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

Optimal. Leaf size=83 \[ \frac{(d+e x)^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e}+\frac{b d \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c}+\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{b d^2 \csc ^{-1}(c x)}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.166665, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {5227, 1568, 1396, 1807, 844, 216, 266, 63, 208} \[ \frac{(d+e x)^2 \left (a+b \csc ^{-1}(c x)\right )}{2 e}+\frac{b d \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{c}+\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{b d^2 \csc ^{-1}(c x)}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a + b*ArcCsc[c*x]),x]

[Out]

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

Rule 5227

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcCsc[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]

Rule 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 1396

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> -Subst[Int[((d + e/x^n)^q*(a +
c/x^(2*n))^p)/x^2, x], x, 1/x] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && ILtQ[n, 0]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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]

Rule 216

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.195894, size = 113, normalized size = 1.36 \[ a d x+\frac{1}{2} a e x^2+\frac{b d x \sqrt{1-\frac{1}{c^2 x^2}} \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{\sqrt{c^2 x^2-1}}+\frac{b e x \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}}{2 c}+b d x \csc ^{-1}(c x)+\frac{1}{2} b e x^2 \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.18, size = 140, normalized size = 1.7 \begin{align*}{\frac{a{x}^{2}e}{2}}+adx+{\frac{b{\rm arccsc} \left (cx\right ){x}^{2}e}{2}}+b{\rm arccsc} \left (cx\right )xd+{\frac{bd}{{c}^{2}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{bex}{2\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{be}{2\,{c}^{3}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00062, size = 124, normalized size = 1.49 \begin{align*} \frac{1}{2} \, a e 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 e + a d x + \frac{{\left (2 \, c x \operatorname{arccsc}\left (c x\right ) + \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right ) - \log \left (-\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )\right )} b d}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e*x^2 + 1/2*(x^2*arccsc(c*x) + x*sqrt(-1/(c^2*x^2) + 1)/c)*b*e + a*d*x + 1/2*(2*c*x*arccsc(c*x) + log(sq
rt(-1/(c^2*x^2) + 1) + 1) - log(-sqrt(-1/(c^2*x^2) + 1) + 1))*b*d/c

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Fricas [A]  time = 2.19668, size = 302, normalized size = 3.64 \begin{align*} \frac{a c^{2} e x^{2} + 2 \, a c^{2} d x - 2 \, b c d \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + \sqrt{c^{2} x^{2} - 1} b e +{\left (b c^{2} e x^{2} + 2 \, b c^{2} d x - 2 \, b c^{2} d - b c^{2} e\right )} \operatorname{arccsc}\left (c x\right ) - 2 \,{\left (2 \, b c^{2} d + b c^{2} e\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*c^2*e*x^2 + 2*a*c^2*d*x - 2*b*c*d*log(-c*x + sqrt(c^2*x^2 - 1)) + sqrt(c^2*x^2 - 1)*b*e + (b*c^2*e*x^2
+ 2*b*c^2*d*x - 2*b*c^2*d - b*c^2*e)*arccsc(c*x) - 2*(2*b*c^2*d + b*c^2*e)*arctan(-c*x + sqrt(c^2*x^2 - 1)))/c
^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{acsc}{\left (c x \right )}\right ) \left (d + e x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}{\left (b \operatorname{arccsc}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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