3.16.56 \(\int (d+e x)^3 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [1556]

Optimal. Leaf size=172 \[ \frac {(b d-a e)^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^4}+\frac {3 e (b d-a e)^2 (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^4}+\frac {e^2 (b d-a e) (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^4}+\frac {e^3 (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^4} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \begin {gather*} \frac {e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5 (b d-a e)}{2 b^4}+\frac {3 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)^2}{5 b^4}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)^3}{4 b^4}+\frac {e^3 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 171, normalized size = 0.99 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (35 a^3 \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+21 a^2 b x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )+7 a 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 )+b^3 x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right )\right )}{140 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(35*a^3*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 21*a^2*b*x*(10*d^3 + 20*d^2*e*x + 1
5*d*e^2*x^2 + 4*e^3*x^3) + 7*a*b^2*x^2*(20*d^3 + 45*d^2*e*x + 36*d*e^2*x^2 + 10*e^3*x^3) + b^3*x^3*(35*d^3 + 8
4*d^2*e*x + 70*d*e^2*x^2 + 20*e^3*x^3)))/(140*(a + b*x))

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Maple [A]
time = 0.69, size = 206, normalized size = 1.20

method	result	size
gosper	\(\frac {x \left (20 b^{3} e^{3} x^{6}+70 x^{5} b^{2} e^{3} a +70 x^{5} b^{3} d \,e^{2}+84 x^{4} a^{2} b \,e^{3}+252 x^{4} a \,b^{2} d \,e^{2}+84 x^{4} d^{2} e \,b^{3}+35 x^{3} e^{3} a^{3}+315 x^{3} a^{2} b d \,e^{2}+315 x^{3} a \,b^{2} d^{2} e +35 x^{3} b^{3} d^{3}+140 a^{3} d \,e^{2} x^{2}+420 a^{2} b \,d^{2} e \,x^{2}+140 a \,b^{2} d^{3} x^{2}+210 x \,a^{3} d^{2} e +210 a^{2} b \,d^{3} x +140 a^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (b x +a \right )^{3}}\)	\(206\)
default	\(\frac {x \left (20 b^{3} e^{3} x^{6}+70 x^{5} b^{2} e^{3} a +70 x^{5} b^{3} d \,e^{2}+84 x^{4} a^{2} b \,e^{3}+252 x^{4} a \,b^{2} d \,e^{2}+84 x^{4} d^{2} e \,b^{3}+35 x^{3} e^{3} a^{3}+315 x^{3} a^{2} b d \,e^{2}+315 x^{3} a \,b^{2} d^{2} e +35 x^{3} b^{3} d^{3}+140 a^{3} d \,e^{2} x^{2}+420 a^{2} b \,d^{2} e \,x^{2}+140 a \,b^{2} d^{3} x^{2}+210 x \,a^{3} d^{2} e +210 a^{2} b \,d^{3} x +140 a^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{140 \left (b x +a \right )^{3}}\)	\(206\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} e^{3} x^{7}}{7 b x +7 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 b^{2} e^{3} a +3 b^{3} d \,e^{2}\right ) x^{6}}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a^{2} b \,e^{3}+9 a \,b^{2} d \,e^{2}+3 d^{2} e \,b^{3}\right ) x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{3} a^{3}+9 a^{2} b d \,e^{2}+9 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a^{3} d \,e^{2}+9 a^{2} b \,d^{2} e +3 a \,b^{2} d^{3}\right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 a^{3} d^{2} e +3 a^{2} b \,d^{3}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} d^{3} x}{b x +a}\)	\(289\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/140*x*(20*b^3*e^3*x^6+70*a*b^2*e^3*x^5+70*b^3*d*e^2*x^5+84*a^2*b*e^3*x^4+252*a*b^2*d*e^2*x^4+84*b^3*d^2*e*x^
4+35*a^3*e^3*x^3+315*a^2*b*d*e^2*x^3+315*a*b^2*d^2*e*x^3+35*b^3*d^3*x^3+140*a^3*d*e^2*x^2+420*a^2*b*d^2*e*x^2+
140*a*b^2*d^3*x^2+210*a^3*d^2*e*x+210*a^2*b*d^3*x+140*a^3*d^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (122) = 244\).
time = 0.35, size = 394, normalized size = 2.29 \begin {gather*} \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{3} x - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{2} x e}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{3}}{4 \, b} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d x e^{2}}{4 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d^{2} e}{4 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} x e^{3}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x^{2} e^{3}}{7 \, b^{2}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} d e^{2}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} d x e^{2}}{2 \, b^{2}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} d^{2} e}{5 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{4} e^{3}}{4 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x e^{3}}{14 \, b^{3}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a d e^{2}}{10 \, b^{3}} + \frac {17 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} e^{3}}{70 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*d^3*x - 3/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d^2*x*e/b + 1/4*(b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*a*d^3/b + 3/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*d*x*e^2/b^2 - 3/4*(b^2*x^2 + 2*a*b*x + a
^2)^(3/2)*a^2*d^2*e/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*x*e^3/b^3 + 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(5
/2)*x^2*e^3/b^2 + 3/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*d*e^2/b^3 + 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*d*x*
e^2/b^2 + 3/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*d^2*e/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^4*e^3/b^4 - 3/
14*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x*e^3/b^3 - 7/10*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*d*e^2/b^3 + 17/70*(b^2
*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*e^3/b^4

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Fricas [A]
time = 2.58, size = 176, normalized size = 1.02 \begin {gather*} \frac {1}{4} \, b^{3} d^{3} x^{4} + a b^{2} d^{3} x^{3} + \frac {3}{2} \, a^{2} b d^{3} x^{2} + a^{3} d^{3} x + \frac {1}{140} \, {\left (20 \, b^{3} x^{7} + 70 \, a b^{2} x^{6} + 84 \, a^{2} b x^{5} + 35 \, a^{3} x^{4}\right )} e^{3} + \frac {1}{20} \, {\left (10 \, b^{3} d x^{6} + 36 \, a b^{2} d x^{5} + 45 \, a^{2} b d x^{4} + 20 \, a^{3} d x^{3}\right )} e^{2} + \frac {3}{20} \, {\left (4 \, b^{3} d^{2} x^{5} + 15 \, a b^{2} d^{2} x^{4} + 20 \, a^{2} b d^{2} x^{3} + 10 \, a^{3} d^{2} x^{2}\right )} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^3*d^3*x^4 + a*b^2*d^3*x^3 + 3/2*a^2*b*d^3*x^2 + a^3*d^3*x + 1/140*(20*b^3*x^7 + 70*a*b^2*x^6 + 84*a^2*b*
x^5 + 35*a^3*x^4)*e^3 + 1/20*(10*b^3*d*x^6 + 36*a*b^2*d*x^5 + 45*a^2*b*d*x^4 + 20*a^3*d*x^3)*e^2 + 3/20*(4*b^3
*d^2*x^5 + 15*a*b^2*d^2*x^4 + 20*a^2*b*d^2*x^3 + 10*a^3*d^2*x^2)*e

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right )^{3} \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((e*x+d)**3*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (122) = 244\).
time = 1.39, size = 280, normalized size = 1.63 \begin {gather*} \frac {1}{7} \, b^{3} x^{7} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b^{3} d x^{6} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, b^{3} d^{2} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, b^{3} d^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a b^{2} x^{6} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {9}{5} \, a b^{2} d x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {9}{4} \, a b^{2} d^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + a b^{2} d^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a^{2} b x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {9}{4} \, a^{2} b d x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d^{2} x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} b d^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a^{3} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{3} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{3} d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{3} d^{3} x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*x^7*e^3*sgn(b*x + a) + 1/2*b^3*d*x^6*e^2*sgn(b*x + a) + 3/5*b^3*d^2*x^5*e*sgn(b*x + a) + 1/4*b^3*d^3*x
^4*sgn(b*x + a) + 1/2*a*b^2*x^6*e^3*sgn(b*x + a) + 9/5*a*b^2*d*x^5*e^2*sgn(b*x + a) + 9/4*a*b^2*d^2*x^4*e*sgn(
b*x + a) + a*b^2*d^3*x^3*sgn(b*x + a) + 3/5*a^2*b*x^5*e^3*sgn(b*x + a) + 9/4*a^2*b*d*x^4*e^2*sgn(b*x + a) + 3*
a^2*b*d^2*x^3*e*sgn(b*x + a) + 3/2*a^2*b*d^3*x^2*sgn(b*x + a) + 1/4*a^3*x^4*e^3*sgn(b*x + a) + a^3*d*x^3*e^2*s
gn(b*x + a) + 3/2*a^3*d^2*x^2*e*sgn(b*x + a) + a^3*d^3*x*sgn(b*x + a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (d+e\,x\right )}^3\,{\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((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

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

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