d^4+8 d^3 e x+28 d^2 e^2 x^2+56 d e^3 x^3+70 e^4 x^4\right )\right )\right )}{280 e^5 (a+b x) (d+e x)^8} \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)^9,x]

[Out]

-1/280*(Sqrt[(a + b*x)^2]*(5*a^3*e^3*(7*A*e + B*(d + 8*e*x)) + 5*a^2*b*e^2*(3*A*e*(d + 8*e*x) + B*(d^2 + 8*d*e
*x + 28*e^2*x^2)) + a*b^2*e*(5*A*e*(d^2 + 8*d*e*x + 28*e^2*x^2) + 3*B*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3
*x^3)) + b^3*(A*e*(d^3 + 8*d^2*e*x + 28*d*e^2*x^2 + 56*e^3*x^3) + B*(d^4 + 8*d^3*e*x + 28*d^2*e^2*x^2 + 56*d*e
^3*x^3 + 70*e^4*x^4))))/(e^5*(a + b*x)*(d + e*x)^8)

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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)^9,x]

[Out]

$Aborted

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fricas [A]  time = 0.44, size = 335, normalized size = 1.12 \begin {gather*} -\frac {70 \, B b^{3} e^{4} x^{4} + B b^{3} d^{4} + 35 \, A a^{3} e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} + 56 \, {\left (B b^{3} d e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 28 \, {\left (B b^{3} d^{2} e^{2} + {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 8 \, {\left (B b^{3} d^{3} e + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 5 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x}{280 \, {\left (e^{13} x^{8} + 8 \, d e^{12} x^{7} + 28 \, d^{2} e^{11} x^{6} + 56 \, d^{3} e^{10} x^{5} + 70 \, d^{4} e^{9} x^{4} + 56 \, d^{5} e^{8} x^{3} + 28 \, d^{6} e^{7} x^{2} + 8 \, d^{7} e^{6} x + d^{8} 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)^9,x, algorithm="fricas")

[Out]

-1/280*(70*B*b^3*e^4*x^4 + B*b^3*d^4 + 35*A*a^3*e^4 + (3*B*a*b^2 + A*b^3)*d^3*e + 5*(B*a^2*b + A*a*b^2)*d^2*e^
2 + 5*(B*a^3 + 3*A*a^2*b)*d*e^3 + 56*(B*b^3*d*e^3 + (3*B*a*b^2 + A*b^3)*e^4)*x^3 + 28*(B*b^3*d^2*e^2 + (3*B*a*
b^2 + A*b^3)*d*e^3 + 5*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 8*(B*b^3*d^3*e + (3*B*a*b^2 + A*b^3)*d^2*e^2 + 5*(B*a^2*
b + A*a*b^2)*d*e^3 + 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)/(e^13*x^8 + 8*d*e^12*x^7 + 28*d^2*e^11*x^6 + 56*d^3*e^10*x^
5 + 70*d^4*e^9*x^4 + 56*d^5*e^8*x^3 + 28*d^6*e^7*x^2 + 8*d^7*e^6*x + d^8*e^5)

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giac [A]  time = 0.19, size = 425, normalized size = 1.43 \begin {gather*} -\frac {{\left (70 \, B b^{3} x^{4} e^{4} \mathrm {sgn}\left (b x + a\right ) + 56 \, B b^{3} d x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, B b^{3} d^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, B b^{3} d^{3} x e \mathrm {sgn}\left (b x + a\right ) + B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) + 168 \, B a b^{2} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 56 \, A b^{3} x^{3} e^{4} \mathrm {sgn}\left (b x + a\right ) + 84 \, B a b^{2} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 28 \, A b^{3} d x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 24 \, B a b^{2} d^{2} x e^{2} \mathrm {sgn}\left (b x + a\right ) + 8 \, 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 ) + 140 \, B a^{2} b x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 140 \, A a b^{2} x^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{2} b d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 40 \, A a b^{2} d x e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{2} b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 5 \, A a b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 40 \, B a^{3} x e^{4} \mathrm {sgn}\left (b x + a\right ) + 120 \, A a^{2} b x e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 15 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + 35 \, A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{280 \, {\left (x e + d\right )}^{8}} \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)^9,x, algorithm="giac")

[Out]

-1/280*(70*B*b^3*x^4*e^4*sgn(b*x + a) + 56*B*b^3*d*x^3*e^3*sgn(b*x + a) + 28*B*b^3*d^2*x^2*e^2*sgn(b*x + a) +
8*B*b^3*d^3*x*e*sgn(b*x + a) + B*b^3*d^4*sgn(b*x + a) + 168*B*a*b^2*x^3*e^4*sgn(b*x + a) + 56*A*b^3*x^3*e^4*sg
n(b*x + a) + 84*B*a*b^2*d*x^2*e^3*sgn(b*x + a) + 28*A*b^3*d*x^2*e^3*sgn(b*x + a) + 24*B*a*b^2*d^2*x*e^2*sgn(b*
x + a) + 8*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) + 140*B*a^2*
b*x^2*e^4*sgn(b*x + a) + 140*A*a*b^2*x^2*e^4*sgn(b*x + a) + 40*B*a^2*b*d*x*e^3*sgn(b*x + a) + 40*A*a*b^2*d*x*e
^3*sgn(b*x + a) + 5*B*a^2*b*d^2*e^2*sgn(b*x + a) + 5*A*a*b^2*d^2*e^2*sgn(b*x + a) + 40*B*a^3*x*e^4*sgn(b*x + a
) + 120*A*a^2*b*x*e^4*sgn(b*x + a) + 5*B*a^3*d*e^3*sgn(b*x + a) + 15*A*a^2*b*d*e^3*sgn(b*x + a) + 35*A*a^3*e^4
*sgn(b*x + a))*e^(-5)/(x*e + d)^8

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maple [A]  time = 0.06, size = 315, normalized size = 1.06 \begin {gather*} -\frac {\left (70 B \,b^{3} e^{4} x^{4}+56 A \,b^{3} e^{4} x^{3}+168 B a \,b^{2} e^{4} x^{3}+56 B \,b^{3} d \,e^{3} x^{3}+140 A a \,b^{2} e^{4} x^{2}+28 A \,b^{3} d \,e^{3} x^{2}+140 B \,a^{2} b \,e^{4} x^{2}+84 B a \,b^{2} d \,e^{3} x^{2}+28 B \,b^{3} d^{2} e^{2} x^{2}+120 A \,a^{2} b \,e^{4} x +40 A a \,b^{2} d \,e^{3} x +8 A \,b^{3} d^{2} e^{2} x +40 B \,a^{3} e^{4} x +40 B \,a^{2} b d \,e^{3} x +24 B a \,b^{2} d^{2} e^{2} x +8 B \,b^{3} d^{3} e x +35 A \,a^{3} e^{4}+15 A \,a^{2} b d \,e^{3}+5 A a \,b^{2} d^{2} e^{2}+A \,b^{3} d^{3} e +5 B \,a^{3} d \,e^{3}+5 B \,a^{2} b \,d^{2} e^{2}+3 B a \,b^{2} d^{3} e +B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{280 \left (e x +d \right )^{8} \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)^9,x)

[Out]

-1/280/e^5*(70*B*b^3*e^4*x^4+56*A*b^3*e^4*x^3+168*B*a*b^2*e^4*x^3+56*B*b^3*d*e^3*x^3+140*A*a*b^2*e^4*x^2+28*A*
b^3*d*e^3*x^2+140*B*a^2*b*e^4*x^2+84*B*a*b^2*d*e^3*x^2+28*B*b^3*d^2*e^2*x^2+120*A*a^2*b*e^4*x+40*A*a*b^2*d*e^3
*x+8*A*b^3*d^2*e^2*x+40*B*a^3*e^4*x+40*B*a^2*b*d*e^3*x+24*B*a*b^2*d^2*e^2*x+8*B*b^3*d^3*e*x+35*A*a^3*e^4+15*A*
a^2*b*d*e^3+5*A*a*b^2*d^2*e^2+A*b^3*d^3*e+5*B*a^3*d*e^3+5*B*a^2*b*d^2*e^2+3*B*a*b^2*d^3*e+B*b^3*d^4)*((b*x+a)^
2)^(3/2)/(e*x+d)^8/(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)^9,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 = 2.27, size = 577, normalized size = 1.94 \begin {gather*} -\frac {\left (\frac {A\,b^3\,e-3\,B\,b^3\,d+3\,B\,a\,b^2\,e}{5\,e^5}-\frac {B\,b^3\,d}{5\,e^5}\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 {A\,a^3}{8\,e}-\frac {d\,\left (\frac {B\,a^3+3\,A\,b\,a^2}{8\,e}+\frac {d\,\left (\frac {d\,\left (\frac {A\,b^3+3\,B\,a\,b^2}{8\,e}-\frac {B\,b^3\,d}{8\,e^2}\right )}{e}-\frac {3\,a\,b\,\left (A\,b+B\,a\right )}{8\,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 )}^8}-\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}{7\,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}{7\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-B\,b\,d\right )}{7\,e^3}-\frac {B\,b^3\,d}{7\,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 )}^7}-\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}{6\,e^5}-\frac {d\,\left (\frac {b^2\,\left (A\,b\,e+3\,B\,a\,e-2\,B\,b\,d\right )}{6\,e^4}-\frac {B\,b^3\,d}{6\,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 )}^6}-\frac {B\,b^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,e^5\,\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4} \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)^9,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**9,x)

[Out]

Timed out

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