3.16 \(\int \frac {(f+g x) (a+b \cos ^{-1}(c x))}{\sqrt {d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac {f \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c \sqrt {d-c^2 d x^2}}-\frac {g \left (1-c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}-\frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}} \]

[Out]

-g*(-c^2*x^2+1)*(a+b*arccos(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*arccos(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 = 0.22, antiderivative size = 127, 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 = {4778, 4764, 4642, 4678, 8} \[ -\frac {f \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c \sqrt {d-c^2 d x^2}}-\frac {g \left (1-c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}-\frac {b g x \sqrt {1-c^2 x^2}}{c \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(a + b*ArcCos[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*ArcCos[c*x]))/(c^2*Sqrt[d - c^2
*d*x^2]) - (f*Sqrt[1 - c^2*x^2]*(a + b*ArcCos[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 4642

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Simp[(a + b*ArcCos[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 4678

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcCos[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*ArcCos[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 4764

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcCos[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 4778

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Dist[(d^IntPart[p]*(d + e*x^2)^FracPart[p])/(1 - c^2*x^2)^FracPart[p], Int[(f + g*x)^m*(1 - c^2*x^2)^p*(a +
b*ArcCos[c*x])^n, x], x] /; FreeQ[{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 \cos ^{-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 \cos ^{-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 \cos ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}+\frac {g x \left (a+b \cos ^{-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 \cos ^{-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 \cos ^{-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 \cos ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}-\frac {f \sqrt {1-c^2 x^2} \left (a+b \cos ^{-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 \cos ^{-1}(c x)\right )}{c^2 \sqrt {d-c^2 d x^2}}-\frac {f \sqrt {1-c^2 x^2} \left (a+b \cos ^{-1}(c x)\right )^2}{2 b c \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 172, normalized size = 1.35 \[ \frac {-2 \sqrt {d} g \left (-a c^2 x^2+a+b c x \sqrt {1-c^2 x^2}\right )-2 a c f \sqrt {d-c^2 d x^2} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )-b c \sqrt {d} f \sqrt {1-c^2 x^2} \cos ^{-1}(c x)^2+2 b \sqrt {d} g \left (c^2 x^2-1\right ) \cos ^{-1}(c x)}{2 c^2 \sqrt {d} \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(a + b*ArcCos[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)*ArcCos[c*x] - b*c*Sqrt[
d]*f*Sqrt[1 - c^2*x^2]*ArcCos[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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (a g x + a f + {\left (b g x + b f\right )} \arccos \left (c x\right )\right )}}{c^{2} d x^{2} - d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x + f\right )} {\left (b \arccos \left (c x\right ) + a\right )}}{\sqrt {-c^{2} d x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.00, size = 235, normalized size = 1.85 \[ -\frac {a g \sqrt {-c^{2} d \,x^{2}+d}}{c^{2} d}+\frac {a f \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\sqrt {c^{2} d}}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arccos \left (c x \right )^{2} f}{2 c d \left (c^{2} x^{2}-1\right )}-\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \arccos \left (c x \right ) x^{2}}{d \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \sqrt {-c^{2} x^{2}+1}\, x}{c d \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, g \arccos \left (c x \right )}{c^{2} d \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))+1/2*b*(-d*(c^2*
x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c/d/(c^2*x^2-1)*arccos(c*x)^2*f-b*(-d*(c^2*x^2-1))^(1/2)*g/d/(c^2*x^2-1)*arcc
os(c*x)*x^2+b*(-d*(c^2*x^2-1))^(1/2)*g/c/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x+b*(-d*(c^2*x^2-1))^(1/2)*g/c^2/d/(
c^2*x^2-1)*arccos(c*x)

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maxima [A]  time = 0.43, size = 108, normalized size = 0.85 \[ \frac {b f \arccos \left (c x\right ) \arcsin \left (c x\right )}{c \sqrt {d}} + \frac {b f \arcsin \left (c x\right )^{2}}{2 \, c \sqrt {d}} - \frac {b g x}{c \sqrt {d}} + \frac {a f \arcsin \left (c x\right )}{c \sqrt {d}} - \frac {\sqrt {-c^{2} d x^{2} + d} b g \arccos \left (c x\right )}{c^{2} d} - \frac {\sqrt {-c^{2} d x^{2} + d} a g}{c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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