3.119 \(\int x^3 \sqrt {d+e x^2} (a+b \csc ^{-1}(c x)) \, dx\)

Optimal. Leaf size=294 \[ -\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {2 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{15 e^2 \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {b x \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}} \]

[Out]

-1/3*d*(e*x^2+d)^(3/2)*(a+b*arccsc(c*x))/e^2+1/5*(e*x^2+d)^(5/2)*(a+b*arccsc(c*x))/e^2+2/15*b*c*d^(5/2)*x*arct
an((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e^2/(c^2*x^2)^(1/2)-1/120*b*(15*c^4*d^2-10*c^2*d*e-9*e^2)*x*arct
anh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^4/e^(3/2)/(c^2*x^2)^(1/2)+1/20*b*x*(e*x^2+d)^(3/2)*(c^2*x^2
-1)^(1/2)/c/e/(c^2*x^2)^(1/2)+1/120*b*(c^2*d+9*e)*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c^3/e/(c^2*x^2)^(1/2)

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Rubi [A]  time = 0.41, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {266, 43, 5239, 12, 573, 154, 157, 63, 217, 206, 93, 204} \[ -\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {2 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{15 e^2 \sqrt {c^2 x^2}}-\frac {b x \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(c^2*d + 9*e)*x*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/(120*c^3*e*Sqrt[c^2*x^2]) + (b*x*Sqrt[-1 + c^2*x^2]*(d
+ e*x^2)^(3/2))/(20*c*e*Sqrt[c^2*x^2]) - (d*(d + e*x^2)^(3/2)*(a + b*ArcCsc[c*x]))/(3*e^2) + ((d + e*x^2)^(5/2
)*(a + b*ArcCsc[c*x]))/(5*e^2) + (2*b*c*d^(5/2)*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(15*e^
2*Sqrt[c^2*x^2]) - (b*(15*c^4*d^2 - 10*c^2*d*e - 9*e^2)*x*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x
^2])])/(120*c^4*e^(3/2)*Sqrt[c^2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 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 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 573

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

Rule 5239

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int x^3 \sqrt {d+e x^2} \left (a+b \csc ^{-1}(c x)\right ) \, dx &=-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {(b c x) \int \frac {\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{15 e^2 x \sqrt {-1+c^2 x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {(b c x) \int \frac {\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{x \sqrt {-1+c^2 x^2}} \, dx}{15 e^2 \sqrt {c^2 x^2}}\\ &=-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {(b c x) \operatorname {Subst}\left (\int \frac {(d+e x)^{3/2} (-2 d+3 e x)}{x \sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{30 e^2 \sqrt {c^2 x^2}}\\ &=\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {(b x) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x} \left (-4 c^2 d^2+\frac {1}{2} e \left (c^2 d+9 e\right ) x\right )}{x \sqrt {-1+c^2 x}} \, dx,x,x^2\right )}{60 c e^2 \sqrt {c^2 x^2}}\\ &=\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {(b x) \operatorname {Subst}\left (\int \frac {-4 c^4 d^3-\frac {1}{4} e \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) x}{x \sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{60 c^3 e^2 \sqrt {c^2 x^2}}\\ &=\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}-\frac {\left (b c d^3 x\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{15 e^2 \sqrt {c^2 x^2}}-\frac {\left (b \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) x\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+c^2 x} \sqrt {d+e x}} \, dx,x,x^2\right )}{240 c^3 e \sqrt {c^2 x^2}}\\ &=\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}-\frac {\left (2 b c d^3 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 )}{15 e^2 \sqrt {c^2 x^2}}-\frac {\left (b \left (15 c^4 d^2-10 c^2 d e-9 e^2\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 )}{120 c^5 e \sqrt {c^2 x^2}}\\ &=\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {2 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{15 e^2 \sqrt {c^2 x^2}}-\frac {\left (b \left (15 c^4 d^2-10 c^2 d e-9 e^2\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 )}{120 c^5 e \sqrt {c^2 x^2}}\\ &=\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \csc ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \csc ^{-1}(c x)\right )}{5 e^2}+\frac {2 b c d^{5/2} x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{15 e^2 \sqrt {c^2 x^2}}-\frac {b \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) x \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt {c^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 326, normalized size = 1.11 \[ \frac {\sqrt {d+e x^2} \left (8 a c^3 \left (-2 d^2+d e x^2+3 e^2 x^4\right )+8 b c^3 \csc ^{-1}(c x) \left (-2 d^2+d e x^2+3 e^2 x^4\right )+b e x \sqrt {1-\frac {1}{c^2 x^2}} \left (c^2 \left (7 d+6 e x^2\right )+9 e\right )\right )}{120 c^3 e^2}-\frac {b x \sqrt {1-\frac {1}{c^2 x^2}} \left (16 c^7 d^{5/2} \sqrt {d+e x^2} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )+\sqrt {c^2} \sqrt {e} \sqrt {c^2 d+e} \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \sqrt {\frac {c^2 \left (d+e x^2\right )}{c^2 d+e}} \sinh ^{-1}\left (\frac {c \sqrt {e} \sqrt {c^2 x^2-1}}{\sqrt {c^2} \sqrt {c^2 d+e}}\right )\right )}{120 c^6 e^2 \sqrt {c^2 x^2-1} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(8*a*c^3*(-2*d^2 + d*e*x^2 + 3*e^2*x^4) + b*e*Sqrt[1 - 1/(c^2*x^2)]*x*(9*e + c^2*(7*d + 6*e*x
^2)) + 8*b*c^3*(-2*d^2 + d*e*x^2 + 3*e^2*x^4)*ArcCsc[c*x]))/(120*c^3*e^2) - (b*Sqrt[1 - 1/(c^2*x^2)]*x*(Sqrt[c
^2]*Sqrt[e]*Sqrt[c^2*d + e]*(15*c^4*d^2 - 10*c^2*d*e - 9*e^2)*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSinh[(c*S
qrt[e]*Sqrt[-1 + c^2*x^2])/(Sqrt[c^2]*Sqrt[c^2*d + e])] + 16*c^7*d^(5/2)*Sqrt[d + e*x^2]*ArcTan[(Sqrt[d]*Sqrt[
-1 + c^2*x^2])/Sqrt[d + e*x^2]]))/(120*c^6*e^2*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])

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fricas [A]  time = 2.22, size = 1379, normalized size = 4.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsc(c*x))*(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/480*(16*b*c^5*sqrt(-d)*d^2*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) - (15*b*c^4*d^2 - 10*b*c^2*d*e - 9*b*e^2)*sq
rt(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*(24*a*c^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + 8*(3*b*c
^5*e^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^2)*arccsc(c*x) + (6*b*c^3*e^2*x^2 + 7*b*c^3*d*e + 9*b*c*e^2)*sqrt(c^2*x
^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2), 1/480*(32*b*c^5*d^(5/2)*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)) - (15*b*c^4*d^2 - 10*b*c^2*d*e - 9*b*e
^2)*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*(24*a*c^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + 8*
(3*b*c^5*e^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^2)*arccsc(c*x) + (6*b*c^3*e^2*x^2 + 7*b*c^3*d*e + 9*b*c*e^2)*sqrt
(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2), 1/240*(8*b*c^5*sqrt(-d)*d^2*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) + (
15*b*c^4*d^2 - 10*b*c^2*d*e - 9*b*e^2)*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*(24*a*c^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*
a*c^5*d^2 + 8*(3*b*c^5*e^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^2)*arccsc(c*x) + (6*b*c^3*e^2*x^2 + 7*b*c^3*d*e + 9
*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2), 1/240*(16*b*c^5*d^(5/2)*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)) + (15*b*c^4*d^2 -
10*b*c^2*d*e - 9*b*e^2)*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*(24*a*c^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + 8*(
3*b*c^5*e^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^2)*arccsc(c*x) + (6*b*c^3*e^2*x^2 + 7*b*c^3*d*e + 9*b*c*e^2)*sqrt(
c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, x)

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maple [F]  time = 6.64, size = 0, normalized size = 0.00 \[ \int x^{3} \left (a +b \,\mathrm {arccsc}\left (c x \right )\right ) \sqrt {e \,x^{2}+d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{15} \, {\left (\frac {3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{2}}{e} - \frac {2 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d}{e^{2}}\right )} a + \frac {{\left (e^{2} \int \frac {{\left (3 \, c^{2} e^{2} x^{5} + c^{2} d e x^{3} - 2 \, c^{2} d^{2} x\right )} e^{\left (\frac {1}{2} \, \log \left (e x^{2} + d\right ) + \frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e^{2} x^{2} + {\left (c^{2} e^{2} x^{2} - e^{2}\right )} {\left (c x + 1\right )} {\left (c x - 1\right )} - e^{2}}\,{d x} + {\left (3 \, e^{2} x^{4} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right ) + d e x^{2} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right ) - 2 \, d^{2} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} \sqrt {e x^{2} + d}\right )} b}{15 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsc(c*x))*(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*(e*x^2 + d)^(3/2)*x^2/e - 2*(e*x^2 + d)^(3/2)*d/e^2)*a + 1/15*(15*e^2*integrate(1/15*(3*c^2*e^2*x^5 +
c^2*d*e*x^3 - 2*c^2*d^2*x)*e^(1/2*log(e*x^2 + d) + 1/2*log(c*x + 1) + 1/2*log(c*x - 1))/(c^2*e^2*x^2 - e^2 + (
c^2*e^2*x^2 - e^2)*e^(log(c*x + 1) + log(c*x - 1))), x) + (3*e^2*x^4*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) +
d*e*x^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) - 2*d^2*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)))*sqrt(e*x^2 +
d))*b/e^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acsc(c*x))*(e*x**2+d)**(1/2),x)

[Out]

Integral(x**3*(a + b*acsc(c*x))*sqrt(d + e*x**2), x)

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