3.1686 \(\int \frac {(a^2+2 a b x+b^2 x^2)^{3/2}}{\sqrt {d+e x}} \, dx\)

Optimal. Leaf size=204 \[ -\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^4 (a+b x)}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^4 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}{e^4 (a+b x)}+\frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^4 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {646, 43} \[ \frac {2 b^3 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^4 (a+b x)}-\frac {6 b^2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^4 (a+b x)}+\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^4 (a+b x)}-\frac {2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}{e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.07, size = 119, normalized size = 0.58 \[ \frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (35 a^3 e^3+35 a^2 b e^2 (e x-2 d)+7 a b^2 e \left (8 d^2-4 d e x+3 e^2 x^2\right )+b^3 \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )\right )}{35 e^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(35*a^3*e^3 + 35*a^2*b*e^2*(-2*d + e*x) + 7*a*b^2*e*(8*d^2 - 4*d*e*x + 3*e^
2*x^2) + b^3*(-16*d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3)))/(35*e^4*(a + b*x))

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fricas [A]  time = 0.95, size = 115, normalized size = 0.56 \[ \frac {2 \, {\left (5 \, b^{3} e^{3} x^{3} - 16 \, b^{3} d^{3} + 56 \, a b^{2} d^{2} e - 70 \, a^{2} b d e^{2} + 35 \, a^{3} e^{3} - 3 \, {\left (2 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + {\left (8 \, b^{3} d^{2} e - 28 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{35 \, e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*e^3*x^3 - 16*b^3*d^3 + 56*a*b^2*d^2*e - 70*a^2*b*d*e^2 + 35*a^3*e^3 - 3*(2*b^3*d*e^2 - 7*a*b^2*e^3
)*x^2 + (8*b^3*d^2*e - 28*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt(e*x + d)/e^4

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giac [A]  time = 0.19, size = 167, normalized size = 0.82 \[ \frac {2}{35} \, {\left (35 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a^{2} b e^{\left (-1\right )} \mathrm {sgn}\left (b x + a\right ) + 7 \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} - 10 \, {\left (x e + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {x e + d} d^{2}\right )} a b^{2} e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) + {\left (5 \, {\left (x e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x e + d} d^{3}\right )} b^{3} e^{\left (-3\right )} \mathrm {sgn}\left (b x + a\right ) + 35 \, \sqrt {x e + d} a^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(35*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a^2*b*e^(-1)*sgn(b*x + a) + 7*(3*(x*e + d)^(5/2) - 10*(x*e + d)
^(3/2)*d + 15*sqrt(x*e + d)*d^2)*a*b^2*e^(-2)*sgn(b*x + a) + (5*(x*e + d)^(7/2) - 21*(x*e + d)^(5/2)*d + 35*(x
*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*b^3*e^(-3)*sgn(b*x + a) + 35*sqrt(x*e + d)*a^3*sgn(b*x + a))*e^(-1)

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maple [A]  time = 0.05, size = 132, normalized size = 0.65 \[ \frac {2 \sqrt {e x +d}\, \left (5 b^{3} e^{3} x^{3}+21 a \,b^{2} e^{3} x^{2}-6 b^{3} d \,e^{2} x^{2}+35 a^{2} b \,e^{3} x -28 a \,b^{2} d \,e^{2} x +8 b^{3} d^{2} e x +35 a^{3} e^{3}-70 a^{2} b d \,e^{2}+56 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{35 \left (b x +a \right )^{3} e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(e*x+d)^(1/2)*(5*b^3*e^3*x^3+21*a*b^2*e^3*x^2-6*b^3*d*e^2*x^2+35*a^2*b*e^3*x-28*a*b^2*d*e^2*x+8*b^3*d^2*e
*x+35*a^3*e^3-70*a^2*b*d*e^2+56*a*b^2*d^2*e-16*b^3*d^3)*((b*x+a)^2)^(3/2)/e^4/(b*x+a)^3

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maxima [A]  time = 1.10, size = 164, normalized size = 0.80 \[ \frac {2 \, {\left (5 \, b^{3} e^{4} x^{4} - 16 \, b^{3} d^{4} + 56 \, a b^{2} d^{3} e - 70 \, a^{2} b d^{2} e^{2} + 35 \, a^{3} d e^{3} - {\left (b^{3} d e^{3} - 21 \, a b^{2} e^{4}\right )} x^{3} + {\left (2 \, b^{3} d^{2} e^{2} - 7 \, a b^{2} d e^{3} + 35 \, a^{2} b e^{4}\right )} x^{2} - {\left (8 \, b^{3} d^{3} e - 28 \, a b^{2} d^{2} e^{2} + 35 \, a^{2} b d e^{3} - 35 \, a^{3} e^{4}\right )} x\right )}}{35 \, \sqrt {e x + d} e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^3*e^4*x^4 - 16*b^3*d^4 + 56*a*b^2*d^3*e - 70*a^2*b*d^2*e^2 + 35*a^3*d*e^3 - (b^3*d*e^3 - 21*a*b^2*e^
4)*x^3 + (2*b^3*d^2*e^2 - 7*a*b^2*d*e^3 + 35*a^2*b*e^4)*x^2 - (8*b^3*d^3*e - 28*a*b^2*d^2*e^2 + 35*a^2*b*d*e^3
 - 35*a^3*e^4)*x)/(sqrt(e*x + d)*e^4)

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mupad [B]  time = 0.98, size = 198, normalized size = 0.97 \[ \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,b^2\,x^4}{7}+\frac {2\,x^2\,\left (35\,a^2\,e^2-7\,a\,b\,d\,e+2\,b^2\,d^2\right )}{35\,e^2}-\frac {-70\,a^3\,d\,e^3+140\,a^2\,b\,d^2\,e^2-112\,a\,b^2\,d^3\,e+32\,b^3\,d^4}{35\,b\,e^4}+\frac {2\,b\,x^3\,\left (21\,a\,e-b\,d\right )}{35\,e}+\frac {x\,\left (70\,a^3\,e^4-70\,a^2\,b\,d\,e^3+56\,a\,b^2\,d^2\,e^2-16\,b^3\,d^3\,e\right )}{35\,b\,e^4}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*b^2*x^4)/7 + (2*x^2*(35*a^2*e^2 + 2*b^2*d^2 - 7*a*b*d*e))/(35*e^2) - (32*
b^3*d^4 - 70*a^3*d*e^3 + 140*a^2*b*d^2*e^2 - 112*a*b^2*d^3*e)/(35*b*e^4) + (2*b*x^3*(21*a*e - b*d))/(35*e) + (
x*(70*a^3*e^4 - 16*b^3*d^3*e + 56*a*b^2*d^2*e^2 - 70*a^2*b*d*e^3))/(35*b*e^4)))/(x*(d + e*x)^(1/2) + (a*(d + e
*x)^(1/2))/b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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