3.1543 \(\int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=92 \[ \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^2 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 92, 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 = {646, 43} \[ \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^2 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

-((b*d - a*e)*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^2*(a + b*x)) + (b*(d + e*x)^6*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/(6*e^2*(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 646

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

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Mathematica [A]  time = 0.04, size = 111, normalized size = 1.21 \[ \frac {x \sqrt {(a+b x)^2} \left (6 a \left (5 d^4+10 d^3 e x+10 d^2 e^2 x^2+5 d e^3 x^3+e^4 x^4\right )+b x \left (15 d^4+40 d^3 e x+45 d^2 e^2 x^2+24 d e^3 x^3+5 e^4 x^4\right )\right )}{30 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x*Sqrt[(a + b*x)^2]*(6*a*(5*d^4 + 10*d^3*e*x + 10*d^2*e^2*x^2 + 5*d*e^3*x^3 + e^4*x^4) + b*x*(15*d^4 + 40*d^3
*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 5*e^4*x^4)))/(30*(a + b*x))

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fricas [A]  time = 0.95, size = 96, normalized size = 1.04 \[ \frac {1}{6} \, b e^{4} x^{6} + a d^{4} x + \frac {1}{5} \, {\left (4 \, b d e^{3} + a e^{4}\right )} x^{5} + \frac {1}{2} \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} x^{4} + \frac {2}{3} \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b d^{4} + 4 \, a d^{3} e\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.16, size = 153, normalized size = 1.66 \[ \frac {1}{6} \, b x^{6} e^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{5} \, b d x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, b d^{2} x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{3} \, b d^{3} x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, a x^{5} e^{4} \mathrm {sgn}\left (b x + a\right ) + a d x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a d^{2} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a d^{3} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a d^{4} x \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 114, normalized size = 1.24 \[ \frac {\left (5 b \,e^{4} x^{5}+6 x^{4} a \,e^{4}+24 x^{4} b d \,e^{3}+30 x^{3} a d \,e^{3}+45 x^{3} b \,d^{2} e^{2}+60 x^{2} a \,d^{2} e^{2}+40 x^{2} b \,d^{3} e +60 x a \,d^{3} e +15 x b \,d^{4}+30 a \,d^{4}\right ) \sqrt {\left (b x +a \right )^{2}}\, x}{30 b x +30 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4*((b*x+a)^2)^(1/2),x)

[Out]

1/30*x*(5*b*e^4*x^5+6*a*e^4*x^4+24*b*d*e^3*x^4+30*a*d*e^3*x^3+45*b*d^2*e^2*x^3+60*a*d^2*e^2*x^2+40*b*d^3*e*x^2
+60*a*d^3*e*x+15*b*d^4*x+30*a*d^4)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 1.28, size = 590, normalized size = 6.41 \[ \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{4} x^{3}}{6 \, b^{2}} + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{4} x - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{3} e x}{b} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{2} e^{2} x}{b^{2}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d e^{3} x}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{4} x}{2 \, b^{4}} + \frac {4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d e^{3} x^{2}}{5 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{4} x^{2}}{10 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{4}}{2 \, b} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{3} e}{b^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d^{2} e^{2}}{b^{3}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} d e^{3}}{b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5} e^{4}}{2 \, b^{5}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{2} e^{2} x}{2 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d e^{3} x}{5 \, b^{3}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{4} x}{5 \, b^{4}} + \frac {4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{3} e}{3 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{2} e^{2}}{2 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d e^{3}}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} e^{4}}{15 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.31, size = 580, normalized size = 6.30 \[ d^4\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {e^4\,x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{6\,b^2}-\frac {a^2\,e^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{24\,b^5}+\frac {3\,d^2\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{2\,b^2}+\frac {4\,d\,e^3\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}-\frac {3\,a\,e^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{40\,b^5}+\frac {d^3\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,b^4}-\frac {7\,a\,d\,e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{15\,b^4}-\frac {3\,a^2\,d^2\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^2}-\frac {5\,a\,d^2\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{16\,b^5}-\frac {a^2\,d\,e^3\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{15\,b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(d + e*x)^4,x)

[Out]

d^4*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (e^4*x^3*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(6*b^2) - (a^2
*e^4*(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))/(24*b^
5) + (3*d^2*e^2*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(2*b^2) + (4*d*e^3*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*
b^2) - (3*a*e^4*(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^5) + (d^3*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))/(6*b^4) - (7*a*d*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))/(15*b^4) - (3*a^2*d^2*e^2*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*
b*x)^(1/2))/(2*b^2) - (5*a*d^2*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))/(16*b^5) - (a^2*d*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))
/(15*b^6)

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sympy [A]  time = 0.13, size = 100, normalized size = 1.09 \[ a d^{4} x + \frac {b e^{4} x^{6}}{6} + x^{5} \left (\frac {a e^{4}}{5} + \frac {4 b d e^{3}}{5}\right ) + x^{4} \left (a d e^{3} + \frac {3 b d^{2} e^{2}}{2}\right ) + x^{3} \left (2 a d^{2} e^{2} + \frac {4 b d^{3} e}{3}\right ) + x^{2} \left (2 a d^{3} e + \frac {b d^{4}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4*((b*x+a)**2)**(1/2),x)

[Out]

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

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