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

Optimal. Leaf size=138 \[ \frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}+\frac {b c d^2 x \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{4 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (2 c^2 d+e\right )}{4 c^3 \sqrt {c^2 x^2}}+\frac {b e x \left (c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt {c^2 x^2}} \]

[Out]

1/4*(e*x^2+d)^2*(a+b*arccsc(c*x))/e+1/12*b*e*x*(c^2*x^2-1)^(3/2)/c^3/(c^2*x^2)^(1/2)+1/4*b*c*d^2*x*arctan((c^2
*x^2-1)^(1/2))/e/(c^2*x^2)^(1/2)+1/4*b*(2*c^2*d+e)*x*(c^2*x^2-1)^(1/2)/c^3/(c^2*x^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {5237, 446, 88, 63, 205} \[ \frac {\left (d+e x^2\right )^2 \left (a+b \csc ^{-1}(c x)\right )}{4 e}+\frac {b c d^2 x \tan ^{-1}\left (\sqrt {c^2 x^2-1}\right )}{4 e \sqrt {c^2 x^2}}+\frac {b x \sqrt {c^2 x^2-1} \left (2 c^2 d+e\right )}{4 c^3 \sqrt {c^2 x^2}}+\frac {b e x \left (c^2 x^2-1\right )^{3/2}}{12 c^3 \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(2*c^2*d + e)*x*Sqrt[-1 + c^2*x^2])/(4*c^3*Sqrt[c^2*x^2]) + (b*e*x*(-1 + c^2*x^2)^(3/2))/(12*c^3*Sqrt[c^2*x
^2]) + ((d + e*x^2)^2*(a + b*ArcCsc[c*x]))/(4*e) + (b*c*d^2*x*ArcTan[Sqrt[-1 + c^2*x^2]])/(4*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 88

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

Rule 205

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

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

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Mathematica [A]  time = 0.10, size = 78, normalized size = 0.57 \[ \frac {x \left (3 a c^3 x \left (2 d+e x^2\right )+3 b c^3 x \csc ^{-1}(c x) \left (2 d+e x^2\right )+b \sqrt {1-\frac {1}{c^2 x^2}} \left (c^2 \left (6 d+e x^2\right )+2 e\right )\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*a*c^3*x*(2*d + e*x^2) + b*Sqrt[1 - 1/(c^2*x^2)]*(2*e + c^2*(6*d + e*x^2)) + 3*b*c^3*x*(2*d + e*x^2)*ArcC
sc[c*x]))/(12*c^3)

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fricas [A]  time = 0.89, size = 85, normalized size = 0.62 \[ \frac {3 \, a c^{4} e x^{4} + 6 \, a c^{4} d x^{2} + 3 \, {\left (b c^{4} e x^{4} + 2 \, b c^{4} d x^{2}\right )} \operatorname {arccsc}\left (c x\right ) + {\left (b c^{2} e x^{2} + 6 \, b c^{2} d + 2 \, b e\right )} \sqrt {c^{2} x^{2} - 1}}{12 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*e*x^4 + 6*a*c^4*d*x^2 + 3*(b*c^4*e*x^4 + 2*b*c^4*d*x^2)*arccsc(c*x) + (b*c^2*e*x^2 + 6*b*c^2*d +
2*b*e)*sqrt(c^2*x^2 - 1))/c^4

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giac [B]  time = 0.22, size = 570, normalized size = 4.13 \[ \frac {1}{192} \, {\left (\frac {3 \, b x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4} \arcsin \left (\frac {1}{c x}\right ) e}{c} + \frac {3 \, a x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4} e}{c} + \frac {24 \, b d x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2} \arcsin \left (\frac {1}{c x}\right )}{c} + \frac {2 \, b x^{3} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{3} e}{c^{2}} + \frac {24 \, a d x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}}{c} + \frac {12 \, b x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2} \arcsin \left (\frac {1}{c x}\right ) e}{c^{3}} + \frac {12 \, a x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2} e}{c^{3}} + \frac {48 \, b d x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}}{c^{2}} + \frac {48 \, b d \arcsin \left (\frac {1}{c x}\right )}{c^{3}} + \frac {18 \, b x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )} e}{c^{4}} + \frac {48 \, a d}{c^{3}} + \frac {18 \, b \arcsin \left (\frac {1}{c x}\right ) e}{c^{5}} + \frac {18 \, a e}{c^{5}} - \frac {48 \, b d}{c^{4} x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}} + \frac {24 \, b d \arcsin \left (\frac {1}{c x}\right )}{c^{5} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} - \frac {18 \, b e}{c^{6} x {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}} + \frac {24 \, a d}{c^{5} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} + \frac {12 \, b \arcsin \left (\frac {1}{c x}\right ) e}{c^{7} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} + \frac {12 \, a e}{c^{7} x^{2} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{2}} - \frac {2 \, b e}{c^{8} x^{3} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{3}} + \frac {3 \, b \arcsin \left (\frac {1}{c x}\right ) e}{c^{9} x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4}} + \frac {3 \, a e}{c^{9} x^{4} {\left (\sqrt {-\frac {1}{c^{2} x^{2}} + 1} + 1\right )}^{4}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(3*b*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*arcsin(1/(c*x))*e/c + 3*a*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4*e/c
+ 24*b*d*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*arcsin(1/(c*x))/c + 2*b*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3*e/c^2
+ 24*a*d*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2/c + 12*b*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*arcsin(1/(c*x))*e/c^3
+ 12*a*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2*e/c^3 + 48*b*d*x*(sqrt(-1/(c^2*x^2) + 1) + 1)/c^2 + 48*b*d*arcsin(1/
(c*x))/c^3 + 18*b*x*(sqrt(-1/(c^2*x^2) + 1) + 1)*e/c^4 + 48*a*d/c^3 + 18*b*arcsin(1/(c*x))*e/c^5 + 18*a*e/c^5
- 48*b*d/(c^4*x*(sqrt(-1/(c^2*x^2) + 1) + 1)) + 24*b*d*arcsin(1/(c*x))/(c^5*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2
) - 18*b*e/(c^6*x*(sqrt(-1/(c^2*x^2) + 1) + 1)) + 24*a*d/(c^5*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) + 12*b*arcsi
n(1/(c*x))*e/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) + 12*a*e/(c^7*x^2*(sqrt(-1/(c^2*x^2) + 1) + 1)^2) - 2*b*
e/(c^8*x^3*(sqrt(-1/(c^2*x^2) + 1) + 1)^3) + 3*b*arcsin(1/(c*x))*e/(c^9*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4) +
3*a*e/(c^9*x^4*(sqrt(-1/(c^2*x^2) + 1) + 1)^4))*c

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maple [A]  time = 0.05, size = 115, normalized size = 0.83 \[ \frac {\frac {a \left (\frac {1}{4} c^{4} e \,x^{4}+\frac {1}{2} c^{4} d \,x^{2}\right )}{c^{2}}+\frac {b \left (\frac {\mathrm {arccsc}\left (c x \right ) e \,c^{4} x^{4}}{4}+\frac {\mathrm {arccsc}\left (c x \right ) c^{4} d \,x^{2}}{2}+\frac {\left (c^{2} x^{2}-1\right ) \left (c^{2} e \,x^{2}+6 c^{2} d +2 e \right )}{12 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c x}\right )}{c^{2}}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(a/c^2*(1/4*c^4*e*x^4+1/2*c^4*d*x^2)+b/c^2*(1/4*arccsc(c*x)*e*c^4*x^4+1/2*arccsc(c*x)*c^4*d*x^2+1/12*(c^
2*x^2-1)*(c^2*e*x^2+6*c^2*d+2*e)/((c^2*x^2-1)/c^2/x^2)^(1/2)/c/x))

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maxima [A]  time = 0.34, size = 98, normalized size = 0.71 \[ \frac {1}{4} \, a e x^{4} + \frac {1}{2} \, a d x^{2} + \frac {1}{2} \, {\left (x^{2} \operatorname {arccsc}\left (c x\right ) + \frac {x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c}\right )} b d + \frac {1}{12} \, {\left (3 \, x^{4} \operatorname {arccsc}\left (c x\right ) + \frac {c^{2} x^{3} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + 3 \, x \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/2*(x^2*arccsc(c*x) + x*sqrt(-1/(c^2*x^2) + 1)/c)*b*d + 1/12*(3*x^4*arccsc(c*x) +
(c^2*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 3*x*sqrt(-1/(c^2*x^2) + 1))/c^3)*b*e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.78, size = 177, normalized size = 1.28 \[ \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} + \frac {b d x^{2} \operatorname {acsc}{\left (c x \right )}}{2} + \frac {b e x^{4} \operatorname {acsc}{\left (c x \right )}}{4} + \frac {b d \left (\begin {cases} \frac {\sqrt {c^{2} x^{2} - 1}}{c} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i \sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right )}{2 c} + \frac {b e \left (\begin {cases} \frac {x^{2} \sqrt {c^{2} x^{2} - 1}}{3 c} + \frac {2 \sqrt {c^{2} x^{2} - 1}}{3 c^{3}} & \text {for}\: \left |{c^{2} x^{2}}\right | > 1 \\\frac {i x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c} + \frac {2 i \sqrt {- c^{2} x^{2} + 1}}{3 c^{3}} & \text {otherwise} \end {cases}\right )}{4 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*acsc(c*x)/2 + b*e*x**4*acsc(c*x)/4 + b*d*Piecewise((sqrt(c**2*x**2 - 1)/c,
Abs(c**2*x**2) > 1), (I*sqrt(-c**2*x**2 + 1)/c, True))/(2*c) + b*e*Piecewise((x**2*sqrt(c**2*x**2 - 1)/(3*c) +
2*sqrt(c**2*x**2 - 1)/(3*c**3), Abs(c**2*x**2) > 1), (I*x**2*sqrt(-c**2*x**2 + 1)/(3*c) + 2*I*sqrt(-c**2*x**2
+ 1)/(3*c**3), True))/(4*c)

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