3.66 \(\int \frac {a+b \sec ^{-1}(c x)}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}-\frac {4 b \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}} \]

[Out]

-2*(a+b*arcsec(c*x))/e/(e*x+d)^(1/2)-4*b*EllipticPi(1/2*(-c*x+1)^(1/2)*2^(1/2),2,2^(1/2)*(e/(c*d+e))^(1/2))*(c
*(e*x+d)/(c*d+e))^(1/2)*(-c^2*x^2+1)^(1/2)/c/e/x/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5226, 1574, 933, 168, 538, 537} \[ -\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}-\frac {4 b \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/(d + e*x)^(3/2),x]

[Out]

(-2*(a + b*ArcSec[c*x]))/(e*Sqrt[d + e*x]) - (4*b*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2
, ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c*e*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x])

Rule 168

Int[1/(((a_.) + (b_.)*(x_))*Sqrt[(c_.) + (d_.)*(x_)]*Sqrt[(e_.) + (f_.)*(x_)]*Sqrt[(g_.) + (h_.)*(x_)]), x_Sym
bol] :> Dist[-2, Subst[Int[1/(Simp[b*c - a*d - b*x^2, x]*Sqrt[Simp[(d*e - c*f)/d + (f*x^2)/d, x]]*Sqrt[Simp[(d
*g - c*h)/d + (h*x^2)/d, x]]), x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && GtQ[(d*e - c
*f)/d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 538

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 +
(d*x^2)/c]/Sqrt[c + d*x^2], Int[1/((a + b*x^2)*Sqrt[1 + (d*x^2)/c]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[c, 0]

Rule 933

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-(c
/a), 2]}, Dist[Sqrt[1 + (c*x^2)/a]/Sqrt[a + c*x^2], Int[1/((d + e*x)*Sqrt[f + g*x]*Sqrt[1 - q*x]*Sqrt[1 + q*x]
), x], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + a*e^2, 0] &&  !GtQ[a, 0]

Rule 1574

Int[(x_)^(m_.)*((a_.) + (c_.)*(x_)^(mn2_.))^(p_)*((d_) + (e_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Dist[(x^(2*n*Fr
acPart[p])*(a + c/x^(2*n))^FracPart[p])/(c + a*x^(2*n))^FracPart[p], Int[x^(m - 2*n*p)*(d + e*x^n)^q*(c + a*x^
(2*n))^p, x], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && EqQ[mn2, -2*n] &&  !IntegerQ[p] &&  !IntegerQ[q] &&
PosQ[n]

Rule 5226

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcSec[c*x]))/(e*(m + 1)), x] - Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x],
x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \sec ^{-1}(c x)}{(d+e x)^{3/2}} \, dx &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}+\frac {(2 b) \int \frac {1}{\sqrt {1-\frac {1}{c^2 x^2}} x^2 \sqrt {d+e x}} \, dx}{c e}\\ &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}+\frac {\left (2 b \sqrt {-\frac {1}{c^2}+x^2}\right ) \int \frac {1}{x \sqrt {d+e x} \sqrt {-\frac {1}{c^2}+x^2}} \, dx}{c e \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}+\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {1}{x \sqrt {1-c x} \sqrt {1+c x} \sqrt {d+e x}} \, dx}{c e \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}-\frac {\left (4 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {2-x^2} \sqrt {d+\frac {e}{c}-\frac {e x^2}{c}}} \, dx,x,\sqrt {1-c x}\right )}{c e \sqrt {1-\frac {1}{c^2 x^2}} x}\\ &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}-\frac {\left (4 b \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {2-x^2} \sqrt {1-\frac {e x^2}{c \left (d+\frac {e}{c}\right )}}} \, dx,x,\sqrt {1-c x}\right )}{c e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ &=-\frac {2 \left (a+b \sec ^{-1}(c x)\right )}{e \sqrt {d+e x}}-\frac {4 b \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 124, normalized size = 1.04 \[ -\frac {2 \left (\left (c^2 x^2-1\right ) \left (a+b \sec ^{-1}(c x)\right )+2 b c x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\sin ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )\right )}{e \left (c^2 x^2-1\right ) \sqrt {d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/(d + e*x)^(3/2),x]

[Out]

(-2*((-1 + c^2*x^2)*(a + b*ArcSec[c*x]) + 2*b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 -
c^2*x^2]*EllipticPi[2, ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)]))/(e*Sqrt[d + e*x]*(-1 + c^2*x^2))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)/(e*x + d)^(3/2), x)

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maple [A]  time = 0.07, size = 217, normalized size = 1.82 \[ \frac {-\frac {2 a}{\sqrt {e x +d}}+2 b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{\sqrt {e x +d}}-\frac {2 \sqrt {-\frac {\left (e x +d \right ) c -d c +e}{d c -e}}\, \sqrt {-\frac {\left (e x +d \right ) c -d c -e}{d c +e}}\, \EllipticPi \left (\sqrt {e x +d}\, \sqrt {\frac {c}{d c -e}}, \frac {d c -e}{c d}, \frac {\sqrt {\frac {c}{d c +e}}}{\sqrt {\frac {c}{d c -e}}}\right )}{c \sqrt {\frac {c^{2} \left (e x +d \right )^{2}-2 c^{2} d \left (e x +d \right )+c^{2} d^{2}-e^{2}}{c^{2} e^{2} x^{2}}}\, x d \sqrt {\frac {c}{d c -e}}}\right )}{e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsec(c*x))/(e*x+d)^(3/2),x)

[Out]

2/e*(-1/(e*x+d)^(1/2)*a+b*(-1/(e*x+d)^(1/2)*arcsec(c*x)-2/c/((c^2*(e*x+d)^2-2*c^2*d*(e*x+d)+c^2*d^2-e^2)/c^2/e
^2/x^2)^(1/2)/x/d/(c/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c+e)/(c*d-e))^(1/2)*(-((e*x+d)*c-d*c-e)/(c*d+e))^(1/2)*Elli
pticPi((e*x+d)^(1/2)*(c/(c*d-e))^(1/2),1/c*(c*d-e)/d,(c/(c*d+e))^(1/2)/(c/(c*d-e))^(1/2))))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c*d-e>0)', see `assume?` for m
ore details)Is c*d-e positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))/(d + e*x)^(3/2),x)

[Out]

int((a + b*acos(1/(c*x)))/(d + e*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asec}{\left (c x \right )}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/(e*x+d)**(3/2),x)

[Out]

Integral((a + b*asec(c*x))/(d + e*x)**(3/2), x)

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