3.1849 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=304 \[ -\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5 (a+b x)}+\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac {2 b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {770, 77} \[ -\frac {2 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5 (a+b x)}+\frac {6 b \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5 (a+b x)}+\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 (a+b x) \sqrt {d+e x}}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 (B d-A e)}{3 e^5 (a+b x) (d+e x)^{3/2}}+\frac {2 b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.18, size = 241, normalized size = 0.79 \[ \frac {2 \sqrt {(a+b x)^2} \left (-5 a^3 e^3 (A e+2 B d+3 B e x)+15 a^2 b e^2 \left (B \left (8 d^2+12 d e x+3 e^2 x^2\right )-A e (2 d+3 e x)\right )+15 a b^2 e \left (A e \left (8 d^2+12 d e x+3 e^2 x^2\right )+B \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )\right )+b^3 \left (5 A e \left (-16 d^3-24 d^2 e x-6 d e^2 x^2+e^3 x^3\right )+B \left (128 d^4+192 d^3 e x+48 d^2 e^2 x^2-8 d e^3 x^3+3 e^4 x^4\right )\right )\right )}{15 e^5 (a+b x) (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(5/2),x]

[Out]

(2*Sqrt[(a + b*x)^2]*(-5*a^3*e^3*(2*B*d + A*e + 3*B*e*x) + 15*a^2*b*e^2*(-(A*e*(2*d + 3*e*x)) + B*(8*d^2 + 12*
d*e*x + 3*e^2*x^2)) + 15*a*b^2*e*(A*e*(8*d^2 + 12*d*e*x + 3*e^2*x^2) + B*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 +
e^3*x^3)) + b^3*(5*A*e*(-16*d^3 - 24*d^2*e*x - 6*d*e^2*x^2 + e^3*x^3) + B*(128*d^4 + 192*d^3*e*x + 48*d^2*e^2
*x^2 - 8*d*e^3*x^3 + 3*e^4*x^4))))/(15*e^5*(a + b*x)*(d + e*x)^(3/2))

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fricas [A]  time = 0.62, size = 284, normalized size = 0.93 \[ \frac {2 \, {\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - {\left (8 \, B b^{3} d e^{3} - 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \, {\left (16 \, B b^{3} d^{2} e^{2} - 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \, {\left (64 \, B b^{3} d^{3} e - 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d
^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2
- 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*
d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^
2*e^5)

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giac [B]  time = 0.25, size = 509, normalized size = 1.67 \[ \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{3} e^{20} \mathrm {sgn}\left (b x + a\right ) - 20 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d e^{20} \mathrm {sgn}\left (b x + a\right ) + 90 \, \sqrt {x e + d} B b^{3} d^{2} e^{20} \mathrm {sgn}\left (b x + a\right ) + 15 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} e^{21} \mathrm {sgn}\left (b x + a\right ) + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} e^{21} \mathrm {sgn}\left (b x + a\right ) - 135 \, \sqrt {x e + d} B a b^{2} d e^{21} \mathrm {sgn}\left (b x + a\right ) - 45 \, \sqrt {x e + d} A b^{3} d e^{21} \mathrm {sgn}\left (b x + a\right ) + 45 \, \sqrt {x e + d} B a^{2} b e^{22} \mathrm {sgn}\left (b x + a\right ) + 45 \, \sqrt {x e + d} A a b^{2} e^{22} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-25\right )} + \frac {2 \, {\left (12 \, {\left (x e + d\right )} B b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - B b^{3} d^{4} \mathrm {sgn}\left (b x + a\right ) - 27 \, {\left (x e + d\right )} B a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 9 \, {\left (x e + d\right )} A b^{3} d^{2} e \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 ) + 18 \, {\left (x e + d\right )} B a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) + 18 \, {\left (x e + d\right )} A a b^{2} d e^{2} \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 ) - 3 \, {\left (x e + d\right )} B a^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 9 \, {\left (x e + d\right )} A a^{2} b e^{3} \mathrm {sgn}\left (b x + a\right ) + B a^{3} d e^{3} \mathrm {sgn}\left (b x + a\right ) + 3 \, A a^{2} b d e^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{3} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \]

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)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*B*b^3*e^20*sgn(b*x + a) - 20*(x*e + d)^(3/2)*B*b^3*d*e^20*sgn(b*x + a) + 90*sqrt(x*e +
d)*B*b^3*d^2*e^20*sgn(b*x + a) + 15*(x*e + d)^(3/2)*B*a*b^2*e^21*sgn(b*x + a) + 5*(x*e + d)^(3/2)*A*b^3*e^21*
sgn(b*x + a) - 135*sqrt(x*e + d)*B*a*b^2*d*e^21*sgn(b*x + a) - 45*sqrt(x*e + d)*A*b^3*d*e^21*sgn(b*x + a) + 45
*sqrt(x*e + d)*B*a^2*b*e^22*sgn(b*x + a) + 45*sqrt(x*e + d)*A*a*b^2*e^22*sgn(b*x + a))*e^(-25) + 2/3*(12*(x*e
+ d)*B*b^3*d^3*sgn(b*x + a) - B*b^3*d^4*sgn(b*x + a) - 27*(x*e + d)*B*a*b^2*d^2*e*sgn(b*x + a) - 9*(x*e + d)*A
*b^3*d^2*e*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) + 18*(x*e + d)*B*a^2*b*d*e^2
*sgn(b*x + a) + 18*(x*e + d)*A*a*b^2*d*e^2*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*s
gn(b*x + a) - 3*(x*e + d)*B*a^3*e^3*sgn(b*x + a) - 9*(x*e + d)*A*a^2*b*e^3*sgn(b*x + a) + B*a^3*d*e^3*sgn(b*x
+ a) + 3*A*a^2*b*d*e^3*sgn(b*x + a) - A*a^3*e^4*sgn(b*x + a))*e^(-5)/(x*e + d)^(3/2)

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maple [A]  time = 0.05, size = 317, normalized size = 1.04 \[ -\frac {2 \left (-3 b^{3} B \,x^{4} e^{4}-5 A \,b^{3} e^{4} x^{3}-15 B a \,b^{2} e^{4} x^{3}+8 B \,b^{3} d \,e^{3} x^{3}-45 A a \,b^{2} e^{4} x^{2}+30 A \,b^{3} d \,e^{3} x^{2}-45 B \,a^{2} b \,e^{4} x^{2}+90 B a \,b^{2} d \,e^{3} x^{2}-48 B \,b^{3} d^{2} e^{2} x^{2}+45 A \,a^{2} b \,e^{4} x -180 A a \,b^{2} d \,e^{3} x +120 A \,b^{3} d^{2} e^{2} x +15 B \,a^{3} e^{4} x -180 B \,a^{2} b d \,e^{3} x +360 B a \,b^{2} d^{2} e^{2} x -192 B \,b^{3} d^{3} e x +5 A \,a^{3} e^{4}+30 A \,a^{2} b d \,e^{3}-120 A a \,b^{2} d^{2} e^{2}+80 A \,b^{3} d^{3} e +10 B \,a^{3} d \,e^{3}-120 B \,a^{2} b \,d^{2} e^{2}+240 B a \,b^{2} d^{3} e -128 B \,b^{3} d^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{15 \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{3} e^{5}} \]

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)^(5/2),x)

[Out]

-2/15/(e*x+d)^(3/2)*(-3*B*b^3*e^4*x^4-5*A*b^3*e^4*x^3-15*B*a*b^2*e^4*x^3+8*B*b^3*d*e^3*x^3-45*A*a*b^2*e^4*x^2+
30*A*b^3*d*e^3*x^2-45*B*a^2*b*e^4*x^2+90*B*a*b^2*d*e^3*x^2-48*B*b^3*d^2*e^2*x^2+45*A*a^2*b*e^4*x-180*A*a*b^2*d
*e^3*x+120*A*b^3*d^2*e^2*x+15*B*a^3*e^4*x-180*B*a^2*b*d*e^3*x+360*B*a*b^2*d^2*e^2*x-192*B*b^3*d^3*e*x+5*A*a^3*
e^4+30*A*a^2*b*d*e^3-120*A*a*b^2*d^2*e^2+80*A*b^3*d^3*e+10*B*a^3*d*e^3-120*B*a^2*b*d^2*e^2+240*B*a*b^2*d^3*e-1
28*B*b^3*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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maxima [A]  time = 0.65, size = 304, normalized size = 1.00 \[ \frac {2 \, {\left (b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 24 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} - a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} - 3 \, {\left (8 \, b^{3} d^{2} e - 12 \, a b^{2} d e^{2} + 3 \, a^{2} b e^{3}\right )} x\right )} A}{3 \, {\left (e^{5} x + d e^{4}\right )} \sqrt {e x + d}} + \frac {2 \, {\left (3 \, b^{3} e^{4} x^{4} + 128 \, b^{3} d^{4} - 240 \, a b^{2} d^{3} e + 120 \, a^{2} b d^{2} e^{2} - 10 \, a^{3} d e^{3} - {\left (8 \, b^{3} d e^{3} - 15 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (16 \, b^{3} d^{2} e^{2} - 30 \, a b^{2} d e^{3} + 15 \, a^{2} b e^{4}\right )} x^{2} + 3 \, {\left (64 \, b^{3} d^{3} e - 120 \, a b^{2} d^{2} e^{2} + 60 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x\right )} B}{15 \, {\left (e^{6} x + d e^{5}\right )} \sqrt {e x + d}} \]

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)^(5/2),x, algorithm="maxima")

[Out]

2/3*(b^3*e^3*x^3 - 16*b^3*d^3 + 24*a*b^2*d^2*e - 6*a^2*b*d*e^2 - a^3*e^3 - 3*(2*b^3*d*e^2 - 3*a*b^2*e^3)*x^2 -
3*(8*b^3*d^2*e - 12*a*b^2*d*e^2 + 3*a^2*b*e^3)*x)*A/((e^5*x + d*e^4)*sqrt(e*x + d)) + 2/15*(3*b^3*e^4*x^4 + 1
28*b^3*d^4 - 240*a*b^2*d^3*e + 120*a^2*b*d^2*e^2 - 10*a^3*d*e^3 - (8*b^3*d*e^3 - 15*a*b^2*e^4)*x^3 + 3*(16*b^3
*d^2*e^2 - 30*a*b^2*d*e^3 + 15*a^2*b*e^4)*x^2 + 3*(64*b^3*d^3*e - 120*a*b^2*d^2*e^2 + 60*a^2*b*d*e^3 - 5*a^3*e
^4)*x)*B/((e^6*x + d*e^5)*sqrt(e*x + d))

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mupad [B]  time = 3.09, size = 362, normalized size = 1.19 \[ \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {x^2\,\left (6\,B\,a^2\,b\,e^4-12\,B\,a\,b^2\,d\,e^3+6\,A\,a\,b^2\,e^4+\frac {32\,B\,b^3\,d^2\,e^2}{5}-4\,A\,b^3\,d\,e^3\right )}{b\,e^6}-\frac {x\,\left (30\,B\,a^3\,e^4-360\,B\,a^2\,b\,d\,e^3+90\,A\,a^2\,b\,e^4+720\,B\,a\,b^2\,d^2\,e^2-360\,A\,a\,b^2\,d\,e^3-384\,B\,b^3\,d^3\,e+240\,A\,b^3\,d^2\,e^2\right )}{15\,b\,e^6}-\frac {\frac {4\,B\,a^3\,d\,e^3}{3}+\frac {2\,A\,a^3\,e^4}{3}-16\,B\,a^2\,b\,d^2\,e^2+4\,A\,a^2\,b\,d\,e^3+32\,B\,a\,b^2\,d^3\,e-16\,A\,a\,b^2\,d^2\,e^2-\frac {256\,B\,b^3\,d^4}{15}+\frac {32\,A\,b^3\,d^3\,e}{3}}{b\,e^6}+\frac {2\,b\,x^3\,\left (5\,A\,b\,e+15\,B\,a\,e-8\,B\,b\,d\right )}{15\,e^3}+\frac {2\,B\,b^2\,x^4}{5\,e^2}\right )}{x^2\,\sqrt {d+e\,x}+\frac {a\,d\,\sqrt {d+e\,x}}{b\,e}+\frac {x\,\left (15\,a\,e^6+15\,b\,d\,e^5\right )\,\sqrt {d+e\,x}}{15\,b\,e^6}} \]

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)^(5/2),x)

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((x^2*(6*A*a*b^2*e^4 + 6*B*a^2*b*e^4 - 4*A*b^3*d*e^3 + (32*B*b^3*d^2*e^2)/5 -
12*B*a*b^2*d*e^3))/(b*e^6) - (x*(30*B*a^3*e^4 + 90*A*a^2*b*e^4 - 384*B*b^3*d^3*e + 240*A*b^3*d^2*e^2 + 720*B*
a*b^2*d^2*e^2 - 360*A*a*b^2*d*e^3 - 360*B*a^2*b*d*e^3))/(15*b*e^6) - ((2*A*a^3*e^4)/3 - (256*B*b^3*d^4)/15 + (
32*A*b^3*d^3*e)/3 + (4*B*a^3*d*e^3)/3 - 16*A*a*b^2*d^2*e^2 - 16*B*a^2*b*d^2*e^2 + 4*A*a^2*b*d*e^3 + 32*B*a*b^2
*d^3*e)/(b*e^6) + (2*b*x^3*(5*A*b*e + 15*B*a*e - 8*B*b*d))/(15*e^3) + (2*B*b^2*x^4)/(5*e^2)))/(x^2*(d + e*x)^(
1/2) + (a*d*(d + e*x)^(1/2))/(b*e) + (x*(15*a*e^6 + 15*b*d*e^5)*(d + e*x)^(1/2))/(15*b*e^6))

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

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)**(5/2),x)

[Out]

Timed out

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