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

Optimal. Leaf size=92 \[ \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5}{5 e^2 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)}{4 e^2 (a+b x)} \]

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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} \begin {gather*} \frac {b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^5}{5 e^2 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)}{4 e^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 89, normalized size = 0.97 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (5 a \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )+b x \left (10 d^3+20 d^2 e x+15 d e^2 x^2+4 e^3 x^3\right )\right )}{20 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(5*a*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + b*x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*x^
2 + 4*e^3*x^3)))/(20*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.94, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 69, normalized size = 0.75 \begin {gather*} \frac {1}{5} \, b e^{3} x^{5} + a d^{3} x + \frac {1}{4} \, {\left (3 \, b d e^{2} + a e^{3}\right )} x^{4} + {\left (b d^{2} e + a d e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b d^{3} + 3 \, a d^{2} e\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 118, normalized size = 1.28 \begin {gather*} \frac {1}{5} \, b x^{5} e^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, b d x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + b d^{2} x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, a x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) + a d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a 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*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.04, size = 90, normalized size = 0.98 \begin {gather*} \frac {\left (4 b \,e^{3} x^{4}+5 x^{3} a \,e^{3}+15 x^{3} b d \,e^{2}+20 a d \,e^{2} x^{2}+20 b \,d^{2} e \,x^{2}+30 x a \,d^{2} e +10 x b \,d^{3}+20 a \,d^{3}\right ) \sqrt {\left (b x +a \right )^{2}}\, x}{20 b x +20 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*x*(4*b*e^3*x^4+5*a*e^3*x^3+15*b*d*e^2*x^3+20*a*d*e^2*x^2+20*b*d^2*e*x^2+30*a*d^2*e*x+10*b*d^3*x+20*a*d^3)
*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 1.24, size = 399, normalized size = 4.34 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{3} x - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{2} e x}{2 \, b} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d e^{2} x}{2 \, b^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e^{3} x}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{3} x^{2}}{5 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{3}}{2 \, b} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{2} e}{2 \, b^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d e^{2}}{2 \, b^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{3}}{2 \, b^{4}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d e^{2} x}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{3} x}{20 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{2} e}{b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d e^{2}}{4 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{3}}{20 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.92, size = 377, normalized size = 4.10 \begin {gather*} d^3\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {e^3\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}-\frac {a^2\,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 {7\,a\,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 )}{60\,b^4}+\frac {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}}{8\,b^4}+\frac {3\,d\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2}-\frac {5\,a\,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}}{32\,b^5}-\frac {3\,a^2\,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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^3*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (e^3*x^2*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) - (a^2
*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) - (7*a*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))/(60*b^4) + (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))/(8*b^4) + (3*d*e^2*x*(a^
2 + b^2*x^2 + 2*a*b*x)^(3/2))/(4*b^2) - (5*a*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))/(32*b^5) - (3*a^2*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 = 0.12, size = 73, normalized size = 0.79 \begin {gather*} a d^{3} x + \frac {b e^{3} x^{5}}{5} + x^{4} \left (\frac {a e^{3}}{4} + \frac {3 b d e^{2}}{4}\right ) + x^{3} \left (a d e^{2} + b d^{2} e\right ) + x^{2} \left (\frac {3 a d^{2} e}{2} + \frac {b d^{3}}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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