3.12 \(\int \frac {(d+e x) (A+B x+C x^2)}{\sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac {\sqrt {d^2-e^2 x^2} \left (3 e (A e+B d)+2 C d^2\right )}{3 e^3}+\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (e (2 A e+B d)+C d^2\right )}{2 e^3}-\frac {x \sqrt {d^2-e^2 x^2} (B e+C d)}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1815, 641, 217, 203} \[ -\frac {\sqrt {d^2-e^2 x^2} \left (3 e (A e+B d)+2 C d^2\right )}{3 e^3}+\frac {d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (e (2 A e+B d)+C d^2\right )}{2 e^3}-\frac {x \sqrt {d^2-e^2 x^2} (B e+C d)}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (A+B x+C x^2\right )}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e}-\frac {\int \frac {-3 A d e^2-e \left (2 C d^2+3 e (B d+A e)\right ) x-3 e^2 (C d+B e) x^2}{\sqrt {d^2-e^2 x^2}} \, dx}{3 e^2}\\ &=-\frac {(C d+B e) x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {\int \frac {3 d e^2 \left (C d^2+e (B d+2 A e)\right )+2 e^3 \left (2 C d^2+3 e (B d+A e)\right ) x}{\sqrt {d^2-e^2 x^2}} \, dx}{6 e^4}\\ &=-\frac {\left (2 C d^2+3 e (B d+A e)\right ) \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {(C d+B e) x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {\left (d \left (C d^2+e (B d+2 A e)\right )\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{2 e^2}\\ &=-\frac {\left (2 C d^2+3 e (B d+A e)\right ) \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {(C d+B e) x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {\left (d \left (C d^2+e (B d+2 A e)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^2}\\ &=-\frac {\left (2 C d^2+3 e (B d+A e)\right ) \sqrt {d^2-e^2 x^2}}{3 e^3}-\frac {(C d+B e) x \sqrt {d^2-e^2 x^2}}{2 e^2}-\frac {C x^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {d \left (C d^2+e (B d+2 A e)\right ) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 103, normalized size = 0.72 \[ \frac {3 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) \left (e (2 A e+B d)+C d^2\right )-\sqrt {d^2-e^2 x^2} \left (3 e (2 A e+2 B d+B e x)+C \left (4 d^2+3 d e x+2 e^2 x^2\right )\right )}{6 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(3*e*(2*B*d + 2*A*e + B*e*x) + C*(4*d^2 + 3*d*e*x + 2*e^2*x^2))) + 3*d*(C*d^2 + e*(B*d
+ 2*A*e))*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(6*e^3)

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fricas [A]  time = 0.97, size = 109, normalized size = 0.76 \[ -\frac {6 \, {\left (C d^{3} + B d^{2} e + 2 \, A d e^{2}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (2 \, C e^{2} x^{2} + 4 \, C d^{2} + 6 \, B d e + 6 \, A e^{2} + 3 \, {\left (C d e + B e^{2}\right )} x\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{6 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(C*d^3 + B*d^2*e + 2*A*d*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (2*C*e^2*x^2 + 4*C*d^2 + 6*B
*d*e + 6*A*e^2 + 3*(C*d*e + B*e^2)*x)*sqrt(-e^2*x^2 + d^2))/e^3

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giac [A]  time = 0.29, size = 97, normalized size = 0.68 \[ \frac {1}{2} \, {\left (C d^{3} + B d^{2} e + 2 \, A d e^{2}\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\relax (d) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left ({\left (2 \, C x e^{\left (-1\right )} + 3 \, {\left (C d e^{3} + B e^{4}\right )} e^{\left (-5\right )}\right )} x + 2 \, {\left (2 \, C d^{2} e^{2} + 3 \, B d e^{3} + 3 \, A e^{4}\right )} e^{\left (-5\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(C*d^3 + B*d^2*e + 2*A*d*e^2)*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/6*sqrt(-x^2*e^2 + d^2)*((2*C*x*e^(-1) + 3*(C
*d*e^3 + B*e^4)*e^(-5))*x + 2*(2*C*d^2*e^2 + 3*B*d*e^3 + 3*A*e^4)*e^(-5))

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maple [A]  time = 0.01, size = 234, normalized size = 1.64 \[ \frac {A d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {B \,d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e}+\frac {C \,d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}\, e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C \,x^{2}}{3 e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, B x}{2 e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, C d x}{2 e^{2}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, A}{e}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, B d}{e^{2}}-\frac {2 \sqrt {-e^{2} x^{2}+d^{2}}\, C \,d^{2}}{3 e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.98, size = 150, normalized size = 1.05 \[ -\frac {\sqrt {-e^{2} x^{2} + d^{2}} C x^{2}}{3 \, e} + \frac {A d \arcsin \left (\frac {e x}{d}\right )}{e} + \frac {{\left (C d + B e\right )} d^{2} \arcsin \left (\frac {e x}{d}\right )}{2 \, e^{3}} - \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} C d^{2}}{3 \, e^{3}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} B d}{e^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} A}{e} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} {\left (C d + B e\right )} x}{2 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.01, size = 270, normalized size = 1.89 \[ \left \{\begin {array}{cl} \frac {2\,C\,d\,x^3+3\,B\,d\,x^2+6\,A\,d\,x}{6\,\sqrt {d^2}} & \text {\ if\ \ }e=0\\ \frac {A\,d\,\ln \left (x\,\sqrt {-e^2}+\sqrt {d^2-e^2\,x^2}\right )}{\sqrt {-e^2}}-\frac {A\,\sqrt {d^2-e^2\,x^2}}{e}-\frac {B\,d\,\sqrt {d^2-e^2\,x^2}}{e^2}-\frac {B\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e}-\frac {C\,\sqrt {d^2-e^2\,x^2}\,\left (2\,d^2+e^2\,x^2\right )}{3\,e^3}-\frac {C\,d^3\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}}-\frac {B\,d^2\,e\,\ln \left (2\,x\,\sqrt {-e^2}+2\,\sqrt {d^2-e^2\,x^2}\right )}{2\,{\left (-e^2\right )}^{3/2}}-\frac {C\,d\,x\,\sqrt {d^2-e^2\,x^2}}{2\,e^2} & \text {\ if\ \ }e\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(e == 0, (6*A*d*x + 3*B*d*x^2 + 2*C*d*x^3)/(6*(d^2)^(1/2)), e ~= 0, - (A*(d^2 - e^2*x^2)^(1/2))/e + (
A*d*log(x*(-e^2)^(1/2) + (d^2 - e^2*x^2)^(1/2)))/(-e^2)^(1/2) - (B*d*(d^2 - e^2*x^2)^(1/2))/e^2 - (B*x*(d^2 -
e^2*x^2)^(1/2))/(2*e) - (C*(d^2 - e^2*x^2)^(1/2)*(2*d^2 + e^2*x^2))/(3*e^3) - (C*d^3*log(2*x*(-e^2)^(1/2) + 2*
(d^2 - e^2*x^2)^(1/2)))/(2*(-e^2)^(3/2)) - (B*d^2*e*log(2*x*(-e^2)^(1/2) + 2*(d^2 - e^2*x^2)^(1/2)))/(2*(-e^2)
^(3/2)) - (C*d*x*(d^2 - e^2*x^2)^(1/2))/(2*e^2))

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sympy [A]  time = 10.17, size = 484, normalized size = 3.38 \[ A d \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 ) + A 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 ) + B d \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 ) + B e \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 ) + C d \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 ) + C e \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)*(C*x**2+B*x+A)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*d*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))/sqr
t(-d**2), (d**2 < 0) & (e**2 < 0))) + A*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**
2)/e**2, True)) + B*d*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e**2, True)) + B*
e*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)), T
rue)) + C*d*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)) + C*e*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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