3.826 \(\int \frac{(d+e x)^5}{\sqrt{d^2-e^2 x^2}} \, dx\)

Optimal. Leaf size=182 \[ -\frac{63 d^4 \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{63 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

[Out]

(-63*d^4*Sqrt[d^2 - e^2*x^2])/(8*e) - (21*d^3*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(8*e) - (21*d^2*(d + e*x)^2*Sqrt[
d^2 - e^2*x^2])/(20*e) - (9*d*(d + e*x)^3*Sqrt[d^2 - e^2*x^2])/(20*e) - ((d + e*x)^4*Sqrt[d^2 - e^2*x^2])/(5*e
) + (63*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

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Rubi [A]  time = 0.078631, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {671, 641, 217, 203} \[ -\frac{63 d^4 \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{63 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^5/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-63*d^4*Sqrt[d^2 - e^2*x^2])/(8*e) - (21*d^3*(d + e*x)*Sqrt[d^2 - e^2*x^2])/(8*e) - (21*d^2*(d + e*x)^2*Sqrt[
d^2 - e^2*x^2])/(20*e) - (9*d*(d + e*x)^3*Sqrt[d^2 - e^2*x^2])/(20*e) - ((d + e*x)^4*Sqrt[d^2 - e^2*x^2])/(5*e
) + (63*d^5*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e)

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
]

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 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 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])

Rubi steps

\begin{align*} \int \frac{(d+e x)^5}{\sqrt{d^2-e^2 x^2}} \, dx &=-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{5} (9 d) \int \frac{(d+e x)^4}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{20} \left (63 d^2\right ) \int \frac{(d+e x)^3}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{4} \left (21 d^3\right ) \int \frac{(d+e x)^2}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{8} \left (63 d^4\right ) \int \frac{d+e x}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{63 d^4 \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{8} \left (63 d^5\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx\\ &=-\frac{63 d^4 \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{1}{8} \left (63 d^5\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{63 d^4 \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^3 (d+e x) \sqrt{d^2-e^2 x^2}}{8 e}-\frac{21 d^2 (d+e x)^2 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{9 d (d+e x)^3 \sqrt{d^2-e^2 x^2}}{20 e}-\frac{(d+e x)^4 \sqrt{d^2-e^2 x^2}}{5 e}+\frac{63 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e}\\ \end{align*}

Mathematica [A]  time = 0.121002, size = 92, normalized size = 0.51 \[ \frac{315 d^5 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )-\sqrt{d^2-e^2 x^2} \left (144 d^2 e^2 x^2+275 d^3 e x+488 d^4+50 d e^3 x^3+8 e^4 x^4\right )}{40 e} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^5/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(488*d^4 + 275*d^3*e*x + 144*d^2*e^2*x^2 + 50*d*e^3*x^3 + 8*e^4*x^4)) + 315*d^5*ArcTan[
(e*x)/Sqrt[d^2 - e^2*x^2]])/(40*e)

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Maple [A]  time = 0.061, size = 144, normalized size = 0.8 \begin{align*} -{\frac{{e}^{3}{x}^{4}}{5}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}-{\frac{18\,e{d}^{2}{x}^{2}}{5}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}-{\frac{61\,{d}^{4}}{5\,e}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}-{\frac{5\,d{e}^{2}{x}^{3}}{4}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}-{\frac{55\,{d}^{3}x}{8}\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}+{\frac{63\,{d}^{5}}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{e}^{2}{x}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^5/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/5*e^3*x^4*(-e^2*x^2+d^2)^(1/2)-18/5*e*d^2*x^2*(-e^2*x^2+d^2)^(1/2)-61/5*d^4*(-e^2*x^2+d^2)^(1/2)/e-5/4*d*e^
2*x^3*(-e^2*x^2+d^2)^(1/2)-55/8*d^3*x*(-e^2*x^2+d^2)^(1/2)+63/8*d^5/(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 = 1.68987, size = 184, normalized size = 1.01 \begin{align*} -\frac{1}{5} \, \sqrt{-e^{2} x^{2} + d^{2}} e^{3} x^{4} - \frac{5}{4} \, \sqrt{-e^{2} x^{2} + d^{2}} d e^{2} x^{3} - \frac{18}{5} \, \sqrt{-e^{2} x^{2} + d^{2}} d^{2} e x^{2} + \frac{63 \, d^{5} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}}} - \frac{55}{8} \, \sqrt{-e^{2} x^{2} + d^{2}} d^{3} x - \frac{61 \, \sqrt{-e^{2} x^{2} + d^{2}} d^{4}}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/5*sqrt(-e^2*x^2 + d^2)*e^3*x^4 - 5/4*sqrt(-e^2*x^2 + d^2)*d*e^2*x^3 - 18/5*sqrt(-e^2*x^2 + d^2)*d^2*e*x^2 +
63/8*d^5*arcsin(e^2*x/sqrt(d^2*e^2))/sqrt(e^2) - 55/8*sqrt(-e^2*x^2 + d^2)*d^3*x - 61/5*sqrt(-e^2*x^2 + d^2)*
d^4/e

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Fricas [A]  time = 2.1493, size = 207, normalized size = 1.14 \begin{align*} -\frac{630 \, d^{5} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (8 \, e^{4} x^{4} + 50 \, d e^{3} x^{3} + 144 \, d^{2} e^{2} x^{2} + 275 \, d^{3} e x + 488 \, d^{4}\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{40 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/40*(630*d^5*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (8*e^4*x^4 + 50*d*e^3*x^3 + 144*d^2*e^2*x^2 + 275*d
^3*e*x + 488*d^4)*sqrt(-e^2*x^2 + d^2))/e

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Sympy [A]  time = 9.07527, size = 644, normalized size = 3.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**5/(-e**2*x**2+d**2)**(1/2),x)

[Out]

d**5*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))) + 5*d**4*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e*
*2*x**2)/e**2, True)) + 10*d**3*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)/Abs(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)) + 10*d**2*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)) + 5*d*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)/Abs(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)) + e**5*Piecewise((-8*d**4*sqrt(d**2 - e**2*x**2)/(15*e**6) - 4*d**2*x**2*sqrt(d**2 -
e**2*x**2)/(15*e**4) - x**4*sqrt(d**2 - e**2*x**2)/(5*e**2), Ne(e, 0)), (x**6/(6*sqrt(d**2)), True))

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Giac [A]  time = 1.27637, size = 99, normalized size = 0.54 \begin{align*} \frac{63}{8} \, d^{5} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-1\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{40} \,{\left (488 \, d^{4} e^{\left (-1\right )} +{\left (275 \, d^{3} + 2 \,{\left (72 \, d^{2} e +{\left (4 \, x e^{3} + 25 \, d e^{2}\right )} x\right )} x\right )} x\right )} \sqrt{-x^{2} e^{2} + d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

63/8*d^5*arcsin(x*e/d)*e^(-1)*sgn(d) - 1/40*(488*d^4*e^(-1) + (275*d^3 + 2*(72*d^2*e + (4*x*e^3 + 25*d*e^2)*x)
*x)*x)*sqrt(-x^2*e^2 + d^2)