3.192 \(\int (a+b \sin ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=59 \[ \frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{d}+\frac {(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}-2 b^2 x \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4803, 4619, 4677, 8} \[ \frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{d}+\frac {(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d}-2 b^2 x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^2,x]

[Out]

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

Rule 8

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

Rule 4619

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

Rule 4677

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 4803

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 61, normalized size = 1.03 \[ \frac {2 b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )+(c+d x) \left (a+b \sin ^{-1}(c+d x)\right )^2-2 b^2 (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 94, normalized size = 1.59 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} d x + {\left (b^{2} d x + b^{2} c\right )} \arcsin \left (d x + c\right )^{2} + 2 \, {\left (a b d x + a b c\right )} \arcsin \left (d x + c\right ) + 2 \, \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1} {\left (b^{2} \arcsin \left (d x + c\right ) + a b\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 111, normalized size = 1.88 \[ \frac {{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2}}{d} + \frac {2 \, {\left (d x + c\right )} a b \arcsin \left (d x + c\right )}{d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right )}{d} + \frac {{\left (d x + c\right )} a^{2}}{d} - \frac {2 \, {\left (d x + c\right )} b^{2}}{d} + \frac {2 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} a b}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 92, normalized size = 1.56 \[ \frac {\left (d x +c \right ) a^{2}+b^{2} \left (\arcsin \left (d x +c \right )^{2} \left (d x +c \right )-2 d x -2 c +2 \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}\right )+2 a b \left (\left (d x +c \right ) \arcsin \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*((d*x+c)*a^2+b^2*(arcsin(d*x+c)^2*(d*x+c)-2*d*x-2*c+2*arcsin(d*x+c)*(1-(d*x+c)^2)^(1/2))+2*a*b*((d*x+c)*ar
csin(d*x+c)+(1-(d*x+c)^2)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ {\left (x \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 2 \, d \int \frac {\sqrt {d x + c + 1} \sqrt {-d x - c + 1} x \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d^{2} x^{2} + 2 \, c d x + c^{2} - 1}\,{d x}\right )} b^{2} + a^{2} x + \frac {2 \, {\left ({\left (d x + c\right )} \arcsin \left (d x + c\right ) + \sqrt {-{\left (d x + c\right )}^{2} + 1}\right )} a b}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 2*d*integrate(sqrt(d*x + c + 1)*sqrt(-d*x - c +
1)*x*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x))*b^2 + a^2*x + 2
*((d*x + c)*arcsin(d*x + c) + sqrt(-(d*x + c)^2 + 1))*a*b/d

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mupad [B]  time = 0.48, size = 88, normalized size = 1.49 \[ a^2\,x+\frac {b^2\,\left ({\mathrm {asin}\left (c+d\,x\right )}^2-2\right )\,\left (c+d\,x\right )}{d}+\frac {2\,a\,b\,\left (\sqrt {1-{\left (c+d\,x\right )}^2}+\mathrm {asin}\left (c+d\,x\right )\,\left (c+d\,x\right )\right )}{d}+\frac {2\,b^2\,\mathrm {asin}\left (c+d\,x\right )\,\sqrt {1-{\left (c+d\,x\right )}^2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))^2,x)

[Out]

a^2*x + (b^2*(asin(c + d*x)^2 - 2)*(c + d*x))/d + (2*a*b*((1 - (c + d*x)^2)^(1/2) + asin(c + d*x)*(c + d*x)))/
d + (2*b^2*asin(c + d*x)*(1 - (c + d*x)^2)^(1/2))/d

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sympy [A]  time = 0.31, size = 143, normalized size = 2.42 \[ \begin {cases} a^{2} x + \frac {2 a b c \operatorname {asin}{\left (c + d x \right )}}{d} + 2 a b x \operatorname {asin}{\left (c + d x \right )} + \frac {2 a b \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{d} + \frac {b^{2} c \operatorname {asin}^{2}{\left (c + d x \right )}}{d} + b^{2} x \operatorname {asin}^{2}{\left (c + d x \right )} - 2 b^{2} x + \frac {2 b^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname {asin}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {asin}{\relax (c )}\right )^{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*x + 2*a*b*c*asin(c + d*x)/d + 2*a*b*x*asin(c + d*x) + 2*a*b*sqrt(-c**2 - 2*c*d*x - d**2*x**2 +
1)/d + b**2*c*asin(c + d*x)**2/d + b**2*x*asin(c + d*x)**2 - 2*b**2*x + 2*b**2*sqrt(-c**2 - 2*c*d*x - d**2*x*
*2 + 1)*asin(c + d*x)/d, Ne(d, 0)), (x*(a + b*asin(c))**2, True))

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