3.120 \(\int x \sqrt {d+e x^2} (a+b \csc ^{-1}(c x)) \, dx\)

Optimal. Leaf size=195 \[ \frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}-\frac {b c d^{3/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {b x \left (3 c^2 d+e\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{6 c^2 \sqrt {e} \sqrt {c^2 x^2}} \]

[Out]

1/3*(e*x^2+d)^(3/2)*(a+b*arccsc(c*x))/e-1/3*b*c*d^(3/2)*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e/
(c^2*x^2)^(1/2)+1/6*b*(3*c^2*d+e)*x*arctanh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^2/e^(1/2)/(c^2*x^2)
^(1/2)+1/6*b*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c/(c^2*x^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5237, 446, 102, 157, 63, 217, 206, 93, 204} \[ \frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}-\frac {b c d^{3/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {b x \left (3 c^2 d+e\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{6 c^2 \sqrt {e} \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[d + e*x^2]*(a + b*ArcCsc[c*x]),x]

[Out]

(b*x*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/(6*c*Sqrt[c^2*x^2]) + ((d + e*x^2)^(3/2)*(a + b*ArcCsc[c*x]))/(3*e) -
(b*c*d^(3/2)*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(3*e*Sqrt[c^2*x^2]) + (b*(3*c^2*d + e)*x
*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/(6*c^2*Sqrt[e]*Sqrt[c^2*x^2])

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 102

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5237

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

Rubi steps

\begin {align*} \int x \sqrt {d+e x^2} \left (a+b \csc ^{-1}(c x)\right ) \, dx &=\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}+\frac {(b c x) \int \frac {\left (d+e x^2\right )^{3/2}}{x \sqrt {-1+c^2 x^2}} \, dx}{3 e \sqrt {c^2 x^2}}\\ &=\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}+\frac {(b c x) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2}}{x \sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{6 e \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}+\frac {(b x) \operatorname {Subst}\left (\int \frac {c^2 d^2+\frac {1}{2} e \left (3 c^2 d+e\right ) x}{x \sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{6 c e \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}+\frac {\left (b c d^2 x\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{6 e \sqrt {c^2 x^2}}+\frac {\left (b \left (3 c^2 d+e\right ) x\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{12 c \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}+\frac {\left (b c d^2 x\right ) \operatorname {Subst}\left (\int \frac {1}{-d-x^2} \, dx,x,\frac {\sqrt {d+e x^2}}{\sqrt {-1+c^2 x^2}}\right )}{3 e \sqrt {c^2 x^2}}+\frac {\left (b \left (3 c^2 d+e\right ) x\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+\frac {e}{c^2}+\frac {e x^2}{c^2}}} \, dx,x,\sqrt {-1+c^2 x^2}\right )}{6 c^3 \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}-\frac {b c d^{3/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{3 e \sqrt {c^2 x^2}}+\frac {\left (b \left (3 c^2 d+e\right ) x\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {e x^2}{c^2}} \, dx,x,\frac {\sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}}\right )}{6 c^3 \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e}-\frac {b c d^{3/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{3 e \sqrt {c^2 x^2}}+\frac {b \left (3 c^2 d+e\right ) x \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^2 \sqrt {e} \sqrt {c^2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.48, size = 265, normalized size = 1.36 \[ \frac {\sqrt {d+e x^2} \left (2 a c \left (d+e x^2\right )+b e x \sqrt {1-\frac {1}{c^2 x^2}}+2 b c \csc ^{-1}(c x) \left (d+e x^2\right )\right )}{6 c e}+\frac {b x \sqrt {1-\frac {1}{c^2 x^2}} \left (\sqrt {c^2} \sqrt {e} \sqrt {c^2 d+e} \left (3 c^2 d+e\right ) \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sinh ^{-1}\left (\frac {c \sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2} \sqrt {c^2 d+e}}\right )+2 c^5 d^{3/2} \sqrt {d+e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )\right )}{6 c^4 e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[d + e*x^2]*(a + b*ArcCsc[c*x]),x]

[Out]

(Sqrt[d + e*x^2]*(b*e*Sqrt[1 - 1/(c^2*x^2)]*x + 2*a*c*(d + e*x^2) + 2*b*c*(d + e*x^2)*ArcCsc[c*x]))/(6*c*e) +
(b*Sqrt[1 - 1/(c^2*x^2)]*x*(Sqrt[c^2]*Sqrt[e]*Sqrt[c^2*d + e]*(3*c^2*d + e)*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)
]*ArcSinh[(c*Sqrt[e]*Sqrt[-1 + c^2*x^2])/(Sqrt[c^2]*Sqrt[c^2*d + e])] + 2*c^5*d^(3/2)*Sqrt[d + e*x^2]*ArcTan[(
Sqrt[d]*Sqrt[-1 + c^2*x^2])/Sqrt[d + e*x^2]]))/(6*c^4*e*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])

________________________________________________________________________________________

fricas [A]  time = 2.84, size = 1098, normalized size = 5.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))*(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/24*(2*b*c^3*sqrt(-d)*d*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*sqrt(c^2*x^2 - 1)*(
(c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) + (3*b*c^2*d + b*e)*sqrt(e)*log(8*c^4*e^2*x^4 +
c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 + 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 +
d)*sqrt(e) + e^2) + 4*(2*a*c^3*e*x^2 + 2*a*c^3*d + sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*arccsc
(c*x))*sqrt(e*x^2 + d))/(c^3*e), -1/24*(4*b*c^3*d^(3/2)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*
sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) - (3*b*c^2*d + b*e)*sqrt(e)*log(8*c^4*e^2*x
^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 + 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*
x^2 + d)*sqrt(e) + e^2) - 4*(2*a*c^3*e*x^2 + 2*a*c^3*d + sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*a
rccsc(c*x))*sqrt(e*x^2 + d))/(c^3*e), 1/12*(b*c^3*sqrt(-d)*d*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2
- d*e)*x^2 + 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) - (3*b*c^2*d
+ b*e)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 -
c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*(2*a*c^3*e*x^2 + 2*a*c^3*d + sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b
*c^3*d)*arccsc(c*x))*sqrt(e*x^2 + d))/(c^3*e), -1/12*(2*b*c^3*d^(3/2)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d -
e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (3*b*c^2*d + b*e)*sqrt(-e)*
arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*d*
e - c*e^2)*x^2)) - 2*(2*a*c^3*e*x^2 + 2*a*c^3*d + sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 + b*c^3*d)*arccsc(c
*x))*sqrt(e*x^2 + d))/(c^3*e)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))*(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*x^2 + d)*(b*arccsc(c*x) + a)*x, x)

________________________________________________________________________________________

maple [F]  time = 5.06, size = 0, normalized size = 0.00 \[ \int x \left (a +b \,\mathrm {arccsc}\left (c x \right )\right ) \sqrt {e \,x^{2}+d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} a}{3 \, e} + \frac {{\left (e \int \frac {{\left (c^{2} e x^{3} + c^{2} d x\right )} e^{\left (\frac {1}{2} \, \log \left (e x^{2} + d\right ) + \frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e x^{2} + {\left (c^{2} e x^{2} - e\right )} {\left (c x + 1\right )} {\left (c x - 1\right )} - e}\,{d x} + {\left (e x^{2} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right ) + d \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} \sqrt {e x^{2} + d}\right )} b}{3 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))*(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/3*(e*x^2 + d)^(3/2)*a/e + 1/3*(3*e*integrate(1/3*(c^2*e*x^3 + c^2*d*x)*e^(1/2*log(e*x^2 + d) + 1/2*log(c*x +
1) + 1/2*log(c*x - 1))/(c^2*e*x^2 + (c^2*e*x^2 - e)*e^(log(c*x + 1) + log(c*x - 1)) - e), x) + (e*x^2*arctan2
(1, sqrt(c*x + 1)*sqrt(c*x - 1)) + d*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)))*sqrt(e*x^2 + d))*b/e

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acsc(c*x))*(e*x**2+d)**(1/2),x)

[Out]

Integral(x*(a + b*acsc(c*x))*sqrt(d + e*x**2), x)

________________________________________________________________________________________