3.17.25 \(\int \frac {a^2+2 a b x+b^2 x^2}{\sqrt {d+e x}} \, dx\) [1625]

Optimal. Leaf size=69 \[ \frac {2 (b d-a e)^2 \sqrt {d+e x}}{e^3}-\frac {4 b (b d-a e) (d+e x)^{3/2}}{3 e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \begin {gather*} -\frac {4 b (d+e x)^{3/2} (b d-a e)}{3 e^3}+\frac {2 \sqrt {d+e x} (b d-a e)^2}{e^3}+\frac {2 b^2 (d+e x)^{5/2}}{5 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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])

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 60, normalized size = 0.87 \begin {gather*} \frac {2 \sqrt {d+e x} \left (15 a^2 e^2+10 a b e (-2 d+e x)+b^2 \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d + e*x]*(15*a^2*e^2 + 10*a*b*e*(-2*d + e*x) + b^2*(8*d^2 - 4*d*e*x + 3*e^2*x^2)))/(15*e^3)

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Maple [A]
time = 0.46, size = 55, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right )^{2} \sqrt {e x +d}}{e^{3}}\)	\(55\)
default	\(\frac {\frac {2 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {4 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{3}+2 \left (a e -b d \right )^{2} \sqrt {e x +d}}{e^{3}}\)	\(55\)
gosper	\(\frac {2 \left (3 b^{2} x^{2} e^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\)	\(63\)
trager	\(\frac {2 \left (3 b^{2} x^{2} e^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\)	\(63\)
risch	\(\frac {2 \left (3 b^{2} x^{2} e^{2}+10 a b \,e^{2} x -4 b^{2} d e x +15 a^{2} e^{2}-20 a b d e +8 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 e^{3}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 85, normalized size = 1.23 \begin {gather*} \frac {2}{15} \, {\left (10 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a b e^{\left (-1\right )} + {\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 )} b^{2} e^{\left (-2\right )} + 15 \, \sqrt {x e + d} a^{2}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(10*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a*b*e^(-1) + (3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqr
t(x*e + d)*d^2)*b^2*e^(-2) + 15*sqrt(x*e + d)*a^2)*e^(-1)

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Fricas [A]
time = 2.27, size = 59, normalized size = 0.86 \begin {gather*} \frac {2}{15} \, {\left (8 \, b^{2} d^{2} + {\left (3 \, b^{2} x^{2} + 10 \, a b x + 15 \, a^{2}\right )} e^{2} - 4 \, {\left (b^{2} d x + 5 \, a b d\right )} e\right )} \sqrt {x e + d} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(8*b^2*d^2 + (3*b^2*x^2 + 10*a*b*x + 15*a^2)*e^2 - 4*(b^2*d*x + 5*a*b*d)*e)*sqrt(x*e + d)*e^(-3)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (63) = 126\).
time = 5.47, size = 236, normalized size = 3.42 \begin {gather*} \begin {cases} \frac {- \frac {2 a^{2} d}{\sqrt {d + e x}} - 2 a^{2} \left (- \frac {d}{\sqrt {d + e x}} - \sqrt {d + e x}\right ) - \frac {4 a b d \left (- \frac {d}{\sqrt {d + e x}} - \sqrt {d + e x}\right )}{e} - \frac {4 a b \left (\frac {d^{2}}{\sqrt {d + e x}} + 2 d \sqrt {d + e x} - \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e} - \frac {2 b^{2} d \left (\frac {d^{2}}{\sqrt {d + e x}} + 2 d \sqrt {d + e x} - \frac {\left (d + e x\right )^{\frac {3}{2}}}{3}\right )}{e^{2}} - \frac {2 b^{2} \left (- \frac {d^{3}}{\sqrt {d + e x}} - 3 d^{2} \sqrt {d + e x} + d \left (d + e x\right )^{\frac {3}{2}} - \frac {\left (d + e x\right )^{\frac {5}{2}}}{5}\right )}{e^{2}}}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{\sqrt {d}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*a**2*d/sqrt(d + e*x) - 2*a**2*(-d/sqrt(d + e*x) - sqrt(d + e*x)) - 4*a*b*d*(-d/sqrt(d + e*x) -
sqrt(d + e*x))/e - 4*a*b*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e - 2*b**2*d*(d**2/sqrt
(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e**2 - 2*b**2*(-d**3/sqrt(d + e*x) - 3*d**2*sqrt(d + e*x)
+ d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**2)/e, Ne(e, 0)), ((a**2*x + a*b*x**2 + b**2*x**3/3)/sqrt(d), Tru
e))

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Giac [A]
time = 1.10, size = 85, normalized size = 1.23 \begin {gather*} \frac {2}{15} \, {\left (10 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {x e + d} d\right )} a b e^{\left (-1\right )} + {\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 )} b^{2} e^{\left (-2\right )} + 15 \, \sqrt {x e + d} a^{2}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(10*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*a*b*e^(-1) + (3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqr
t(x*e + d)*d^2)*b^2*e^(-2) + 15*sqrt(x*e + d)*a^2)*e^(-1)

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Mupad [B]
time = 0.06, size = 68, normalized size = 0.99 \begin {gather*} \frac {2\,\sqrt {d+e\,x}\,\left (3\,b^2\,{\left (d+e\,x\right )}^2+15\,a^2\,e^2+15\,b^2\,d^2-10\,b^2\,d\,\left (d+e\,x\right )+10\,a\,b\,e\,\left (d+e\,x\right )-30\,a\,b\,d\,e\right )}{15\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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