3.1.65 \(\int \frac {x (a+b \csc ^{-1}(c x))}{(d+e x)^{3/2}} \, dx\) [65]

Optimal. Leaf size=238 \[ \frac {2 d \left (a+b \csc ^{-1}(c x)\right )}{e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a+b \csc ^{-1}(c x)\right )}{e^2}-\frac {4 b \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} F\left (\text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c^2 e \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}}-\frac {8 b d \sqrt {\frac {c (d+e x)}{c d+e}} \sqrt {1-c^2 x^2} \Pi \left (2;\text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c e^2 \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {d+e x}} \]

[Out]

2*d*(a+b*arccsc(c*x))/e^2/(e*x+d)^(1/2)+2*(a+b*arccsc(c*x))*(e*x+d)^(1/2)/e^2-4*b*EllipticF(1/2*(-c*x+1)^(1/2)
*2^(1/2),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^2/e/x/(1-1/c^2/x^2)^(1/2)/(
e*x+d)^(1/2)-8*b*d*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^2/x/(1-1/c^2/x^2)^(1/2)/(e*x+d)^(1/2)

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Rubi [A]
time = 1.20, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {45, 5355, 12, 6853, 6874, 733, 430, 946, 174, 552, 551} \begin {gather*} \frac {2 \sqrt {d+e x} \left (a+b \csc ^{-1}(c x)\right )}{e^2}+\frac {2 d \left (a+b \csc ^{-1}(c x)\right )}{e^2 \sqrt {d+e x}}-\frac {8 b d \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} \Pi \left (2;\text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c e^2 x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}}-\frac {4 b \sqrt {1-c^2 x^2} \sqrt {\frac {c (d+e x)}{c d+e}} F\left (\text {ArcSin}\left (\frac {\sqrt {1-c x}}{\sqrt {2}}\right )|\frac {2 e}{c d+e}\right )}{c^2 e x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d*(a + b*ArcCsc[c*x]))/(e^2*Sqrt[d + e*x]) + (2*Sqrt[d + e*x]*(a + b*ArcCsc[c*x]))/e^2 - (4*b*Sqrt[(c*(d +
e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticF[ArcSin[Sqrt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c^2*e*Sqrt[1 -
1/(c^2*x^2)]*x*Sqrt[d + e*x]) - (8*b*d*Sqrt[(c*(d + e*x))/(c*d + e)]*Sqrt[1 - c^2*x^2]*EllipticPi[2, ArcSin[Sq
rt[1 - c*x]/Sqrt[2]], (2*e)/(c*d + e)])/(c*e^2*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[d + e*x])

Rule 12

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

Rule 45

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 174

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 430

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

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], 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 552

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

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2*a*Rt[-c/a, 2]*(d + e*x)^m*(Sqrt[1
+ c*(x^2/a)]/(c*Sqrt[a + c*x^2]*(c*((d + e*x)/(c*d - a*e*Rt[-c/a, 2])))^m)), Subst[Int[(1 + 2*a*e*Rt[-c/a, 2]*
(x^2/(c*d - a*e*Rt[-c/a, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(1 - Rt[-c/a, 2]*x)/2]], x] /; FreeQ[{a, c, d, e},
x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m^2, 1/4]

Rule 946

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> With[{q = Rt[-c/
a, 2]}, Dist[1/Sqrt[a], 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 5355

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide[u, x]}, Dist[a + b*ArcCsc[c*x], v,
x] + Dist[b/c, Int[SimplifyIntegrand[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]]
/; FreeQ[{a, b, c}, x]

Rule 6853

Int[(u_.)*((a_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[b^IntPart[p]*((a + b*x^n)^FracPart[p]/(x^(n*FracPa
rt[p])*(1 + a*(1/(x^n*b)))^FracPart[p])), Int[u*x^(n*p)*(1 + a*(1/(x^n*b)))^p, x], x] /; FreeQ[{a, b, p}, x] &
&  !IntegerQ[p] && ILtQ[n, 0] &&  !RationalFunctionQ[u, x] && IntegerQ[p + 1/2]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 11.22, size = 226, normalized size = 0.95 \begin {gather*} \frac {2 \left (\frac {a (2 d+e x)}{\sqrt {d+e x}}+\frac {b (2 d+e x) \csc ^{-1}(c x)}{\sqrt {d+e x}}-\frac {2 i b \sqrt {\frac {e (1+c x)}{-c d+e}} \sqrt {\frac {e-c e x}{c d+e}} \left (F\left (i \sinh ^{-1}\left (\sqrt {-\frac {c}{c d+e}} \sqrt {d+e x}\right )|\frac {c d+e}{c d-e}\right )-2 \Pi \left (1+\frac {e}{c d};i \sinh ^{-1}\left (\sqrt {-\frac {c}{c d+e}} \sqrt {d+e x}\right )|\frac {c d+e}{c d-e}\right )\right )}{c \sqrt {-\frac {c}{c d+e}} \sqrt {1-\frac {1}{c^2 x^2}} x}\right )}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*(2*d + e*x))/Sqrt[d + e*x] + (b*(2*d + e*x)*ArcCsc[c*x])/Sqrt[d + e*x] - ((2*I)*b*Sqrt[(e*(1 + c*x))/(-
(c*d) + e)]*Sqrt[(e - c*e*x)/(c*d + e)]*(EllipticF[I*ArcSinh[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c
*d - e)] - 2*EllipticPi[1 + e/(c*d), I*ArcSinh[Sqrt[-(c/(c*d + e))]*Sqrt[d + e*x]], (c*d + e)/(c*d - e)]))/(c*
Sqrt[-(c/(c*d + e))]*Sqrt[1 - 1/(c^2*x^2)]*x)))/e^2

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Maple [A]
time = 0.51, size = 282, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {-2 a \left (-\sqrt {e x +d}-\frac {d}{\sqrt {e x +d}}\right )-2 b \left (-\sqrt {e x +d}\, \mathrm {arccsc}\left (c x \right )-\frac {\mathrm {arccsc}\left (c x \right ) d}{\sqrt {e x +d}}-\frac {2 \sqrt {\frac {-c \left (e x +d \right )+c d -e}{c d -e}}\, \sqrt {\frac {-c \left (e x +d \right )+c d +e}{c d +e}}\, \left (\EllipticF \left (\sqrt {e x +d}\, \sqrt {\frac {c}{c d -e}}, \sqrt {\frac {c d -e}{c d +e}}\right )-2 \EllipticPi \left (\sqrt {e x +d}\, \sqrt {\frac {c}{c d -e}}, \frac {c d -e}{c d}, \frac {\sqrt {\frac {c}{c d +e}}}{\sqrt {\frac {c}{c d -e}}}\right )\right )}{c \sqrt {\frac {c^{2} \left (e x +d \right )^{2}-2 c^{2} d \left (e x +d \right )+c^{2} d^{2}-e^{2}}{c^{2} e^{2} x^{2}}}\, x \sqrt {\frac {c}{c d -e}}}\right )}{e^{2}}\)	\(282\)
default	\(\frac {-2 a \left (-\sqrt {e x +d}-\frac {d}{\sqrt {e x +d}}\right )-2 b \left (-\sqrt {e x +d}\, \mathrm {arccsc}\left (c x \right )-\frac {\mathrm {arccsc}\left (c x \right ) d}{\sqrt {e x +d}}-\frac {2 \sqrt {\frac {-c \left (e x +d \right )+c d -e}{c d -e}}\, \sqrt {\frac {-c \left (e x +d \right )+c d +e}{c d +e}}\, \left (\EllipticF \left (\sqrt {e x +d}\, \sqrt {\frac {c}{c d -e}}, \sqrt {\frac {c d -e}{c d +e}}\right )-2 \EllipticPi \left (\sqrt {e x +d}\, \sqrt {\frac {c}{c d -e}}, \frac {c d -e}{c d}, \frac {\sqrt {\frac {c}{c d +e}}}{\sqrt {\frac {c}{c d -e}}}\right )\right )}{c \sqrt {\frac {c^{2} \left (e x +d \right )^{2}-2 c^{2} d \left (e x +d \right )+c^{2} d^{2}-e^{2}}{c^{2} e^{2} x^{2}}}\, x \sqrt {\frac {c}{c d -e}}}\right )}{e^{2}}\)	\(282\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(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(%e-c*d>0)', see `assume?` for
more details

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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