3.1.46 \(\int \frac {(f+g x) (a+b \text {ArcSin}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\) [46]

Optimal. Leaf size=126 \[ \frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}}-\frac {g \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c^2 \sqrt {d-c^2 d x^2}}+\frac {f \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c \sqrt {d-c^2 d x^2}} \]

[Out]

-g*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c^2/(-c^2*d*x^2+d)^(1/2)+b*g*x*(-c^2*x^2+1)^(1/2)/c/(-c^2*d*x^2+d)^(1/2)+1/2
*f*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {4861, 4847, 4737, 4767, 8} \begin {gather*} \frac {f \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c \sqrt {d-c^2 d x^2}}-\frac {g \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c^2 \sqrt {d-c^2 d x^2}}+\frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(b*g*x*Sqrt[1 - c^2*x^2])/(c*Sqrt[d - c^2*d*x^2]) - (g*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c^2*Sqrt[d - c^2*d*
x^2]) + (f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c*Sqrt[d - c^2*d*x^2])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4767

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p
], Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*
d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4847

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rule 4861

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] /; F
reeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && IntegerQ[p - 1/2] &&  !GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {(f+g x) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {\sqrt {1-c^2 x^2} \int \left (\frac {f \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {g x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}\right ) \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {\left (f \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d-c^2 d x^2}}+\frac {\left (g \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}+\frac {f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt {d-c^2 d x^2}}+\frac {\left (b g \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{c \sqrt {d-c^2 d x^2}}\\ &=\frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}}-\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}+\frac {f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 172, normalized size = 1.37 \begin {gather*} \frac {2 \sqrt {d} g \left (-a+a c^2 x^2+b c x \sqrt {1-c^2 x^2}\right )+2 b \sqrt {d} g \left (-1+c^2 x^2\right ) \text {ArcSin}(c x)+b c \sqrt {d} f \sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2-2 a c f \sqrt {d-c^2 d x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )}{2 c^2 \sqrt {d} \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(a + b*ArcSin[c*x]))/Sqrt[d - c^2*d*x^2],x]

[Out]

(2*Sqrt[d]*g*(-a + a*c^2*x^2 + b*c*x*Sqrt[1 - c^2*x^2]) + 2*b*Sqrt[d]*g*(-1 + c^2*x^2)*ArcSin[c*x] + b*c*Sqrt[
d]*f*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 - 2*a*c*f*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(
-1 + c^2*x^2))])/(2*c^2*Sqrt[d]*Sqrt[d - c^2*d*x^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.49, size = 247, normalized size = 1.96

method	result	size
default	\(-\frac {a g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}+\frac {a f \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}+b \left (-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} f}{2 c d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (c^{2} x^{2}-i \sqrt {-c^{2} x^{2}+1}\, x c -1\right ) g \left (\arcsin \left (c x \right )+i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}-\frac {\sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \sqrt {-c^{2} x^{2}+1}\, x c +c^{2} x^{2}-1\right ) g \left (\arcsin \left (c x \right )-i\right )}{2 c^{2} d \left (c^{2} x^{2}-1\right )}\right )\)	\(247\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 90, normalized size = 0.71 \begin {gather*} \frac {b f \arcsin \left (c x\right )^{2}}{2 \, c \sqrt {d}} + \frac {b g x}{c \sqrt {d}} + \frac {a f \arcsin \left (c x\right )}{c \sqrt {d}} - \frac {\sqrt {-c^{2} d x^{2} + d} b g \arcsin \left (c x\right )}{c^{2} d} - \frac {\sqrt {-c^{2} d x^{2} + d} a g}{c^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

1/2*b*f*arcsin(c*x)^2/(c*sqrt(d)) + b*g*x/(c*sqrt(d)) + a*f*arcsin(c*x)/(c*sqrt(d)) - sqrt(-c^2*d*x^2 + d)*b*g
*arcsin(c*x)/(c^2*d) - sqrt(-c^2*d*x^2 + d)*a*g/(c^2*d)

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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((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c^2*d*x^2 + d)*(a*g*x + a*f + (b*g*x + b*f)*arcsin(c*x))/(c^2*d*x^2 - d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(a+b*asin(c*x))/(-c**2*d*x**2+d)**(1/2),x)

[Out]

Exception raised: TypeError >> Invalid comparison of non-real zoo

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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((g*x+f)*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((g*x + f)*(b*arcsin(c*x) + a)/sqrt(-c^2*d*x^2 + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{\sqrt {d-c^2\,d\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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