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

Optimal. Leaf size=298 \[ -\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^9 (-3 a B e-A b e+4 b B d)}{9 e^5 (a+b x)}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^8 (b d-a e) (-a B e-A b e+2 b B d)}{8 e^5 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^7 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6 (b d-a e)^3 (B d-A e)}{6 e^5 (a+b x)}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{10}}{10 e^5 (a+b x)} \]

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Rubi [A]  time = 0.51, 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} (d+e x)^9 (-3 a B e-A b e+4 b B d)}{9 e^5 (a+b x)}+\frac {3 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^8 (b d-a e) (-a B e-A b e+2 b B d)}{8 e^5 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^7 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{7 e^5 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6 (b d-a e)^3 (B d-A e)}{6 e^5 (a+b x)}+\frac {b^3 B \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{10}}{10 e^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^3*(B*d - A*e)*(d + e*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*e^5*(a + b*x)) - ((b*d - a*e)^2*(4*b*
B*d - 3*A*b*e - a*B*e)*(d + e*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x)) + (3*b*(b*d - a*e)*(2*b*B*
d - A*b*e - a*B*e)*(d + e*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*e^5*(a + b*x)) - (b^2*(4*b*B*d - A*b*e - 3*a*
B*e)*(d + e*x)^9*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*e^5*(a + b*x)) + (b^3*B*(d + e*x)^10*Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/(10*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 (A+B x) (d+e x)^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^3 (A+B x) (d+e x)^5 \, 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) (d+e x)^5}{e^4}+\frac {b^3 (b d-a e)^2 (-4 b B d+3 A b e+a B e) (d+e x)^6}{e^4}-\frac {3 b^4 (b d-a e) (-2 b B d+A b e+a B e) (d+e x)^7}{e^4}+\frac {b^5 (-4 b B d+A b e+3 a B e) (d+e x)^8}{e^4}+\frac {b^6 B (d+e x)^9}{e^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {(b d-a e)^3 (B d-A e) (d+e x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 e^5 (a+b x)}-\frac {(b d-a e)^2 (4 b B d-3 A b e-a B e) (d+e x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x)}+\frac {3 b (b d-a e) (2 b B d-A b e-a B e) (d+e x)^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 e^5 (a+b x)}-\frac {b^2 (4 b B d-A b e-3 a B e) (d+e x)^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^5 (a+b x)}+\frac {b^3 B (d+e x)^{10} \sqrt {a^2+2 a b x+b^2 x^2}}{10 e^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 496, normalized size = 1.66 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (60 a^3 \left (7 A \left (6 d^5+15 d^4 e x+20 d^3 e^2 x^2+15 d^2 e^3 x^3+6 d e^4 x^4+e^5 x^5\right )+B x \left (21 d^5+70 d^4 e x+105 d^3 e^2 x^2+84 d^2 e^3 x^3+35 d e^4 x^4+6 e^5 x^5\right )\right )+45 a^2 b x \left (4 A \left (21 d^5+70 d^4 e x+105 d^3 e^2 x^2+84 d^2 e^3 x^3+35 d e^4 x^4+6 e^5 x^5\right )+B x \left (56 d^5+210 d^4 e x+336 d^3 e^2 x^2+280 d^2 e^3 x^3+120 d e^4 x^4+21 e^5 x^5\right )\right )+15 a b^2 x^2 \left (3 A \left (56 d^5+210 d^4 e x+336 d^3 e^2 x^2+280 d^2 e^3 x^3+120 d e^4 x^4+21 e^5 x^5\right )+B x \left (126 d^5+504 d^4 e x+840 d^3 e^2 x^2+720 d^2 e^3 x^3+315 d e^4 x^4+56 e^5 x^5\right )\right )+b^3 x^3 \left (5 A \left (126 d^5+504 d^4 e x+840 d^3 e^2 x^2+720 d^2 e^3 x^3+315 d e^4 x^4+56 e^5 x^5\right )+2 B x \left (252 d^5+1050 d^4 e x+1800 d^3 e^2 x^2+1575 d^2 e^3 x^3+700 d e^4 x^4+126 e^5 x^5\right )\right )\right )}{2520 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(60*a^3*(7*A*(6*d^5 + 15*d^4*e*x + 20*d^3*e^2*x^2 + 15*d^2*e^3*x^3 + 6*d*e^4*x^4 + e^5*x^
5) + B*x*(21*d^5 + 70*d^4*e*x + 105*d^3*e^2*x^2 + 84*d^2*e^3*x^3 + 35*d*e^4*x^4 + 6*e^5*x^5)) + 45*a^2*b*x*(4*
A*(21*d^5 + 70*d^4*e*x + 105*d^3*e^2*x^2 + 84*d^2*e^3*x^3 + 35*d*e^4*x^4 + 6*e^5*x^5) + B*x*(56*d^5 + 210*d^4*
e*x + 336*d^3*e^2*x^2 + 280*d^2*e^3*x^3 + 120*d*e^4*x^4 + 21*e^5*x^5)) + 15*a*b^2*x^2*(3*A*(56*d^5 + 210*d^4*e
*x + 336*d^3*e^2*x^2 + 280*d^2*e^3*x^3 + 120*d*e^4*x^4 + 21*e^5*x^5) + B*x*(126*d^5 + 504*d^4*e*x + 840*d^3*e^
2*x^2 + 720*d^2*e^3*x^3 + 315*d*e^4*x^4 + 56*e^5*x^5)) + b^3*x^3*(5*A*(126*d^5 + 504*d^4*e*x + 840*d^3*e^2*x^2
+ 720*d^2*e^3*x^3 + 315*d*e^4*x^4 + 56*e^5*x^5) + 2*B*x*(252*d^5 + 1050*d^4*e*x + 1800*d^3*e^2*x^2 + 1575*d^2
*e^3*x^3 + 700*d*e^4*x^4 + 126*e^5*x^5))))/(2520*(a + b*x))

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IntegrateAlgebraic [F]  time = 6.65, size = 0, normalized size = 0.00 \begin {gather*} \int (A+B x) (d+e x)^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 518, normalized size = 1.74 \begin {gather*} \frac {1}{10} \, B b^{3} e^{5} x^{10} + A a^{3} d^{5} x + \frac {1}{9} \, {\left (5 \, B b^{3} d e^{4} + {\left (3 \, B a b^{2} + A b^{3}\right )} e^{5}\right )} x^{9} + \frac {1}{8} \, {\left (10 \, B b^{3} d^{2} e^{3} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d e^{4} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e^{5}\right )} x^{8} + \frac {1}{7} \, {\left (10 \, B b^{3} d^{3} e^{2} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{3} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} d e^{4} + {\left (B a^{3} + 3 \, A a^{2} b\right )} e^{5}\right )} x^{7} + \frac {1}{6} \, {\left (5 \, B b^{3} d^{4} e + A a^{3} e^{5} + 10 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e^{2} + 30 \, {\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{3} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{4}\right )} x^{6} + \frac {1}{5} \, {\left (B b^{3} d^{5} + 5 \, A a^{3} d e^{4} + 5 \, {\left (3 \, B a b^{2} + A b^{3}\right )} d^{4} e + 30 \, {\left (B a^{2} b + A a b^{2}\right )} d^{3} e^{2} + 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d^{2} e^{3}\right )} x^{5} + \frac {1}{4} \, {\left (10 \, A a^{3} d^{2} e^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} d^{5} + 15 \, {\left (B a^{2} b + A a b^{2}\right )} d^{4} e + 10 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d^{3} e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (10 \, A a^{3} d^{3} e^{2} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} d^{5} + 5 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} d^{4} e\right )} x^{3} + \frac {1}{2} \, {\left (5 \, A a^{3} d^{4} e + {\left (B a^{3} + 3 \, A a^{2} b\right )} d^{5}\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.22, size = 922, normalized size = 3.09

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 676, normalized size = 2.27 \begin {gather*} \frac {\left (252 b^{3} B \,e^{5} x^{9}+280 x^{8} A \,b^{3} e^{5}+840 x^{8} B a \,b^{2} e^{5}+1400 x^{8} b^{3} B d \,e^{4}+945 x^{7} A a \,b^{2} e^{5}+1575 x^{7} A \,b^{3} d \,e^{4}+945 x^{7} B \,a^{2} b \,e^{5}+4725 x^{7} B a \,b^{2} d \,e^{4}+3150 x^{7} b^{3} B \,d^{2} e^{3}+1080 x^{6} A \,a^{2} b \,e^{5}+5400 x^{6} A a \,b^{2} d \,e^{4}+3600 x^{6} A \,b^{3} d^{2} e^{3}+360 x^{6} B \,a^{3} e^{5}+5400 x^{6} B \,a^{2} b d \,e^{4}+10800 x^{6} B a \,b^{2} d^{2} e^{3}+3600 x^{6} b^{3} B \,d^{3} e^{2}+420 x^{5} A \,a^{3} e^{5}+6300 x^{5} A \,a^{2} b d \,e^{4}+12600 x^{5} A a \,b^{2} d^{2} e^{3}+4200 x^{5} A \,b^{3} d^{3} e^{2}+2100 x^{5} B \,a^{3} d \,e^{4}+12600 x^{5} B \,a^{2} b \,d^{2} e^{3}+12600 x^{5} B a \,b^{2} d^{3} e^{2}+2100 x^{5} b^{3} B \,d^{4} e +2520 x^{4} A \,a^{3} d \,e^{4}+15120 x^{4} A \,a^{2} b \,d^{2} e^{3}+15120 x^{4} A a \,b^{2} d^{3} e^{2}+2520 x^{4} A \,b^{3} d^{4} e +5040 x^{4} B \,a^{3} d^{2} e^{3}+15120 x^{4} B \,a^{2} b \,d^{3} e^{2}+7560 x^{4} B a \,b^{2} d^{4} e +504 x^{4} b^{3} B \,d^{5}+6300 x^{3} A \,a^{3} d^{2} e^{3}+18900 x^{3} A \,a^{2} b \,d^{3} e^{2}+9450 x^{3} A a \,b^{2} d^{4} e +630 x^{3} A \,b^{3} d^{5}+6300 x^{3} B \,a^{3} d^{3} e^{2}+9450 x^{3} B \,a^{2} b \,d^{4} e +1890 x^{3} B a \,b^{2} d^{5}+8400 x^{2} A \,a^{3} d^{3} e^{2}+12600 x^{2} A \,a^{2} b \,d^{4} e +2520 x^{2} A a \,b^{2} d^{5}+4200 x^{2} B \,a^{3} d^{4} e +2520 x^{2} B \,a^{2} b \,d^{5}+6300 x A \,a^{3} d^{4} e +3780 x A \,a^{2} b \,d^{5}+1260 x B \,a^{3} d^{5}+2520 A \,a^{3} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{2520 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2520*x*(252*B*b^3*e^5*x^9+280*A*b^3*e^5*x^8+840*B*a*b^2*e^5*x^8+1400*B*b^3*d*e^4*x^8+945*A*a*b^2*e^5*x^7+157
5*A*b^3*d*e^4*x^7+945*B*a^2*b*e^5*x^7+4725*B*a*b^2*d*e^4*x^7+3150*B*b^3*d^2*e^3*x^7+1080*A*a^2*b*e^5*x^6+5400*
A*a*b^2*d*e^4*x^6+3600*A*b^3*d^2*e^3*x^6+360*B*a^3*e^5*x^6+5400*B*a^2*b*d*e^4*x^6+10800*B*a*b^2*d^2*e^3*x^6+36
00*B*b^3*d^3*e^2*x^6+420*A*a^3*e^5*x^5+6300*A*a^2*b*d*e^4*x^5+12600*A*a*b^2*d^2*e^3*x^5+4200*A*b^3*d^3*e^2*x^5
+2100*B*a^3*d*e^4*x^5+12600*B*a^2*b*d^2*e^3*x^5+12600*B*a*b^2*d^3*e^2*x^5+2100*B*b^3*d^4*e*x^5+2520*A*a^3*d*e^
4*x^4+15120*A*a^2*b*d^2*e^3*x^4+15120*A*a*b^2*d^3*e^2*x^4+2520*A*b^3*d^4*e*x^4+5040*B*a^3*d^2*e^3*x^4+15120*B*
a^2*b*d^3*e^2*x^4+7560*B*a*b^2*d^4*e*x^4+504*B*b^3*d^5*x^4+6300*A*a^3*d^2*e^3*x^3+18900*A*a^2*b*d^3*e^2*x^3+94
50*A*a*b^2*d^4*e*x^3+630*A*b^3*d^5*x^3+6300*B*a^3*d^3*e^2*x^3+9450*B*a^2*b*d^4*e*x^3+1890*B*a*b^2*d^5*x^3+8400
*A*a^3*d^3*e^2*x^2+12600*A*a^2*b*d^4*e*x^2+2520*A*a*b^2*d^5*x^2+4200*B*a^3*d^4*e*x^2+2520*B*a^2*b*d^5*x^2+6300
*A*a^3*d^4*e*x+3780*A*a^2*b*d^5*x+1260*B*a^3*d^5*x+2520*A*a^3*d^5)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.63, size = 1330, normalized size = 4.46

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/10*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*e^5*x^5/b^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a*e^5*x^4/b^3 + 5/2
4*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^2*e^5*x^3/b^4 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*A*d^5*x + 1/4*(b^2*x
^2 + 2*a*b*x + a^2)^(3/2)*B*a^6*e^5*x/b^6 - 13/56*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^3*e^5*x^2/b^5 + 1/4*(b^2
*x^2 + 2*a*b*x + a^2)^(3/2)*A*a*d^5/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^7*e^5/b^7 + 41/168*(b^2*x^2 +
2*a*b*x + a^2)^(5/2)*B*a^4*e^5*x/b^6 - 209/840*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*B*a^5*e^5/b^7 + 1/9*(5*B*d*e^4
+ A*e^5)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^4/b^2 - 13/72*(5*B*d*e^4 + A*e^5)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a
*x^3/b^3 + 5/8*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^3/b^2 - 1/4*(5*B*d*e^4 + A*e^5)*(b^2*
x^2 + 2*a*b*x + a^2)^(3/2)*a^5*x/b^5 + 5/4*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4*x/b^4 -
5/2*(B*d^3*e^2 + A*d^2*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*x/b^3 + 5/4*(B*d^4*e + 2*A*d^3*e^2)*(b^2*x^2
+ 2*a*b*x + a^2)^(3/2)*a^2*x/b^2 - 1/4*(B*d^5 + 5*A*d^4*e)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*x/b + 37/168*(5*B
*d*e^4 + A*e^5)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*x^2/b^4 - 55/56*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*a*x^2/b^3 + 10/7*(B*d^3*e^2 + A*d^2*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^2/b^2 - 1/4*(5*B*d*e^
4 + A*e^5)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^6/b^6 + 5/4*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(3/
2)*a^5/b^5 - 5/2*(B*d^3*e^2 + A*d^2*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4/b^4 + 5/4*(B*d^4*e + 2*A*d^3*e^2)
*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3/b^3 - 1/4*(B*d^5 + 5*A*d^4*e)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2/b^2 - 1
21/504*(5*B*d*e^4 + A*e^5)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*x/b^5 + 65/56*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2
+ 2*a*b*x + a^2)^(5/2)*a^2*x/b^4 - 15/7*(B*d^3*e^2 + A*d^2*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x/b^3 + 5/6*
(B*d^4*e + 2*A*d^3*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x/b^2 + 125/504*(5*B*d*e^4 + A*e^5)*(b^2*x^2 + 2*a*b*x
+ a^2)^(5/2)*a^4/b^6 - 69/56*(2*B*d^2*e^3 + A*d*e^4)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3/b^5 + 17/7*(B*d^3*e^
2 + A*d^2*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2/b^4 - 7/6*(B*d^4*e + 2*A*d^3*e^2)*(b^2*x^2 + 2*a*b*x + a^2)
^(5/2)*a/b^3 + 1/5*(B*d^5 + 5*A*d^4*e)*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (A+B\,x\right )\,{\left (d+e\,x\right )}^5\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (A + B x\right ) \left (d + e x\right )^{5} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**5*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**5*((a + b*x)**2)**(3/2), x)

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