3.9.29 \(\int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\) [829]

Optimal. Leaf size=74 \[ -\frac {2 d \sqrt {d^2-e^2 x^2}}{e}-\frac {1}{2} x \sqrt {d^2-e^2 x^2}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

3/2*d^2*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-2*d*(-e^2*x^2+d^2)^(1/2)/e-1/2*x*(-e^2*x^2+d^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 83, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {685, 655, 223, 209} \begin {gather*} \frac {3 d^2 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\frac {3 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*d*Sqrt[d^2 - e^2*x^2])/(2*e) - ((d + e*x)*Sqrt[d^2 - e^2*x^2])/(2*e) + (3*d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*
x^2]])/(2*e)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}+\frac {1}{2} (3 d) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {3 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}+\frac {1}{2} \left (3 d^2\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {3 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}+\frac {1}{2} \left (3 d^2\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=-\frac {3 d \sqrt {d^2-e^2 x^2}}{2 e}-\frac {(d+e x) \sqrt {d^2-e^2 x^2}}{2 e}+\frac {3 d^2 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 78, normalized size = 1.05 \begin {gather*} \frac {(-4 d-e x) \sqrt {d^2-e^2 x^2}}{2 e}-\frac {3 d^2 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 113, normalized size = 1.53

method	result	size
risch	\(-\frac {\left (e x +4 d \right ) \sqrt {-e^{2} x^{2}+d^{2}}}{2 e}+\frac {3 d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\)	\(60\)
default	\(e^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {2 d \sqrt {-e^{2} x^{2}+d^{2}}}{e}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^2*(-1/2*x/e^2*(-e^2*x^2+d^2)^(1/2)+1/2*d^2/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2)))-2*d*(
-e^2*x^2+d^2)^(1/2)/e+d^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [A]
time = 0.51, size = 50, normalized size = 0.68 \begin {gather*} \frac {3}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} d e^{\left (-1\right )} - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*d^2*arcsin(x*e/d)*e^(-1) - 2*sqrt(-x^2*e^2 + d^2)*d*e^(-1) - 1/2*sqrt(-x^2*e^2 + d^2)*x

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Fricas [A]
time = 2.42, size = 57, normalized size = 0.77 \begin {gather*} -\frac {1}{2} \, {\left (6 \, d^{2} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + \sqrt {-x^{2} e^{2} + d^{2}} {\left (x e + 4 \, d\right )}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(6*d^2*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + sqrt(-x^2*e^2 + d^2)*(x*e + 4*d))*e^(-1)

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Sympy [A]
time = 2.28, size = 269, normalized size = 3.64 \begin {gather*} d^{2} \left (\begin {cases} \frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {asin}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac {\sqrt {- \frac {d^{2}}{e^{2}}} \operatorname {asinh}{\left (x \sqrt {- \frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {acosh}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: d^{2} < 0 \wedge e^{2} < 0 \end {cases}\right ) + 2 d e \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d^{2}}} & \text {for}\: e^{2} = 0 \\- \frac {\sqrt {d^{2} - e^{2} x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} + \frac {i d x}{2 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*Piecewise((sqrt(d**2/e**2)*asin(x*sqrt(e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 > 0)), (sqrt(-d**2/e**2
)*asinh(x*sqrt(-e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 < 0)), (sqrt(d**2/e**2)*acosh(x*sqrt(e**2/d**2))/sq
rt(-d**2), (d**2 < 0) & (e**2 < 0))) + 2*d*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*
x**2)/e**2, True)) + e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e**3) + I*d*x/(2*e**2*sqrt(-1 + e**2*x**2/d**2))
- I*x**3/(2*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x*sqrt(1 -
e**2*x**2/d**2)/(2*e**2), True))

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Giac [A]
time = 1.83, size = 40, normalized size = 0.54 \begin {gather*} \frac {3}{2} \, d^{2} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (4 \, d e^{\left (-1\right )} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*d^2*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/2*sqrt(-x^2*e^2 + d^2)*(4*d*e^(-1) + x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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