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

Optimal. Leaf size=142 \[ \frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {b e x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {e \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-2 b^2 d x-\frac {1}{4} b^2 e x^2 \]

[Out]

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

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Rubi [A]  time = 0.31, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4743, 4763, 4641, 4677, 8, 4707, 30} \[ \frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {b e x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {e \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-2 b^2 d x-\frac {1}{4} b^2 e x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*b^2*d*x - (b^2*e*x^2)/4 + (2*b*d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/c + (b*e*x*Sqrt[1 - c^2*x^2]*(a + b
*ArcSin[c*x]))/(2*c) - (d^2*(a + b*ArcSin[c*x])^2)/(2*e) - (e*(a + b*ArcSin[c*x])^2)/(4*c^2) + ((d + e*x)^2*(a
+ b*ArcSin[c*x])^2)/(2*e)

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4641

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

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 4707

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

Rule 4743

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

Rule 4763

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))

Rubi steps

\begin {align*} \int (d+e x) \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-\frac {(b c) \int \frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{e}\\ &=\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-\frac {(b c) \int \left (\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {2 d e x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}\right ) \, dx}{e}\\ &=\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-(2 b c d) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx-\frac {\left (b c d^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{e}-(b c e) \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {b e x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}+\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-\left (2 b^2 d\right ) \int 1 \, dx-\frac {1}{2} \left (b^2 e\right ) \int x \, dx-\frac {(b e) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 c}\\ &=-2 b^2 d x-\frac {1}{4} b^2 e x^2+\frac {2 b d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c}+\frac {b e x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c}-\frac {d^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}-\frac {e \left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2}+\frac {(d+e x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 e}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 147, normalized size = 1.04 \[ \frac {c \left (2 a^2 c x (2 d+e x)+2 a b \sqrt {1-c^2 x^2} (4 d+e x)-b^2 c x (8 d+e x)\right )+2 b \sin ^{-1}(c x) \left (4 a c^2 d x+a e \left (2 c^2 x^2-1\right )+b c \sqrt {1-c^2 x^2} (4 d+e x)\right )+b^2 \sin ^{-1}(c x)^2 \left (4 c^2 d x+e \left (2 c^2 x^2-1\right )\right )}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(2*a^2*c*x*(2*d + e*x) - b^2*c*x*(8*d + e*x) + 2*a*b*(4*d + e*x)*Sqrt[1 - c^2*x^2]) + 2*b*(4*a*c^2*d*x + b*
c*(4*d + e*x)*Sqrt[1 - c^2*x^2] + a*e*(-1 + 2*c^2*x^2))*ArcSin[c*x] + b^2*(4*c^2*d*x + e*(-1 + 2*c^2*x^2))*Arc
Sin[c*x]^2)/(4*c^2)

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fricas [A]  time = 0.67, size = 156, normalized size = 1.10 \[ \frac {{\left (2 \, a^{2} - b^{2}\right )} c^{2} e x^{2} + 4 \, {\left (a^{2} - 2 \, b^{2}\right )} c^{2} d x + {\left (2 \, b^{2} c^{2} e x^{2} + 4 \, b^{2} c^{2} d x - b^{2} e\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (2 \, a b c^{2} e x^{2} + 4 \, a b c^{2} d x - a b e\right )} \arcsin \left (c x\right ) + 2 \, {\left (a b c e x + 4 \, a b c d + {\left (b^{2} c e x + 4 \, b^{2} c d\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 253, normalized size = 1.78 \[ b^{2} d x \arcsin \left (c x\right )^{2} + 2 \, a b d x \arcsin \left (c x\right ) + \frac {\sqrt {-c^{2} x^{2} + 1} b^{2} x \arcsin \left (c x\right ) e}{2 \, c} + a^{2} d x - 2 \, b^{2} d x + \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} \arcsin \left (c x\right )^{2} e}{2 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} b^{2} d \arcsin \left (c x\right )}{c} + \frac {\sqrt {-c^{2} x^{2} + 1} a b x e}{2 \, c} + \frac {{\left (c^{2} x^{2} - 1\right )} a b \arcsin \left (c x\right ) e}{c^{2}} + \frac {b^{2} \arcsin \left (c x\right )^{2} e}{4 \, c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1} a b d}{c} + \frac {{\left (c^{2} x^{2} - 1\right )} a^{2} e}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - 1\right )} b^{2} e}{4 \, c^{2}} + \frac {a b \arcsin \left (c x\right ) e}{2 \, c^{2}} - \frac {b^{2} e}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*d*x*arcsin(c*x)^2 + 2*a*b*d*x*arcsin(c*x) + 1/2*sqrt(-c^2*x^2 + 1)*b^2*x*arcsin(c*x)*e/c + a^2*d*x - 2*b^2
*d*x + 1/2*(c^2*x^2 - 1)*b^2*arcsin(c*x)^2*e/c^2 + 2*sqrt(-c^2*x^2 + 1)*b^2*d*arcsin(c*x)/c + 1/2*sqrt(-c^2*x^
2 + 1)*a*b*x*e/c + (c^2*x^2 - 1)*a*b*arcsin(c*x)*e/c^2 + 1/4*b^2*arcsin(c*x)^2*e/c^2 + 2*sqrt(-c^2*x^2 + 1)*a*
b*d/c + 1/2*(c^2*x^2 - 1)*a^2*e/c^2 - 1/4*(c^2*x^2 - 1)*b^2*e/c^2 + 1/2*a*b*arcsin(c*x)*e/c^2 - 1/8*b^2*e/c^2

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maple [A]  time = 0.09, size = 198, normalized size = 1.39 \[ \frac {\frac {a^{2} \left (\frac {1}{2} x^{2} c^{2} e +c^{2} d x \right )}{c}+\frac {b^{2} \left (\frac {e \left (2 \arcsin \left (c x \right )^{2} c^{2} x^{2}+2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x -\arcsin \left (c x \right )^{2}-c^{2} x^{2}\right )}{4}+d c \left (c x \arcsin \left (c x \right )^{2}-2 c x +2 \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\right )\right )}{c}+\frac {2 a b \left (\frac {\arcsin \left (c x \right ) c^{2} x^{2} e}{2}+\arcsin \left (c x \right ) d \,c^{2} x -\frac {e \left (-\frac {c x \sqrt {-c^{2} x^{2}+1}}{2}+\frac {\arcsin \left (c x \right )}{2}\right )}{2}+d c \sqrt {-c^{2} x^{2}+1}\right )}{c}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{2} d x \arcsin \left (c x\right )^{2} + \frac {1}{2} \, a^{2} e x^{2} + \frac {1}{2} \, {\left (2 \, x^{2} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x}{c^{2}} - \frac {\arcsin \left (c x\right )}{c^{3}}\right )}\right )} a b e + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, c \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{2} x^{2} - 1}\,{d x}\right )} b^{2} e - 2 \, b^{2} d {\left (x - \frac {\sqrt {-c^{2} x^{2} + 1} \arcsin \left (c x\right )}{c}\right )} + a^{2} d x + \frac {2 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} a b d}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*d*x*arcsin(c*x)^2 + 1/2*a^2*e*x^2 + 1/2*(2*x^2*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x/c^2 - arcsin(c*x)/c^3
))*a*b*e + 1/2*(x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*c*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*
x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^2*x^2 - 1), x))*b^2*e - 2*b^2*d*(x - sqrt(-c^2*x^2 + 1)*arcs
in(c*x)/c) + a^2*d*x + 2*(c*x*arcsin(c*x) + sqrt(-c^2*x^2 + 1))*a*b*d/c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^2*(d + e*x),x)

[Out]

int((a + b*asin(c*x))^2*(d + e*x), x)

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sympy [A]  time = 0.67, size = 233, normalized size = 1.64 \[ \begin {cases} a^{2} d x + \frac {a^{2} e x^{2}}{2} + 2 a b d x \operatorname {asin}{\left (c x \right )} + a b e x^{2} \operatorname {asin}{\left (c x \right )} + \frac {2 a b d \sqrt {- c^{2} x^{2} + 1}}{c} + \frac {a b e x \sqrt {- c^{2} x^{2} + 1}}{2 c} - \frac {a b e \operatorname {asin}{\left (c x \right )}}{2 c^{2}} + b^{2} d x \operatorname {asin}^{2}{\left (c x \right )} - 2 b^{2} d x + \frac {b^{2} e x^{2} \operatorname {asin}^{2}{\left (c x \right )}}{2} - \frac {b^{2} e x^{2}}{4} + \frac {2 b^{2} d \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{c} + \frac {b^{2} e x \sqrt {- c^{2} x^{2} + 1} \operatorname {asin}{\left (c x \right )}}{2 c} - \frac {b^{2} e \operatorname {asin}^{2}{\left (c x \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\a^{2} \left (d x + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*d*x + a**2*e*x**2/2 + 2*a*b*d*x*asin(c*x) + a*b*e*x**2*asin(c*x) + 2*a*b*d*sqrt(-c**2*x**2 + 1
)/c + a*b*e*x*sqrt(-c**2*x**2 + 1)/(2*c) - a*b*e*asin(c*x)/(2*c**2) + b**2*d*x*asin(c*x)**2 - 2*b**2*d*x + b**
2*e*x**2*asin(c*x)**2/2 - b**2*e*x**2/4 + 2*b**2*d*sqrt(-c**2*x**2 + 1)*asin(c*x)/c + b**2*e*x*sqrt(-c**2*x**2
+ 1)*asin(c*x)/(2*c) - b**2*e*asin(c*x)**2/(4*c**2), Ne(c, 0)), (a**2*(d*x + e*x**2/2), True))

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