3.827 \(\int \frac {(d+e x)^4}{\sqrt {d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e} \]

[Out]

35/8*d^4*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e-35/8*d^3*(-e^2*x^2+d^2)^(1/2)/e-35/24*d^2*(e*x+d)*(-e^2*x^2+d^2)^(
1/2)/e-7/12*d*(e*x+d)^2*(-e^2*x^2+d^2)^(1/2)/e-1/4*(e*x+d)^3*(-e^2*x^2+d^2)^(1/2)/e

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Rubi [A]  time = 0.06, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {671, 641, 217, 203} \[ -\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-35*d^3*Sqrt[d^2 - e^2*x^2])/(8*e) - (35*d^2*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(24*e) - (7*d*(d + e*x)^2*Sqrt[d^
2 - e^2*x^2])/(12*e) - ((d + e*x)^3*Sqrt[d^2 - e^2*x^2])/(4*e) + (35*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8
*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)^4}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{4} (7 d) \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{12} \left (35 d^2\right ) \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^3\right ) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^4\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {1}{8} \left (35 d^4\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 {35 d^3 \sqrt {d^2-e^2 x^2}}{8 e}-\frac {35 d^2 (d+e x) \sqrt {d^2-e^2 x^2}}{24 e}-\frac {7 d (d+e x)^2 \sqrt {d^2-e^2 x^2}}{12 e}-\frac {(d+e x)^3 \sqrt {d^2-e^2 x^2}}{4 e}+\frac {35 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{8 e}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 81, normalized size = 0.54 \[ \frac {105 d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\sqrt {d^2-e^2 x^2} \left (160 d^3+81 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )}{24 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(160*d^3 + 81*d^2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)) + 105*d^4*ArcTan[(e*x)/Sqrt[d^2 - e^
2*x^2]])/(24*e)

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fricas [A]  time = 0.73, size = 83, normalized size = 0.56 \[ -\frac {210 \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (6 \, e^{3} x^{3} + 32 \, d e^{2} x^{2} + 81 \, d^{2} e x + 160 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{24 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/24*(210*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (6*e^3*x^3 + 32*d*e^2*x^2 + 81*d^2*e*x + 160*d^3)*s
qrt(-e^2*x^2 + d^2))/e

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giac [A]  time = 0.29, size = 63, normalized size = 0.42 \[ \frac {35}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \frac {1}{24} \, {\left (160 \, d^{3} e^{\left (-1\right )} + {\left (81 \, d^{2} + 2 \, {\left (3 \, x e^{2} + 16 \, d e\right )} x\right )} x\right )} \sqrt {-x^{2} e^{2} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

35/8*d^4*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/24*(160*d^3*e^(-1) + (81*d^2 + 2*(3*x*e^2 + 16*d*e)*x)*x)*sqrt(-x^2*e
^2 + d^2)

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maple [A]  time = 0.06, size = 119, normalized size = 0.80 \[ \frac {35 d^{4} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x^{3}}{4}-\frac {4 \sqrt {-e^{2} x^{2}+d^{2}}\, d e \,x^{2}}{3}-\frac {27 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} x}{8}-\frac {20 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}}{3 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/4*e^2*x^3*(-e^2*x^2+d^2)^(1/2)-27/8*d^2*x*(-e^2*x^2+d^2)^(1/2)+35/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^
2*x^2+d^2)^(1/2)*x)-4/3*e*d*x^2*(-e^2*x^2+d^2)^(1/2)-20/3*d^3*(-e^2*x^2+d^2)^(1/2)/e

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maxima [A]  time = 2.95, size = 101, normalized size = 0.68 \[ -\frac {1}{4} \, \sqrt {-e^{2} x^{2} + d^{2}} e^{2} x^{3} - \frac {4}{3} \, \sqrt {-e^{2} x^{2} + d^{2}} d e x^{2} + \frac {35 \, d^{4} \arcsin \left (\frac {e x}{d}\right )}{8 \, e} - \frac {27}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} x - \frac {20 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}}{3 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/4*sqrt(-e^2*x^2 + d^2)*e^2*x^3 - 4/3*sqrt(-e^2*x^2 + d^2)*d*e*x^2 + 35/8*d^4*arcsin(e*x/d)/e - 27/8*sqrt(-e
^2*x^2 + d^2)*d^2*x - 20/3*sqrt(-e^2*x^2 + d^2)*d^3/e

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^4}{\sqrt {d^2-e^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(d^2 - e^2*x^2)^(1/2),x)

[Out]

int((d + e*x)^4/(d^2 - e^2*x^2)^(1/2), x)

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sympy [A]  time = 9.62, size = 546, normalized size = 3.66 \[ d^{4} \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 ) + 4 d^{3} 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 ) + 6 d^{2} 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 ) + 4 d 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 ) + e^{4} \left (\begin {cases} - \frac {3 i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{5}} + \frac {3 i d^{3} x}{8 e^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i d x^{3}}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{5}}{4 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {3 d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{5}} - \frac {3 d^{3} x}{8 e^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {d x^{3}}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**4*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))) + 4*d**3*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e*
*2*x**2)/e**2, True)) + 6*d**2*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)) + 4*d*e**3*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*s
qrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4/(4*sqrt(d**2)), True)) + e**4*Piecewise((-3*I*d**4*acosh(e*x/
d)/(8*e**5) + 3*I*d**3*x/(8*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d*x**3/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - I*
x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2/d**2) > 1), (3*d**4*asin(e*x/d)/(8*e**5) - 3*d**3*x/(8*e**
4*sqrt(1 - e**2*x**2/d**2)) + d*x**3/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + x**5/(4*d*sqrt(1 - e**2*x**2/d**2)),
True))

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