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

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

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Rubi [A]  time = 0.08, antiderivative size = 182, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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 {gather*}

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 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)^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*}

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Mathematica [A]  time = 0.12, size = 92, normalized size = 0.51 \begin {gather*} \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 (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\right )}{40 e} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.38, size = 114, normalized size = 0.63 \begin {gather*} \frac {63 d^5 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e^2}+\frac {\sqrt {d^2-e^2 x^2} \left (-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\right )}{40 e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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))/(40*e) + (63*d^5*S
qrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e^2)

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fricas [A]  time = 0.42, size = 94, normalized size = 0.52 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.27, size = 73, normalized size = 0.40 \begin {gather*} \frac {63}{8} \, d^{5} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\relax (d) - \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 {gather*}

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)

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maple [A]  time = 0.07, size = 144, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, e^{3} x^{4}}{5}+\frac {63 d^{5} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}-\frac {5 \sqrt {-e^{2} x^{2}+d^{2}}\, d \,e^{2} x^{3}}{4}-\frac {18 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e \,x^{2}}{5}-\frac {55 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{3} x}{8}-\frac {61 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{4}}{5 e} \end {gather*}

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

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maxima [A]  time = 2.96, size = 126, normalized size = 0.69 \begin {gather*} -\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 x}{d}\right )}{8 \, e} - \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 {gather*}

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*x/d)/e - 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^5}{\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)^5/(d^2 - e^2*x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 11.02, size = 641, normalized size = 3.52 \begin {gather*} d^{5} \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 ) + 5 d^{4} 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 ) + 10 d^{3} 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 ) + 10 d^{2} 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 ) + 5 d 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 ) + e^{5} \left (\begin {cases} - \frac {8 d^{4} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{6}} - \frac {4 d^{2} x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{15 e^{4}} - \frac {x^{4} \sqrt {d^{2} - e^{2} x^{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \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)**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/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*a
cosh(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)) + 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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