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

Optimal. Leaf size=113 \[ \frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {815, 844, 217, 203, 266, 63, 208} \begin {gather*} \frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(8*d + 3*e*x)*Sqrt[d^2 - e^2*x^2])/8 + ((4*d + 3*e*x)*(d^2 - e^2*x^2)^(3/2))/12 + (3*d^4*ArcTan[(e*x)/Sqr
t[d^2 - e^2*x^2]])/8 - d^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 208

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

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (d^2-e^2 x^2\right )^{3/2}}{x} \, dx &=\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}-\frac {\int \frac {\left (-4 d^3 e^2-3 d^2 e^3 x\right ) \sqrt {d^2-e^2 x^2}}{x} \, dx}{4 e^2}\\ &=\frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {\int \frac {8 d^5 e^4+3 d^4 e^5 x}{x \sqrt {d^2-e^2 x^2}} \, dx}{8 e^4}\\ &=\frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+d^5 \int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx+\frac {1}{8} \left (3 d^4 e\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=\frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {1}{2} d^5 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )+\frac {1}{8} \left (3 d^4 e\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 {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-\frac {d^5 \operatorname {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{e^2}\\ &=\frac {1}{8} d^2 (8 d+3 e x) \sqrt {d^2-e^2 x^2}+\frac {1}{12} (4 d+3 e x) \left (d^2-e^2 x^2\right )^{3/2}+\frac {3}{8} d^4 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )-d^4 \tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 124, normalized size = 1.10 \begin {gather*} d^4 \left (-\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )\right )+\frac {3 d^3 \sqrt {d^2-e^2 x^2} \sin ^{-1}\left (\frac {e x}{d}\right )}{8 \sqrt {1-\frac {e^2 x^2}{d^2}}}+\frac {1}{24} \sqrt {d^2-e^2 x^2} \left (32 d^3+15 d^2 e x-8 d e^2 x^2-6 e^3 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(32*d^3 + 15*d^2*e*x - 8*d*e^2*x^2 - 6*e^3*x^3))/24 + (3*d^3*Sqrt[d^2 - e^2*x^2]*ArcSin[(
e*x)/d])/(8*Sqrt[1 - (e^2*x^2)/d^2]) - d^4*ArcTanh[Sqrt[d^2 - e^2*x^2]/d]

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IntegrateAlgebraic [A]  time = 0.51, size = 142, normalized size = 1.26 \begin {gather*} \frac {3 d^4 \sqrt {-e^2} \log \left (\sqrt {d^2-e^2 x^2}-\sqrt {-e^2} x\right )}{8 e}+2 d^4 \tanh ^{-1}\left (\frac {\sqrt {-e^2} x}{d}-\frac {\sqrt {d^2-e^2 x^2}}{d}\right )+\frac {1}{24} \sqrt {d^2-e^2 x^2} \left (32 d^3+15 d^2 e x-8 d e^2 x^2-6 e^3 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d^2 - e^2*x^2]*(32*d^3 + 15*d^2*e*x - 8*d*e^2*x^2 - 6*e^3*x^3))/24 + 2*d^4*ArcTanh[(Sqrt[-e^2]*x)/d - Sq
rt[d^2 - e^2*x^2]/d] + (3*d^4*Sqrt[-e^2]*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(8*e)

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fricas [A]  time = 0.41, size = 107, normalized size = 0.95 \begin {gather*} -\frac {3}{4} \, d^{4} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + d^{4} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - \frac {1}{24} \, {\left (6 \, e^{3} x^{3} + 8 \, d e^{2} x^{2} - 15 \, d^{2} e x - 32 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x,x, algorithm="fricas")

[Out]

-3/4*d^4*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + d^4*log(-(d - sqrt(-e^2*x^2 + d^2))/x) - 1/24*(6*e^3*x^3
+ 8*d*e^2*x^2 - 15*d^2*e*x - 32*d^3)*sqrt(-e^2*x^2 + d^2)

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giac [A]  time = 0.27, size = 99, normalized size = 0.88 \begin {gather*} \frac {3}{8} \, d^{4} \arcsin \left (\frac {x e}{d}\right ) \mathrm {sgn}\relax (d) - d^{4} \log \left (\frac {{\left | -2 \, d e - 2 \, \sqrt {-x^{2} e^{2} + d^{2}} e \right |} e^{\left (-2\right )}}{2 \, {\left | x \right |}}\right ) + \frac {1}{24} \, {\left (32 \, d^{3} + {\left (15 \, d^{2} e - 2 \, {\left (3 \, x e^{3} + 4 \, d e^{2}\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)*(-e^2*x^2+d^2)^(3/2)/x,x, algorithm="giac")

[Out]

3/8*d^4*arcsin(x*e/d)*sgn(d) - d^4*log(1/2*abs(-2*d*e - 2*sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/abs(x)) + 1/24*(32*d^
3 + (15*d^2*e - 2*(3*x*e^3 + 4*d*e^2)*x)*x)*sqrt(-x^2*e^2 + d^2)

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maple [A]  time = 0.02, size = 151, normalized size = 1.34 \begin {gather*} -\frac {d^{5} \ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{\sqrt {d^{2}}}+\frac {3 d^{4} e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{8 \sqrt {e^{2}}}+\frac {3 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2} e x}{8}+\sqrt {-e^{2} x^{2}+d^{2}}\, d^{3}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} e x}{4}+\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}} d}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x,x)

[Out]

1/4*e*x*(-e^2*x^2+d^2)^(3/2)+3/8*e*d^2*x*(-e^2*x^2+d^2)^(1/2)+3/8*e*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)/(-e^2*x
^2+d^2)^(1/2)*x)+1/3*d*(-e^2*x^2+d^2)^(3/2)+d^3*(-e^2*x^2+d^2)^(1/2)-d^5/(d^2)^(1/2)*ln((2*d^2+2*(d^2)^(1/2)*(
-e^2*x^2+d^2)^(1/2))/x)

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maxima [A]  time = 0.99, size = 124, normalized size = 1.10 \begin {gather*} \frac {3}{8} \, d^{4} \arcsin \left (\frac {e x}{d}\right ) - d^{4} \log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right ) + \frac {3}{8} \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} e x + \sqrt {-e^{2} x^{2} + d^{2}} d^{3} + \frac {1}{4} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e x + \frac {1}{3} \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e^2*x^2+d^2)^(3/2)/x,x, algorithm="maxima")

[Out]

3/8*d^4*arcsin(e*x/d) - d^4*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x)) + 3/8*sqrt(-e^2*x^2 + d^2)*d^2
*e*x + sqrt(-e^2*x^2 + d^2)*d^3 + 1/4*(-e^2*x^2 + d^2)^(3/2)*e*x + 1/3*(-e^2*x^2 + d^2)^(3/2)*d

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mupad [B]  time = 2.90, size = 107, normalized size = 0.95 \begin {gather*} \frac {d\,{\left (d^2-e^2\,x^2\right )}^{3/2}}{3}-d^4\,\mathrm {atanh}\left (\frac {\sqrt {d^2-e^2\,x^2}}{d}\right )+d^3\,\sqrt {d^2-e^2\,x^2}+\frac {e\,x\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ \frac {e^2\,x^2}{d^2}\right )}{{\left (1-\frac {e^2\,x^2}{d^2}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d^2 - e^2*x^2)^(3/2)*(d + e*x))/x,x)

[Out]

(d*(d^2 - e^2*x^2)^(3/2))/3 - d^4*atanh((d^2 - e^2*x^2)^(1/2)/d) + d^3*(d^2 - e^2*x^2)^(1/2) + (e*x*(d^2 - e^2
*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, (e^2*x^2)/d^2))/(1 - (e^2*x^2)/d^2)^(3/2)

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sympy [C]  time = 22.41, size = 469, normalized size = 4.15 \begin {gather*} d^{3} \left (\begin {cases} \frac {d^{2}}{e x \sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} - d \operatorname {acosh}{\left (\frac {d}{e x} \right )} - \frac {e x}{\sqrt {\frac {d^{2}}{e^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {d^{2}}{e^{2} x^{2}}}\right | > 1 \\- \frac {i d^{2}}{e x \sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} + i d \operatorname {asin}{\left (\frac {d}{e x} \right )} + \frac {i e x}{\sqrt {- \frac {d^{2}}{e^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) + d^{2} e \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e} - \frac {i d x}{2 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{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} + \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \frac {x^{2} \sqrt {d^{2}}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\left (d^{2} - e^{2} x^{2}\right )^{\frac {3}{2}}}{3 e^{2}} & \text {otherwise} \end {cases}\right ) - e^{3} \left (\begin {cases} - \frac {i d^{4} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{8 e^{3}} + \frac {i d^{3} x}{8 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {3 i d x^{3}}{8 \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {i e^{2} 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 {d^{4} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{8 e^{3}} - \frac {d^{3} x}{8 e^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} + \frac {3 d x^{3}}{8 \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {e^{2} x^{5}}{4 d \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(-e**2*x**2+d**2)**(3/2)/x,x)

[Out]

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

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