3.651 \(\int \frac {a+b \sin ^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=70 \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {191, 4665, 12, 444, 63, 217, 203} \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}+\frac {b \tan ^{-1}\left (\frac {\sqrt {e} \sqrt {1-c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{d \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSin[c*x]))/(d*Sqrt[d + e*x^2]) + (b*ArcTan[(Sqrt[e]*Sqrt[1 - c^2*x^2])/(c*Sqrt[d + e*x^2])])/(d*S
qrt[e])

Rule 12

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

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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 4665

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2)
^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

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

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Mathematica [C]  time = 0.17, size = 74, normalized size = 1.06 \[ \frac {x \left (2 \left (a+b \sin ^{-1}(c x)\right )-b c x \sqrt {\frac {e x^2}{d}+1} F_1\left (1;\frac {1}{2},\frac {1}{2};2;c^2 x^2,-\frac {e x^2}{d}\right )\right )}{2 d \sqrt {d+e x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-(b*c*x*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, c^2*x^2, -((e*x^2)/d)]) + 2*(a + b*ArcSin[c*x])))/(2*
d*Sqrt[d + e*x^2])

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fricas [B]  time = 1.06, size = 294, normalized size = 4.20 \[ \left [-\frac {{\left (b e x^{2} + b d\right )} \sqrt {-e} \log \left (8 \, c^{4} e^{2} x^{4} + c^{4} d^{2} - 6 \, c^{2} d e + 8 \, {\left (c^{4} d e - c^{2} e^{2}\right )} x^{2} + 4 \, {\left (2 \, c^{3} e x^{2} + c^{3} d - c e\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {e x^{2} + d} \sqrt {-e} + e^{2}\right ) - 4 \, {\left (b e x \arcsin \left (c x\right ) + a e x\right )} \sqrt {e x^{2} + d}}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}, \frac {{\left (b e x^{2} + b d\right )} \sqrt {e} \arctan \left (\frac {{\left (2 \, c^{2} e x^{2} + c^{2} d - e\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {e x^{2} + d} \sqrt {e}}{2 \, {\left (c^{3} e^{2} x^{4} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) + 2 \, {\left (b e x \arcsin \left (c x\right ) + a e x\right )} \sqrt {e x^{2} + d}}{2 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((b*e*x^2 + b*d)*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*(b*e*x*arcsin(c*x) + a*e*x)*sqrt(
e*x^2 + d))/(d*e^2*x^2 + d^2*e), 1/2*((b*e*x^2 + b*d)*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*(b*e*x*arcsin(c*x) + a*e*x)
*sqrt(e*x^2 + d))/(d*e^2*x^2 + d^2*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsin \left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(e*x^2+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^2*d>0)', see `assume?` for
more details)Is e-c^2*d zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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