3.1956 \(\int (a+b x) (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=146 \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^3 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)^2}{4 e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^3 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)^2}{4 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.04, size = 130, normalized size = 0.89 \[ \frac {x \sqrt {(a+b x)^2} \left (15 a^2 \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+6 a b x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )+b^2 x^2 \left (20 d^3+45 d^2 e x+36 d e^2 x^2+10 e^3 x^3\right )\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x*Sqrt[(a + b*x)^2]*(15*a^2*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 6*a*b*x*(10*d^3 + 20*d^2*e*x + 15*d
*e^2*x^2 + 4*e^3*x^3) + b^2*x^2*(20*d^3 + 45*d^2*e*x + 36*d*e^2*x^2 + 10*e^3*x^3)))/(60*(a + b*x))

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fricas [A]  time = 1.13, size = 124, normalized size = 0.85 \[ \frac {1}{6} \, b^{2} e^{3} x^{6} + a^{2} d^{3} x + \frac {1}{5} \, {\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{5} + \frac {1}{4} \, {\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} x^{4} + \frac {1}{3} \, {\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 199, normalized size = 1.36 \[ \frac {1}{6} \, b^{2} x^{6} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, b^{2} d x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, b^{2} d^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, b^{2} d^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, a b x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a b d x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a b d^{2} x^{3} e \mathrm {sgn}\left (b x + a\right ) + a b d^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{2} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{2} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{2} d^{3} x \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*b^2*x^6*e^3*sgn(b*x + a) + 3/5*b^2*d*x^5*e^2*sgn(b*x + a) + 3/4*b^2*d^2*x^4*e*sgn(b*x + a) + 1/3*b^2*d^3*x
^3*sgn(b*x + a) + 2/5*a*b*x^5*e^3*sgn(b*x + a) + 3/2*a*b*d*x^4*e^2*sgn(b*x + a) + 2*a*b*d^2*x^3*e*sgn(b*x + a)
+ a*b*d^3*x^2*sgn(b*x + a) + 1/4*a^2*x^4*e^3*sgn(b*x + a) + a^2*d*x^3*e^2*sgn(b*x + a) + 3/2*a^2*d^2*x^2*e*sg
n(b*x + a) + a^2*d^3*x*sgn(b*x + a)

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maple [A]  time = 0.05, size = 148, normalized size = 1.01 \[ \frac {\left (10 b^{2} e^{3} x^{5}+24 x^{4} a b \,e^{3}+36 x^{4} b^{2} d \,e^{2}+15 x^{3} a^{2} e^{3}+90 x^{3} a b d \,e^{2}+45 x^{3} b^{2} d^{2} e +60 x^{2} a^{2} d \,e^{2}+120 x^{2} a b \,d^{2} e +20 x^{2} b^{2} d^{3}+90 x \,a^{2} d^{2} e +60 x a b \,d^{3}+60 a^{2} d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}\, x}{60 b x +60 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x)

[Out]

1/60*x*(10*b^2*e^3*x^5+24*a*b*e^3*x^4+36*b^2*d*e^2*x^4+15*a^2*e^3*x^3+90*a*b*d*e^2*x^3+45*b^2*d^2*e*x^3+60*a^2
*d*e^2*x^2+120*a*b*d^2*e*x^2+20*b^2*d^3*x^2+90*a^2*d^2*e*x+60*a*b*d^3*x+60*a^2*d^3)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 0.61, size = 693, normalized size = 4.75 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{3} x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{3} x + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{3} x}{2 \, b^{3}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{3} x^{2}}{10 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{3}}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5} e^{3}}{2 \, b^{4}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{3} x}{5 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} e^{3}}{15 \, b^{4}} - \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} x}{2 \, b^{3}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x}{2 \, b^{2}} - \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a x}{2 \, b} + \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, b^{2}} - \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}}{2 \, b^{4}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}}{2 \, b^{3}} - \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{2 \, b^{2}} - \frac {7 \, {\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x}{20 \, b^{3}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x}{4 \, b^{2}} + \frac {9 \, {\left (3 \, b d e^{2} + a e^{3}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}}{20 \, b^{4}} - \frac {5 \, {\left (b d^{2} e + a d e^{2}\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a}{4 \, b^{3}} + \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}{3 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*e^3*x^3/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d^3*x + 1/2*sqrt(b^2*x^2 +
2*a*b*x + a^2)*a^4*e^3*x/b^3 - 3/10*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*e^3*x^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*
x + a^2)*a^2*d^3/b + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^5*e^3/b^4 + 2/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*e
^3*x/b^3 - 7/15*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*e^3/b^4 - 1/2*(3*b*d*e^2 + a*e^3)*sqrt(b^2*x^2 + 2*a*b*x +
a^2)*a^3*x/b^3 + 3/2*(b*d^2*e + a*d*e^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x/b^2 - 1/2*(b*d^3 + 3*a*d^2*e)*sq
rt(b^2*x^2 + 2*a*b*x + a^2)*a*x/b + 1/5*(3*b*d*e^2 + a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*x^2/b^2 - 1/2*(3*b
*d*e^2 + a*e^3)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4/b^4 + 3/2*(b*d^2*e + a*d*e^2)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*
a^3/b^3 - 1/2*(b*d^3 + 3*a*d^2*e)*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2/b^2 - 7/20*(3*b*d*e^2 + a*e^3)*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*a*x/b^3 + 3/4*(b*d^2*e + a*d*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*x/b^2 + 9/20*(3*b*d*e^2
+ a*e^3)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2/b^4 - 5/4*(b*d^2*e + a*d*e^2)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a/
b^3 + 1/3*(b*d^3 + 3*a*d^2*e)*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)/b^2

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mupad [B]  time = 3.01, size = 734, normalized size = 5.03 \[ a\,d^3\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {d^3\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^3}+\frac {e^3\,x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{6\,b}-\frac {19\,a^2\,e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{120\,b^4}-\frac {a^3\,e^3\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{60\,b^6}+\frac {a\,e^3\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}+\frac {3\,d\,e^2\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b}-\frac {3\,a\,e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{40\,b^4}+\frac {3\,d^2\,e\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b}-\frac {7\,a\,d\,e^2\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{20\,b^3}-\frac {a\,d^2\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{32\,b^4}+\frac {3\,a\,d\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {33\,a^2\,d\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{160\,b^5}-\frac {3\,a^2\,d^2\,e\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b}-\frac {3\,a^3\,d\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(a + b*x)*(d + e*x)^3,x)

[Out]

a*d^3*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (d^3*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*
(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^3) + (e^3*x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(6*b) - (19*a^2*e^3*(a^2
+ b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(120*b^4) - (a^
3*e^3*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(60*b^6) + (a*e^3*x^2*
(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) + (3*d*e^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b) - (3*a*e^3*(a^2
+ b^2*x^2 + 2*a*b*x)^(1/2)*(4*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) - a^4 + 9*a^2*b^2*x^2 + 8*a^3*b*x - 7*a*b*x*(
a^2 + b^2*x^2 + 2*a*b*x)))/(40*b^4) + (3*d^2*e*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b) - (7*a*d*e^2*(a^2 + b^
2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(20*b^3) - (a*d^2*e*
(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(32*b^4) + (3*a*d*e^2*x*(a^2
+ b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (33*a^2*d*e^2*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b
^2*x^2 + 2*a*b*x)^(1/2))/(160*b^5) - (3*a^2*d^2*e*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b) - (3*
a^3*d*e^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(4*b^2)

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sympy [A]  time = 0.13, size = 133, normalized size = 0.91 \[ a^{2} d^{3} x + \frac {b^{2} e^{3} x^{6}}{6} + x^{5} \left (\frac {2 a b e^{3}}{5} + \frac {3 b^{2} d e^{2}}{5}\right ) + x^{4} \left (\frac {a^{2} e^{3}}{4} + \frac {3 a b d e^{2}}{2} + \frac {3 b^{2} d^{2} e}{4}\right ) + x^{3} \left (a^{2} d e^{2} + 2 a b d^{2} e + \frac {b^{2} d^{3}}{3}\right ) + x^{2} \left (\frac {3 a^{2} d^{2} e}{2} + a b d^{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**3*((b*x+a)**2)**(1/2),x)

[Out]

a**2*d**3*x + b**2*e**3*x**6/6 + x**5*(2*a*b*e**3/5 + 3*b**2*d*e**2/5) + x**4*(a**2*e**3/4 + 3*a*b*d*e**2/2 +
3*b**2*d**2*e/4) + x**3*(a**2*d*e**2 + 2*a*b*d**2*e + b**2*d**3/3) + x**2*(3*a**2*d**2*e/2 + a*b*d**3)

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