3.828 \(\int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx\)

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, 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 = {671, 641, 217, 203} \[ -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]

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 203

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

Rule 217

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 641

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 671

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 ) \operatorname {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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 70, normalized size = 0.60 \[ \frac {15 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (22 d^2+9 d e x+2 e^2 x^2\right )}{6 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(22*d^2 + 9*d*e*x + 2*e^2*x^2)) + 15*d^3*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(6*e)

________________________________________________________________________________________

fricas [A]  time = 1.02, size = 72, normalized size = 0.62 \[ -\frac {30 \, d^{3} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, e^{2} x^{2} + 9 \, d e x + 22 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e} \]

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(-e^2*x^2 + d^2))/(e*x)) + (2*e^2*x^2 + 9*d*e*x + 22*d^2)*sqrt(-e^2*x^2 + d^2))/
e

________________________________________________________________________________________

giac [A]  time = 0.31, size = 52, normalized size = 0.45 \[ \frac {5}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (22 \, d^{2} e^{\left (-1\right )} + {\left (2 \, x e + 9 \, d\right )} x\right )} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 94, normalized size = 0.81 \[ \frac {5 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e \,x^{2}}{3}-\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d x}{2}-\frac {11 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}}{3 e} \]

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)

[Out]

-1/3*e*x^2*(-e^2*x^2+d^2)^(1/2)-11/3*d^2*(-e^2*x^2+d^2)^(1/2)/e-3/2*d*x*(-e^2*x^2+d^2)^(1/2)+5/2*d^3/(e^2)^(1/
2)*arctan((e^2)^(1/2)/(-e^2*x^2+d^2)^(1/2)*x)

________________________________________________________________________________________

maxima [A]  time = 2.94, size = 76, normalized size = 0.66 \[ -\frac {1}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} e x^{2} + \frac {5 \, d^{3} \arcsin \left (\frac {e x}{d}\right )}{2 \, e} - \frac {3}{2} \, \sqrt {-e^{2} x^{2} + d^{2}} d x - \frac {11 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e} \]

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]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^3}{\sqrt {d^2-e^2\,x^2}} \,d x \]

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)

________________________________________________________________________________________

sympy [A]  time = 5.58, size = 337, normalized size = 2.91 \[ d^{3} \left (\begin {cases} \frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {asin}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac {\sqrt {- \frac {d^{2}}{e^{2}}} \operatorname {asinh}{\left (x \sqrt {- \frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {acosh}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: d^{2} < 0 \wedge e^{2} < 0 \end {cases}\right ) + 3 d^{2} e \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d^{2}}} & \text {for}\: e^{2} = 0 \\- \frac {\sqrt {d^{2} - e^{2} x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {i d x \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x}{2 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{3}}{2 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \]

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*sqrt(-1 + e**2*x**2/d**2)/(2
*e**2), Abs(e**2*x**2/d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2
*d*sqrt(1 - e**2*x**2/d**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))

________________________________________________________________________________________