3.9.28 \(\int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx\) [828]

Optimal. Leaf size=116 \[ -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

[Out]

5/2*d^3*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-5/2*d^2*(-e^2*x^2+d^2)^(1/2)/e-5/6*d*(e*x+d)*(-e^2*x^2+d^2)^(1/2)/e
-1/3*(e*x+d)^2*(-e^2*x^2+d^2)^(1/2)/e

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Rubi [A]
time = 0.03, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {5 d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*d^2*Sqrt[d^2 - e^2*x^2])/(2*e) - (5*d*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(6*e) - ((d + e*x)^2*Sqrt[d^2 - e^2*x
^2])/(3*e) + (5*d^3*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)^3}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{3} (5 d) \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^2\right ) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\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 {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \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 = 0.22, size = 89, normalized size = 0.77 \begin {gather*} \frac {\left (-22 d^2-9 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {5 d^3 \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)^3/Sqrt[d^2 - e^2*x^2],x]

[Out]

((-22*d^2 - 9*d*e*x - 2*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(6*e) - (5*d^3*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]
])/(2*Sqrt[-e^2])

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Maple [A]
time = 0.48, size = 166, normalized size = 1.43

method	result	size
risch	\(-\frac {\left (2 e^{2} x^{2}+9 d x e +22 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e}+\frac {5 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\)	\(72\)
default	\(e^{3} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+3 d \,e^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {3 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(-e^2*x^2+d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

e^3*(-1/3*x^2/e^2*(-e^2*x^2+d^2)^(1/2)-2/3*d^2/e^4*(-e^2*x^2+d^2)^(1/2))+3*d*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)))-3*d^2*(-e^2*x^2+d^2)^(1/2)/e+d^3/(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 = 0.56, size = 73, normalized size = 0.63 \begin {gather*} \frac {5}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} x^{2} e - \frac {11}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-1\right )} - \frac {3}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*d^3*arcsin(x*e/d)*e^(-1) - 1/3*sqrt(-x^2*e^2 + d^2)*x^2*e - 11/3*sqrt(-x^2*e^2 + d^2)*d^2*e^(-1) - 3/2*sqr
t(-x^2*e^2 + d^2)*d*x

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Fricas [A]
time = 1.98, size = 68, normalized size = 0.59 \begin {gather*} -\frac {1}{6} \, {\left (30 \, d^{3} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (2 \, x^{2} e^{2} + 9 \, d x e + 22 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(30*d^3*arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) + (2*x^2*e^2 + 9*d*x*e + 22*d^2)*sqrt(-x^2*e^2 + d^2
))*e^(-1)

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Sympy [A]
time = 2.80, size = 337, normalized size = 2.91 \begin {gather*} d^{3} \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 ) + 3 d^{2} 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 ) + 3 d 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 ) + e^{3} \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*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))) + 3*d**2*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e*
*2*x**2)/e**2, True)) + 3*d*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*sq
rt(1 - e**2*x**2/d**2)/(2*e**2), True)) + e**3*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(
d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4/(4*sqrt(d**2)), True))

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Giac [A]
time = 2.65, size = 52, normalized size = 0.45 \begin {gather*} \frac {5}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (22 \, d^{2} e^{\left (-1\right )} + {\left (2 \, x e + 9 \, d\right )} x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*d^3*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/6*sqrt(-x^2*e^2 + d^2)*(22*d^2*e^(-1) + (2*x*e + 9*d)*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 )}^3}{\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)^3/(d^2 - e^2*x^2)^(1/2),x)

[Out]

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

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