3.1798 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^2}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=214 \[ -\frac{2 b^3 (d+e x)^{5/2} (-4 a B e-A b e+5 b B d)}{5 e^6}+\frac{4 b^2 (d+e x)^{3/2} (b d-a e) (-3 a B e-2 A b e+5 b B d)}{3 e^6}-\frac{4 b \sqrt{d+e x} (b d-a e)^2 (-2 a B e-3 A b e+5 b B d)}{e^6}-\frac{2 (b d-a e)^3 (-a B e-4 A b e+5 b B d)}{e^6 \sqrt{d+e x}}+\frac{2 (b d-a e)^4 (B d-A e)}{3 e^6 (d+e x)^{3/2}}+\frac{2 b^4 B (d+e x)^{7/2}}{7 e^6} \]

[Out]

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

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Rubi [A]  time = 0.0942208, antiderivative size = 214, normalized size of antiderivative = 1., 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 = {27, 77} \[ -\frac{2 b^3 (d+e x)^{5/2} (-4 a B e-A b e+5 b B d)}{5 e^6}+\frac{4 b^2 (d+e x)^{3/2} (b d-a e) (-3 a B e-2 A b e+5 b B d)}{3 e^6}-\frac{4 b \sqrt{d+e x} (b d-a e)^2 (-2 a B e-3 A b e+5 b B d)}{e^6}-\frac{2 (b d-a e)^3 (-a B e-4 A b e+5 b B d)}{e^6 \sqrt{d+e x}}+\frac{2 (b d-a e)^4 (B d-A e)}{3 e^6 (d+e x)^{3/2}}+\frac{2 b^4 B (d+e x)^{7/2}}{7 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.113058, size = 183, normalized size = 0.86 \[ \frac{2 \left (-21 b^3 (d+e x)^4 (-4 a B e-A b e+5 b B d)+70 b^2 (d+e x)^3 (b d-a e) (-3 a B e-2 A b e+5 b B d)-210 b (d+e x)^2 (b d-a e)^2 (-2 a B e-3 A b e+5 b B d)-105 (d+e x) (b d-a e)^3 (-a B e-4 A b e+5 b B d)+35 (b d-a e)^4 (B d-A e)+15 b^4 B (d+e x)^5\right )}{105 e^6 (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(35*(b*d - a*e)^4*(B*d - A*e) - 105*(b*d - a*e)^3*(5*b*B*d - 4*A*b*e - a*B*e)*(d + e*x) - 210*b*(b*d - a*e)
^2*(5*b*B*d - 3*A*b*e - 2*a*B*e)*(d + e*x)^2 + 70*b^2*(b*d - a*e)*(5*b*B*d - 2*A*b*e - 3*a*B*e)*(d + e*x)^3 -
21*b^3*(5*b*B*d - A*b*e - 4*a*B*e)*(d + e*x)^4 + 15*b^4*B*(d + e*x)^5))/(105*e^6*(d + e*x)^(3/2))

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Maple [B]  time = 0.009, size = 469, normalized size = 2.2 \begin{align*} -{\frac{-30\,{b}^{4}B{x}^{5}{e}^{5}-42\,A{b}^{4}{e}^{5}{x}^{4}-168\,Ba{b}^{3}{e}^{5}{x}^{4}+60\,B{b}^{4}d{e}^{4}{x}^{4}-280\,Aa{b}^{3}{e}^{5}{x}^{3}+112\,A{b}^{4}d{e}^{4}{x}^{3}-420\,B{a}^{2}{b}^{2}{e}^{5}{x}^{3}+448\,Ba{b}^{3}d{e}^{4}{x}^{3}-160\,B{b}^{4}{d}^{2}{e}^{3}{x}^{3}-1260\,A{a}^{2}{b}^{2}{e}^{5}{x}^{2}+1680\,Aa{b}^{3}d{e}^{4}{x}^{2}-672\,A{b}^{4}{d}^{2}{e}^{3}{x}^{2}-840\,B{a}^{3}b{e}^{5}{x}^{2}+2520\,B{a}^{2}{b}^{2}d{e}^{4}{x}^{2}-2688\,Ba{b}^{3}{d}^{2}{e}^{3}{x}^{2}+960\,B{b}^{4}{d}^{3}{e}^{2}{x}^{2}+840\,A{a}^{3}b{e}^{5}x-5040\,A{a}^{2}{b}^{2}d{e}^{4}x+6720\,Aa{b}^{3}{d}^{2}{e}^{3}x-2688\,A{b}^{4}{d}^{3}{e}^{2}x+210\,B{a}^{4}{e}^{5}x-3360\,B{a}^{3}bd{e}^{4}x+10080\,B{a}^{2}{b}^{2}{d}^{2}{e}^{3}x-10752\,Ba{b}^{3}{d}^{3}{e}^{2}x+3840\,B{b}^{4}{d}^{4}ex+70\,A{a}^{4}{e}^{5}+560\,Ad{a}^{3}b{e}^{4}-3360\,A{d}^{2}{a}^{2}{b}^{2}{e}^{3}+4480\,Aa{b}^{3}{d}^{3}{e}^{2}-1792\,A{b}^{4}{d}^{4}e+140\,B{a}^{4}d{e}^{4}-2240\,B{d}^{2}{a}^{3}b{e}^{3}+6720\,B{d}^{3}{a}^{2}{b}^{2}{e}^{2}-7168\,Ba{b}^{3}{d}^{4}e+2560\,{b}^{4}B{d}^{5}}{105\,{e}^{6}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

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)^2/(e*x+d)^(5/2),x)

[Out]

-2/105*(-15*B*b^4*e^5*x^5-21*A*b^4*e^5*x^4-84*B*a*b^3*e^5*x^4+30*B*b^4*d*e^4*x^4-140*A*a*b^3*e^5*x^3+56*A*b^4*
d*e^4*x^3-210*B*a^2*b^2*e^5*x^3+224*B*a*b^3*d*e^4*x^3-80*B*b^4*d^2*e^3*x^3-630*A*a^2*b^2*e^5*x^2+840*A*a*b^3*d
*e^4*x^2-336*A*b^4*d^2*e^3*x^2-420*B*a^3*b*e^5*x^2+1260*B*a^2*b^2*d*e^4*x^2-1344*B*a*b^3*d^2*e^3*x^2+480*B*b^4
*d^3*e^2*x^2+420*A*a^3*b*e^5*x-2520*A*a^2*b^2*d*e^4*x+3360*A*a*b^3*d^2*e^3*x-1344*A*b^4*d^3*e^2*x+105*B*a^4*e^
5*x-1680*B*a^3*b*d*e^4*x+5040*B*a^2*b^2*d^2*e^3*x-5376*B*a*b^3*d^3*e^2*x+1920*B*b^4*d^4*e*x+35*A*a^4*e^5+280*A
*a^3*b*d*e^4-1680*A*a^2*b^2*d^2*e^3+2240*A*a*b^3*d^3*e^2-896*A*b^4*d^4*e+70*B*a^4*d*e^4-1120*B*a^3*b*d^2*e^3+3
360*B*a^2*b^2*d^3*e^2-3584*B*a*b^3*d^4*e+1280*B*b^4*d^5)/(e*x+d)^(3/2)/e^6

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Maxima [B]  time = 1.04097, size = 560, normalized size = 2.62 \begin{align*} \frac{2 \,{\left (\frac{15 \,{\left (e x + d\right )}^{\frac{7}{2}} B b^{4} - 21 \,{\left (5 \, B b^{4} d -{\left (4 \, B a b^{3} + A b^{4}\right )} e\right )}{\left (e x + d\right )}^{\frac{5}{2}} + 70 \,{\left (5 \, B b^{4} d^{2} - 2 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d e +{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{2}\right )}{\left (e x + d\right )}^{\frac{3}{2}} - 210 \,{\left (5 \, B b^{4} d^{3} - 3 \,{\left (4 \, 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 (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{3}\right )} \sqrt{e x + d}}{e^{5}} + \frac{35 \,{\left (B b^{4} d^{5} - A a^{4} e^{5} -{\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} - 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} +{\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} - 3 \,{\left (5 \, B b^{4} d^{4} - 4 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e + 6 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{2} - 4 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{3} +{\left (B a^{4} + 4 \, A a^{3} b\right )} e^{4}\right )}{\left (e x + d\right )}\right )}}{{\left (e x + d\right )}^{\frac{3}{2}} e^{5}}\right )}}{105 \, e} \end{align*}

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)^2/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

2/105*((15*(e*x + d)^(7/2)*B*b^4 - 21*(5*B*b^4*d - (4*B*a*b^3 + A*b^4)*e)*(e*x + d)^(5/2) + 70*(5*B*b^4*d^2 -
2*(4*B*a*b^3 + A*b^4)*d*e + (3*B*a^2*b^2 + 2*A*a*b^3)*e^2)*(e*x + d)^(3/2) - 210*(5*B*b^4*d^3 - 3*(4*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 - (2*B*a^3*b + 3*A*a^2*b^2)*e^3)*sqrt(e*x + d))/e^5 + 35*(B*
b^4*d^5 - A*a^4*e^5 - (4*B*a*b^3 + A*b^4)*d^4*e + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*d^3*e^2 - 2*(2*B*a^3*b + 3*A*a^2
*b^2)*d^2*e^3 + (B*a^4 + 4*A*a^3*b)*d*e^4 - 3*(5*B*b^4*d^4 - 4*(4*B*a*b^3 + A*b^4)*d^3*e + 6*(3*B*a^2*b^2 + 2*
A*a*b^3)*d^2*e^2 - 4*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^3 + (B*a^4 + 4*A*a^3*b)*e^4)*(e*x + d))/((e*x + d)^(3/2)*e^
5))/e

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Fricas [B]  time = 1.37709, size = 933, normalized size = 4.36 \begin{align*} \frac{2 \,{\left (15 \, B b^{4} e^{5} x^{5} - 1280 \, B b^{4} d^{5} - 35 \, A a^{4} e^{5} + 896 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{4} e - 1120 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{3} e^{2} + 560 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d^{2} e^{3} - 70 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} d e^{4} - 3 \,{\left (10 \, B b^{4} d e^{4} - 7 \,{\left (4 \, B a b^{3} + A b^{4}\right )} e^{5}\right )} x^{4} + 2 \,{\left (40 \, B b^{4} d^{2} e^{3} - 28 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d e^{4} + 35 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} e^{5}\right )} x^{3} - 6 \,{\left (80 \, B b^{4} d^{3} e^{2} - 56 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{2} e^{3} + 70 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d e^{4} - 35 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} e^{5}\right )} x^{2} - 3 \,{\left (640 \, B b^{4} d^{4} e - 448 \,{\left (4 \, B a b^{3} + A b^{4}\right )} d^{3} e^{2} + 560 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} d^{2} e^{3} - 280 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} d e^{4} + 35 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} e^{5}\right )} x\right )} \sqrt{e x + d}}{105 \,{\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} \end{align*}

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)^2/(e*x+d)^(5/2),x, algorithm="fricas")

[Out]

2/105*(15*B*b^4*e^5*x^5 - 1280*B*b^4*d^5 - 35*A*a^4*e^5 + 896*(4*B*a*b^3 + A*b^4)*d^4*e - 1120*(3*B*a^2*b^2 +
2*A*a*b^3)*d^3*e^2 + 560*(2*B*a^3*b + 3*A*a^2*b^2)*d^2*e^3 - 70*(B*a^4 + 4*A*a^3*b)*d*e^4 - 3*(10*B*b^4*d*e^4
- 7*(4*B*a*b^3 + A*b^4)*e^5)*x^4 + 2*(40*B*b^4*d^2*e^3 - 28*(4*B*a*b^3 + A*b^4)*d*e^4 + 35*(3*B*a^2*b^2 + 2*A*
a*b^3)*e^5)*x^3 - 6*(80*B*b^4*d^3*e^2 - 56*(4*B*a*b^3 + A*b^4)*d^2*e^3 + 70*(3*B*a^2*b^2 + 2*A*a*b^3)*d*e^4 -
35*(2*B*a^3*b + 3*A*a^2*b^2)*e^5)*x^2 - 3*(640*B*b^4*d^4*e - 448*(4*B*a*b^3 + A*b^4)*d^3*e^2 + 560*(3*B*a^2*b^
2 + 2*A*a*b^3)*d^2*e^3 - 280*(2*B*a^3*b + 3*A*a^2*b^2)*d*e^4 + 35*(B*a^4 + 4*A*a^3*b)*e^5)*x)*sqrt(e*x + d)/(e
^8*x^2 + 2*d*e^7*x + d^2*e^6)

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Sympy [A]  time = 111.143, size = 304, normalized size = 1.42 \begin{align*} \frac{2 B b^{4} \left (d + e x\right )^{\frac{7}{2}}}{7 e^{6}} + \frac{\left (d + e x\right )^{\frac{5}{2}} \left (2 A b^{4} e + 8 B a b^{3} e - 10 B b^{4} d\right )}{5 e^{6}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (8 A a b^{3} e^{2} - 8 A b^{4} d e + 12 B a^{2} b^{2} e^{2} - 32 B a b^{3} d e + 20 B b^{4} d^{2}\right )}{3 e^{6}} + \frac{\sqrt{d + e x} \left (12 A a^{2} b^{2} e^{3} - 24 A a b^{3} d e^{2} + 12 A b^{4} d^{2} e + 8 B a^{3} b e^{3} - 36 B a^{2} b^{2} d e^{2} + 48 B a b^{3} d^{2} e - 20 B b^{4} d^{3}\right )}{e^{6}} - \frac{2 \left (a e - b d\right )^{3} \left (4 A b e + B a e - 5 B b d\right )}{e^{6} \sqrt{d + e x}} + \frac{2 \left (- A e + B d\right ) \left (a e - b d\right )^{4}}{3 e^{6} \left (d + e x\right )^{\frac{3}{2}}} \end{align*}

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)**2/(e*x+d)**(5/2),x)

[Out]

2*B*b**4*(d + e*x)**(7/2)/(7*e**6) + (d + e*x)**(5/2)*(2*A*b**4*e + 8*B*a*b**3*e - 10*B*b**4*d)/(5*e**6) + (d
+ e*x)**(3/2)*(8*A*a*b**3*e**2 - 8*A*b**4*d*e + 12*B*a**2*b**2*e**2 - 32*B*a*b**3*d*e + 20*B*b**4*d**2)/(3*e**
6) + sqrt(d + e*x)*(12*A*a**2*b**2*e**3 - 24*A*a*b**3*d*e**2 + 12*A*b**4*d**2*e + 8*B*a**3*b*e**3 - 36*B*a**2*
b**2*d*e**2 + 48*B*a*b**3*d**2*e - 20*B*b**4*d**3)/e**6 - 2*(a*e - b*d)**3*(4*A*b*e + B*a*e - 5*B*b*d)/(e**6*s
qrt(d + e*x)) + 2*(-A*e + B*d)*(a*e - b*d)**4/(3*e**6*(d + e*x)**(3/2))

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Giac [B]  time = 1.15605, size = 765, normalized size = 3.57 \begin{align*} \frac{2}{105} \,{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} B b^{4} e^{36} - 105 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{4} d e^{36} + 350 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{4} d^{2} e^{36} - 1050 \, \sqrt{x e + d} B b^{4} d^{3} e^{36} + 84 \,{\left (x e + d\right )}^{\frac{5}{2}} B a b^{3} e^{37} + 21 \,{\left (x e + d\right )}^{\frac{5}{2}} A b^{4} e^{37} - 560 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{3} d e^{37} - 140 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{4} d e^{37} + 2520 \, \sqrt{x e + d} B a b^{3} d^{2} e^{37} + 630 \, \sqrt{x e + d} A b^{4} d^{2} e^{37} + 210 \,{\left (x e + d\right )}^{\frac{3}{2}} B a^{2} b^{2} e^{38} + 140 \,{\left (x e + d\right )}^{\frac{3}{2}} A a b^{3} e^{38} - 1890 \, \sqrt{x e + d} B a^{2} b^{2} d e^{38} - 1260 \, \sqrt{x e + d} A a b^{3} d e^{38} + 420 \, \sqrt{x e + d} B a^{3} b e^{39} + 630 \, \sqrt{x e + d} A a^{2} b^{2} e^{39}\right )} e^{\left (-42\right )} - \frac{2 \,{\left (15 \,{\left (x e + d\right )} B b^{4} d^{4} - B b^{4} d^{5} - 48 \,{\left (x e + d\right )} B a b^{3} d^{3} e - 12 \,{\left (x e + d\right )} A b^{4} d^{3} e + 4 \, B a b^{3} d^{4} e + A b^{4} d^{4} e + 54 \,{\left (x e + d\right )} B a^{2} b^{2} d^{2} e^{2} + 36 \,{\left (x e + d\right )} A a b^{3} d^{2} e^{2} - 6 \, B a^{2} b^{2} d^{3} e^{2} - 4 \, A a b^{3} d^{3} e^{2} - 24 \,{\left (x e + d\right )} B a^{3} b d e^{3} - 36 \,{\left (x e + d\right )} A a^{2} b^{2} d e^{3} + 4 \, B a^{3} b d^{2} e^{3} + 6 \, A a^{2} b^{2} d^{2} e^{3} + 3 \,{\left (x e + d\right )} B a^{4} e^{4} + 12 \,{\left (x e + d\right )} A a^{3} b e^{4} - B a^{4} d e^{4} - 4 \, A a^{3} b d e^{4} + A a^{4} e^{5}\right )} e^{\left (-6\right )}}{3 \,{\left (x e + d\right )}^{\frac{3}{2}}} \end{align*}

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)^2/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2/105*(15*(x*e + d)^(7/2)*B*b^4*e^36 - 105*(x*e + d)^(5/2)*B*b^4*d*e^36 + 350*(x*e + d)^(3/2)*B*b^4*d^2*e^36 -
1050*sqrt(x*e + d)*B*b^4*d^3*e^36 + 84*(x*e + d)^(5/2)*B*a*b^3*e^37 + 21*(x*e + d)^(5/2)*A*b^4*e^37 - 560*(x*
e + d)^(3/2)*B*a*b^3*d*e^37 - 140*(x*e + d)^(3/2)*A*b^4*d*e^37 + 2520*sqrt(x*e + d)*B*a*b^3*d^2*e^37 + 630*sqr
t(x*e + d)*A*b^4*d^2*e^37 + 210*(x*e + d)^(3/2)*B*a^2*b^2*e^38 + 140*(x*e + d)^(3/2)*A*a*b^3*e^38 - 1890*sqrt(
x*e + d)*B*a^2*b^2*d*e^38 - 1260*sqrt(x*e + d)*A*a*b^3*d*e^38 + 420*sqrt(x*e + d)*B*a^3*b*e^39 + 630*sqrt(x*e
+ d)*A*a^2*b^2*e^39)*e^(-42) - 2/3*(15*(x*e + d)*B*b^4*d^4 - B*b^4*d^5 - 48*(x*e + d)*B*a*b^3*d^3*e - 12*(x*e
+ d)*A*b^4*d^3*e + 4*B*a*b^3*d^4*e + A*b^4*d^4*e + 54*(x*e + d)*B*a^2*b^2*d^2*e^2 + 36*(x*e + d)*A*a*b^3*d^2*e
^2 - 6*B*a^2*b^2*d^3*e^2 - 4*A*a*b^3*d^3*e^2 - 24*(x*e + d)*B*a^3*b*d*e^3 - 36*(x*e + d)*A*a^2*b^2*d*e^3 + 4*B
*a^3*b*d^2*e^3 + 6*A*a^2*b^2*d^2*e^3 + 3*(x*e + d)*B*a^4*e^4 + 12*(x*e + d)*A*a^3*b*e^4 - B*a^4*d*e^4 - 4*A*a^
3*b*d*e^4 + A*a^4*e^5)*e^(-6)/(x*e + d)^(3/2)